Commutative monoids and their spectra

Written by Claude Opus 5.5 (Anthropic), September 2026, revised October 2026. The September text was spot-checked by Claude Opus 5.5 in a separate session; the October revision links the AI Integrated Stacks Project citations and is self-checked by the writing AI. The October revision also corrects points found by GPT-6 Astra (OpenAI), Ultra, in a separate review session. Public domain (CC0).

Introduction

Take a commutative ring and forget its addition. What is left is a set with a commutative multiplication, a unit \(1\) and an absorbing element \(0\): a commutative monoid with zero. The simplest geometry over the field with one element replaces rings by such monoids. A monoid \(M\) plays the role of a ring over \(\mathbb F_1\), and the ring \(\mathbb Z[M]\) spanned by \(M\) is its base change to the integers. This lesson develops the commutative algebra of monoids that the course uses: ideals and prime ideals, the spectrum, localization, the structure sheaf, quotients, finiteness conditions, and the passage from \(M\) to \(\mathbb Z[M]\). The next lesson, Monoid schemes, glues spectra together.

Most definitions are copied from ring theory. The interest lies in what changes. The lesson proves the following statements. The first six have no counterpart for rings, and the seventh relates a monoid to its ring.

1. A non-zero monoid has exactly one maximal ideal, the set of its non-units. A union of ideals is an ideal, and a union of prime ideals is a prime ideal (Section 2).
2. For every multiplicative set \(S\) that does not contain \(0\) there is a largest ideal disjoint from \(S\). It is prime and it is given by a formula, so prime ideals are found without Zorn's lemma (Theorem 2.8).
3. A non-empty open subset of the spectrum is a basic open set \(D(f)\) exactly when it has a largest point, and every non-zero localization of a monoid is the localization at a prime ideal (Theorems 3.4 and 4.3).
4. A section of the structure sheaf over any open set is a compatible family of germs. The formulas \(\mathcal O(D(f))=M_f\) and \(\mathcal O(\operatorname{Spec}M)=M\) follow in two lines (Section 5).
5. Quotients of a monoid are described by congruences, and only some congruences come from ideals (Section 6).
6. A finitely generated monoid has finitely many prime ideals, and its spectrum is a finite partially ordered set (Section 7).
7. The map \(\operatorname{Spec}\mathbb Z[M]\to\operatorname{Spec}M\) is surjective, and its fibres are spectra of group rings. The ring \(\mathbb Z[M]\) is a domain exactly when \(M\) is cancellative and torsion-free (Section 8).

Section 9 works out free monoids and the monoids of lattice points of cones. Section 10 compares the monoids with zero used here with the monoids without zero of [Deitmar 2005]. Section 11 contains exercises with solutions.

What is assumed. Commutative rings, ideals, prime ideals and localization [Stacks, Tags 00AR and 00CM]; the spectrum of a ring with its Zariski topology [Stacks, Tag 00DY]; sheaves on a topological space and their stalks [Stacks, Tags 006S and 0078]. Section 9 uses linear algebra over \(\mathbb Q\) and \(\mathbb R\). The lessons Counting over finite fields and the limit q → 1 and Weil's proof for curves and what is missing over the integers explain why one looks for a geometry below the integers; they are not used in the proofs.

Basic references are [Deitmar 2005], [Chu–Lorscheid–Santhanam 2012], [Cortiñas–Haesemeyer–Walker–Weibel 2015] and [Connes–Consani 2010b].

Conventions. Rings are commutative with \(1\). \(\mathbb N=\{0,1,2,\dots\}\). The word monoid always has the meaning of Definition 1.1: commutative, written multiplicatively, with \(1\) and with \(0\). A commutative monoid without zero is a monoid in the ordinary sense, in which no zero is required. For elements \(a,b\) of a monoid, "\(a\) divides \(b\)" means \(b=ac\) for some \(c\).

1. Monoids with zero

Definition 1.1. A monoid is a set \(M\) with an associative and commutative multiplication, an element \(1\) such that \(1a=a\) for all \(a\in M\), and an element \(0\) such that \(0a=0\) for all \(a\in M\). A morphism of monoids is a map \(\varphi:M\to N\) with \(\varphi(ab)=\varphi(a)\varphi(b)\), \(\varphi(1)=1\) and \(\varphi(0)=0\). The set of morphisms is written \(\operatorname{Hom}(M,N)\). A submonoid of \(M\) is a subset that contains \(0\) and \(1\) and is closed under multiplication.

The elements \(0\) and \(1\) are unique. If \(0=1\) then \(a=1a=0a=0\) for all \(a\), so \(M=\{0\}\). This is the zero monoid, written \(0\).

Definition 1.2. An element \(a\in M\) is a unit if \(ab=1\) for some \(b\in M\). The units form an abelian group \(M^\times\). If \(M\neq0\), then \(0\) is not a unit, because \(0b=0\neq1\).

Examples 1.3.

(a) \(\mathbb F_1=\{0,1\}\). For every monoid \(M\) there is exactly one morphism \(\mathbb F_1\to M\) and exactly one morphism \(M\to0\).

(b) If \(R\) is a ring, forgetting the addition gives the monoid \((R,\cdot)\). A ring homomorphism is a morphism of the multiplicative monoids.

(c) Let \(A\) be a commutative monoid without zero. Then \(A_0=A\sqcup\{0\}\), with \(0a=0\) for all \(a\), is a monoid. If \(A=G\) is an abelian group, every non-zero element of \(G_0\) is a unit; we call \(G_0\) a group with zero. For a field \(k\) we have \((k,\cdot)=(k^\times)_0\). For \(n\ge1\) the monoid \(\mathbb F_{1^n}=(\mu_n)_0\) consists of \(0\) and the \(n\)-th roots of unity.

(d) Let \(P\) be a commutative monoid without zero, written additively. We write \(\mathbb F_1[P]\) for the monoid \(P_0\) in multiplicative notation: its elements are \(0\) and the symbols \(\chi^m\), \(m\in P\), with \(\chi^m\chi^{m'}=\chi^{m+m'}\) and \(1=\chi^0\). For \(P=\mathbb N^n\) put \(x_i=\chi^{e_i}\). Then \(\mathbb F_1[x_1,\dots,x_n]:=\mathbb F_1[\mathbb N^n]\) consists of \(0\) and the monomials \(x^a=x_1^{a_1}\cdots x_n^{a_n}\). It is the free monoid on \(x_1,\dots,x_n\). For \(P=\mathbb Z^n\) we get the group with zero \(\mathbb F_1[x_1^{\pm1},\dots,x_n^{\pm1}]\) of Laurent monomials.

(e) The product \(M\times N\) of two monoids, with componentwise multiplication, is a monoid with zero \((0,0)\) and one \((1,1)\).

(f) The set \(\{0,1,\varepsilon\}\) with \(\varepsilon^2=0\) is a monoid. So is \(\{0,e,1\}\) with \(e^2=e\).

(g) For readers who know adèles: let \(K\) be a global field with adèle ring \(\mathbb A_K\). The group \(K^\times\) acts on \(\mathbb A_K\) by multiplication, and the product of \(\mathbb A_K\) descends to the set of orbits \(\mathbb A_K/K^\times\). This is the monoid of adèle classes of [Connes–Consani 2010b, Section 3.2]. Its group of units is the idèle class group \(\mathbb A_K^\times/K^\times\).

Proposition 1.4 (free monoids). Let \(N\) be a monoid and \(b_1,\dots,b_n\in N\). There is exactly one morphism \(\varphi:\mathbb F_1[x_1,\dots,x_n]\to N\) with \(\varphi(x_i)=b_i\) for all \(i\).

Proof. A morphism with \(\varphi(x_i)=b_i\) must send \(x^a\) to \(b_1^{a_1}\cdots b_n^{a_n}\) and \(0\) to \(0\). This map is a morphism, because exponents add. \(\square\)

Definition 1.5. A monoid \(M\) is without zero divisors if \(M\neq0\) and \(ab=0\) implies \(a=0\) or \(b=0\).

Proposition 1.6 (adjoining a zero).

(a) Let \(A\) be a commutative monoid without zero and \(N\) a monoid. Restriction to \(A\) is a bijection from \(\operatorname{Hom}(A_0,N)\) to the set of maps \(\psi:A\to N\) with \(\psi(ab)=\psi(a)\psi(b)\) and \(\psi(1)=1\).

(b) A monoid \(M\) is isomorphic to \(A_0\) for some commutative monoid \(A\) without zero if and only if \(M\) is without zero divisors. In that case \(A\cong M\setminus\{0\}\).

Proof. (a) Extend \(\psi\) by \(0\mapsto0\). The extension is multiplicative, and it is the only extension that is a morphism. (b) In \(A_0\) a product of two elements of \(A\) lies in \(A\), and \(1\neq0\). Conversely, if \(M\) is without zero divisors, then \(M\setminus\{0\}\) contains \(1\) and is closed under multiplication, and \(M=(M\setminus\{0\})_0\). \(\square\)

Several monoids of Examples 1.3 have zero divisors, so they are not of the form \(A_0\): \(\varepsilon\cdot\varepsilon=0\) in \(\{0,1,\varepsilon\}\); \((1,0)(0,1)=(0,0)\) in \(M\times N\) when \(M,N\neq0\); and in the monoid of adèle classes the classes of two non-zero adèles with disjoint supports have product \(0\). The monoid \((R,\cdot)\) of a ring is without zero divisors exactly when \(R\) is a domain. Zero divisors are the reason for working with a zero from the start; Section 10 returns to this.

2. Ideals, prime ideals and faces

Definition 2.1. An ideal of a monoid \(M\) is a subset \(I\subseteq M\) with \(0\in I\) and \(ab\in I\) for all \(a\in I\), \(b\in M\). The ideal generated by a subset \(T\subseteq M\) is \(\langle T\rangle=\{tb:t\in T,\ b\in M\}\cup\{0\}\); for one element \(f\) it is \(fM\). An ideal \(I\) is proper if \(I\neq M\). This holds if and only if \(1\notin I\), and if and only if \(I\) contains no unit.

Lemma 2.2.

(a) The union and the intersection of a non-empty family of ideals of \(M\) are ideals.

(b) If \(M\neq0\), then \(\mathfrak m_M:=M\setminus M^\times\) is an ideal, and it contains every proper ideal.

Proof. (a) Both contain \(0\) and are stable under multiplication by elements of \(M\). (b) We have \(0\in\mathfrak m_M\) by Definition 1.2. Let \(a\in\mathfrak m_M\) and \(b\in M\). If \(ab\) were a unit, say \((ab)c=1\), then \(a(bc)=1\) and \(a\) would be a unit. So \(ab\in\mathfrak m_M\). A proper ideal contains no unit, so it lies in \(\mathfrak m_M\). \(\square\)

So a non-zero monoid has exactly one maximal ideal, \(\mathfrak m_M\). A ring with exactly one maximal ideal is called local [Stacks, Tag 07BI]; in this sense every non-zero monoid is local. The difference from rings is that an ideal of a monoid has no addition to respect. In a ring the non-units are stable under multiplication by ring elements, but not under addition: in \(\mathbb Z\) the elements \(2\) and \(3\) are non-units and \(3-2=1\). For the same reason the union of two ideals of a monoid is an ideal, which is rarely true in a ring.

Definition 2.3. An ideal \(\mathfrak p\) of \(M\) is prime if \(\mathfrak p\neq M\) and \(ab\in\mathfrak p\) implies \(a\in\mathfrak p\) or \(b\in\mathfrak p\). The set of prime ideals of \(M\) is written \(\operatorname{Spec}M\). A face of \(M\) is a subset \(F\subseteq M\) with \(1\in F\), \(0\notin F\), and such that for all \(a,b\in M\):

Lemma 2.4. A subset \(\mathfrak p\subseteq M\) is a prime ideal if and only if \(M\setminus\mathfrak p\) is a face. So \(\mathfrak p\mapsto M\setminus\mathfrak p\) is a bijection from \(\operatorname{Spec}M\) to the set of faces of \(M\), and it reverses inclusions.

Proof. Let \(\mathfrak p\) be prime and \(F=M\setminus\mathfrak p\). Then \(0\notin F\), and \(1\in F\) because \(\mathfrak p\) is proper. If \(a,b\in F\) then \(ab\in F\), because \(\mathfrak p\) is prime. If \(ab\in F\) then \(a,b\in F\), because \(\mathfrak p\) is an ideal. Conversely let \(F\) be a face and \(\mathfrak p=M\setminus F\). Then \(0\in\mathfrak p\) and \(1\notin\mathfrak p\). If \(a\in\mathfrak p\) and \(b\in M\), then \(ab\notin F\), since \(ab\in F\) would force \(a\in F\). So \(\mathfrak p\) is a proper ideal. If \(ab\in\mathfrak p\) and \(a,b\notin\mathfrak p\), then \(a,b\in F\) and \(ab\in F\), a contradiction. \(\square\)

Examples 2.5. Let \(M\neq0\). The maximal ideal \(\mathfrak m_M\) is prime; its face \(M^\times\) is the smallest face. The ideal \(\{0\}\) is prime if and only if \(M\) is without zero divisors; its face \(M\setminus\{0\}\) is then the largest face. The zero monoid has no prime ideal.

Lemma 2.6. Let \(\varphi:M\to N\) be a morphism and \(\mathfrak q\) a prime ideal of \(N\). Then \(\varphi^{-1}(\mathfrak q)\) is a prime ideal of \(M\).

Proof. It contains \(0\), it is stable under multiplication by \(M\), and it does not contain \(1\), because \(\varphi(1)=1\notin\mathfrak q\). If \(ab\in\varphi^{-1}(\mathfrak q)\), then \(\varphi(a)\varphi(b)\in\mathfrak q\), so \(\varphi(a)\in\mathfrak q\) or \(\varphi(b)\in\mathfrak q\). \(\square\)

So a morphism \(\varphi:M\to N\) induces a map \(\varphi^*:\operatorname{Spec}N\to\operatorname{Spec}M\), \(\mathfrak q\mapsto\varphi^{-1}(\mathfrak q)\). A morphism between non-zero monoids is called local if \(\varphi^{-1}(\mathfrak m_N)=\mathfrak m_M\), that is, if \(\varphi(a)\) is a unit only when \(a\) is a unit.

Proposition 2.7 (points are morphisms to \(\mathbb F_1\)). For a prime ideal \(\mathfrak p\) of \(M\) define \(\chi_{\mathfrak p}:M\to\mathbb F_1\) by \(\chi_{\mathfrak p}(a)=0\) if \(a\in\mathfrak p\) and \(\chi_{\mathfrak p}(a)=1\) if \(a\notin\mathfrak p\). Then \(\mathfrak p\mapsto\chi_{\mathfrak p}\) is a bijection

with inverse \(\chi\mapsto\chi^{-1}(0)\).

Proof. The map \(\chi_{\mathfrak p}\) sends \(0\) to \(0\) and \(1\) to \(1\). It is multiplicative by Lemma 2.4: \(ab\notin\mathfrak p\) holds if and only if \(a\notin\mathfrak p\) and \(b\notin\mathfrak p\). Conversely \(\{0\}\) is a prime ideal of \(\mathbb F_1\), so \(\chi^{-1}(0)\) is a prime ideal by Lemma 2.6. The two constructions are inverse to each other. \(\square\)

For a ring one needs homomorphisms to all fields to see all points of the spectrum. For a monoid the single target \(\mathbb F_1\) is enough. Reference: [Connes–Consani 2010b, Theorem 3.1] recalls the corresponding statement for the schemes built from monoids.

A subset \(S\subseteq M\) is multiplicative if \(1\in S\) and \(st\in S\) for all \(s,t\in S\). A face is a multiplicative subset.

Theorem 2.8 (the prime ideal of a multiplicative set). Let \(S\) be a multiplicative subset of \(M\) with \(0\notin S\). Put

(a) \(S^{\mathrm{sat}}\) is a face. It is the smallest face that contains \(S\).

(b) \(\mathfrak p_S\) is a prime ideal with \(\mathfrak p_S\cap S=\emptyset\), and it contains every ideal \(I\) of \(M\) with \(I\cap S=\emptyset\).

Proof. (a) We have \(1\in S\subseteq S^{\mathrm{sat}}\), and \(0\notin S^{\mathrm{sat}}\) because \(0b=0\notin S\). If \(ab\in S\) and \(a'b'\in S\), then \((aa')(bb')\in S\), so \(S^{\mathrm{sat}}\) is closed under multiplication. If \(aa'\in S^{\mathrm{sat}}\), say \((aa')b\in S\), then \(a(a'b)\in S\) and \(a'(ab)\in S\), so \(a,a'\in S^{\mathrm{sat}}\). Hence \(S^{\mathrm{sat}}\) is a face. A face \(F\supseteq S\) contains every divisor of every element of \(S\), so \(F\supseteq S^{\mathrm{sat}}\).

(b) By (a) and Lemma 2.4, \(\mathfrak p_S\) is a prime ideal, and it is disjoint from \(S\). Let \(I\) be an ideal with \(I\cap S=\emptyset\) and let \(a\in I\). Then \(ab\in I\) for all \(b\), so \(ab\notin S\) for all \(b\). Hence \(a\notin S^{\mathrm{sat}}\), that is, \(a\in\mathfrak p_S\). \(\square\)

In a ring, a prime ideal that is disjoint from \(S\) and contains a given ideal is found with Zorn's lemma, and there is no largest one in general. Here the prime ideal is given by a formula: \(\mathfrak p_S\) is the set of elements that divide no element of \(S\).

Corollary 2.9. Let \(M\) be a monoid.

(a) If \(M\neq0\), then \(\operatorname{Spec}M\neq\emptyset\).

(b) Let \(f\in M\) be not nilpotent. Then \(\mathfrak p_f:=\{a\in M: a\ \text{divides no power}\ f^n,\ n\ge0\}\) is a prime ideal, and it is the largest prime ideal that does not contain \(f\).

(c) For an ideal \(I\) put \(\sqrt I=\{a\in M:a^n\in I\ \text{for some}\ n\ge1\}\). Then \(\sqrt I\) is the intersection of all prime ideals containing \(I\). In particular the set of nilpotent elements of \(M\) is the intersection of all prime ideals.

Proof. (a) \(\mathfrak m_M\) is prime. (b) The set \(S=\{f^n:n\ge0\}\) is multiplicative and \(0\notin S\). By Theorem 2.8, \(\mathfrak p_f=\mathfrak p_S\) is prime and \(f\notin\mathfrak p_f\). A prime ideal \(\mathfrak q\) with \(f\notin\mathfrak q\) contains no power of \(f\), so it is disjoint from \(S\) and \(\mathfrak q\subseteq\mathfrak p_S\). (c) If \(a^n\in I\subseteq\mathfrak p\) with \(\mathfrak p\) prime, then \(a\in\mathfrak p\). Conversely let \(a\notin\sqrt I\) and \(S=\{a^n:n\ge0\}\). No power \(a^n\) with \(n\ge1\) lies in \(I\), and \(1\notin I\), because otherwise \(I=M\) would contain \(a\). So \(S\cap I=\emptyset\), and \(0\notin S\) because \(0\in I\). By Theorem 2.8 the prime ideal \(\mathfrak p_S\) contains \(I\) and not \(a\). (If \(I=M\), both sides are \(M\), the intersection of the empty family.) \(\square\)

Lemma 2.10. The union of a non-empty family of prime ideals of \(M\) is a prime ideal.

Proof. The complement of the union is the intersection of the faces \(F_i=M\setminus\mathfrak p_i\). It contains \(1\) and not \(0\), and \(ab\) lies in all \(F_i\) if and only if \(a\) and \(b\) lie in all \(F_i\). So it is a face, and Lemma 2.4 applies. \(\square\)

So any two prime ideals \(\mathfrak p,\mathfrak q\) are contained in a smallest prime ideal, namely \(\mathfrak p\cup\mathfrak q\). In \(\mathbb Z\) the union \(2\mathbb Z\cup3\mathbb Z\) is not an ideal of the ring.

3. The spectrum as a topological space

Definition 3.1. For a subset \(T\subseteq M\) and an element \(f\in M\) put

Proposition 3.2. Let \(M\) be a monoid.

(a) \(V(T)=V(\langle T\rangle)\) for every subset \(T\).

(b) \(V(\{0\})=\operatorname{Spec}M\) and \(V(M)=\emptyset\). For ideals \(I,J\) we have \(V(I)\cup V(J)=V(I\cap J)\). For a non-empty family of ideals \(I_\alpha\) we have \(\bigcap_\alpha V(I_\alpha)=V(\bigcup_\alpha I_\alpha)\).

(c) The sets \(V(I)\), \(I\) an ideal, are the closed sets of a topology on \(\operatorname{Spec}M\). The sets \(D(f)\) are open and form a basis of this topology, and \(D(f)\cap D(g)=D(fg)\).

(d) \(D(f)=\emptyset\) if and only if \(f\) is nilpotent. If \(M\neq0\), then \(D(f)=\operatorname{Spec}M\) if and only if \(f\) is a unit.

(e) \(D(f)\subseteq D(g)\) if and only if \(g\) divides \(f^n\) for some \(n\ge1\).

Proof. (a) A prime ideal that contains \(T\) contains \(\langle T\rangle\). (b) Every prime ideal contains \(0\) and none contains \(1\). A prime ideal that contains \(I\) or \(J\) contains \(I\cap J\). If \(\mathfrak p\) contains neither, take \(a\in I\setminus\mathfrak p\) and \(b\in J\setminus\mathfrak p\); then \(ab\in I\cap J\) and \(ab\notin\mathfrak p\). The last formula is clear, and \(\bigcup_\alpha I_\alpha\) is an ideal by Lemma 2.2. (c) Part (b) gives the axioms for closed sets. \(D(f)\) is the complement of \(V(fM)\), and the complement of \(V(I)\) is \(\bigcup_{f\in I}D(f)\). The formula \(D(f)\cap D(g)=D(fg)\) restates Lemma 2.4. (d) \(D(f)\) is empty if and only if \(f\) lies in all prime ideals, that is, \(f\) is nilpotent (Corollary 2.9(c)). Every prime ideal lies in \(\mathfrak m_M\). So \(D(f)=\operatorname{Spec}M\) if and only if \(f\notin\mathfrak m_M\). (e) If \(f^n=gh\) and \(g\in\mathfrak p\), then \(f^n\in\mathfrak p\) and \(f\in\mathfrak p\). Conversely suppose that no power \(f^n\), \(n\ge1\), lies in \(gM\). Then \(1\notin gM\) as well, since otherwise \(gM=M\) would contain \(f\). So \(S=\{f^n:n\ge0\}\) is disjoint from the ideal \(gM\), and \(0\notin S\). By Theorem 2.8 the prime ideal \(\mathfrak p_S\) contains \(g\) and not \(f\). It lies in \(D(f)\) and not in \(D(g)\). \(\square\)

The topology of (c) is the Zariski topology. It is the topology generated by the sets \(D(f)\). From now on \(\operatorname{Spec}M\) carries this topology. For a morphism \(\varphi:M\to N\) the map \(\varphi^*\) of Lemma 2.6 is continuous, because \((\varphi^*)^{-1}(D(f))=D(\varphi(f))\).

Proposition 3.3 (order and the closed point). Let \(M\neq0\) and \(X=\operatorname{Spec}M\).

(a) The closure of a point \(\mathfrak p\) is \(V(\mathfrak p)\). So \(\mathfrak q\) lies in the closure of \(\mathfrak p\) if and only if \(\mathfrak p\subseteq\mathfrak q\), and different points have different closures.

(b) Every open set \(U\) is stable under generization: if \(\mathfrak p\in U\) and \(\mathfrak q\subseteq\mathfrak p\) is prime, then \(\mathfrak q\in U\).

(c) The point \(\mathfrak m_M\) lies in every non-empty closed subset of \(X\), and \(X\) is the only open set that contains \(\mathfrak m_M\).

(d) \(X\) is quasi-compact and connected.

Proof. (a) \(V(\mathfrak p)\) is closed and contains \(\mathfrak p\). A closed set \(V(I)\) that contains \(\mathfrak p\) has \(I\subseteq\mathfrak p\), hence \(V(\mathfrak p)\subseteq V(I)\). (b) \(U\) is a union of sets \(D(f)\), and \(f\notin\mathfrak p\) implies \(f\notin\mathfrak q\). (c) If \(V(I)\neq\emptyset\) then \(I\) is proper, so \(I\subseteq\mathfrak m_M\) by Lemma 2.2. The complement of an open set that contains \(\mathfrak m_M\) is a closed set that does not contain \(\mathfrak m_M\), so it is empty. (d) In an open cover of \(X\) one member contains \(\mathfrak m_M\), so it is \(X\). If \(X=U\sqcup U'\) with \(U,U'\) open, the one that contains \(\mathfrak m_M\) is \(X\). \(\square\)

The spectrum of a ring is quasi-compact too [Stacks, Tag 00E8], but it need not be connected: the spectrum of a product of two rings is the disjoint union of the two spectra [Stacks, Tag 00ED]. The spectrum of a product of two non-zero monoids is connected. Exercise 2 describes it.

For a prime ideal \(\mathfrak p\) write

for the set of its generizations. By Proposition 3.3(b) every open set that contains \(\mathfrak p\) contains \(\Lambda(\mathfrak p)\). In fact \(\Lambda(\mathfrak p)\) is the intersection of all open sets containing \(\mathfrak p\): if \(\mathfrak q\not\subseteq\mathfrak p\), take \(f\in\mathfrak q\setminus\mathfrak p\); then \(D(f)\) contains \(\mathfrak p\) and not \(\mathfrak q\).

Theorem 3.4 (basic open sets). For a non-empty open subset \(U\) of \(\operatorname{Spec}M\) the following are equivalent.

(i) \(U=D(f)\) for some \(f\in M\).

(ii) \(U\) has a largest element for inclusion.

(iii) \(U=\Lambda(\mathfrak p)\) for some prime ideal \(\mathfrak p\).

If \(U=D(f)\), then the largest element of \(U\) is \(\mathfrak p_f\).

Proof. (i)⇒(iii): \(D(f)\neq\emptyset\), so \(f\) is not nilpotent. By Corollary 2.9(b) every \(\mathfrak q\in D(f)\) lies in \(\mathfrak p_f\). Conversely \(\mathfrak q\subseteq\mathfrak p_f\) implies \(f\notin\mathfrak q\). So \(D(f)=\Lambda(\mathfrak p_f)\). (iii)⇒(ii) is clear. (ii)⇒(i): let \(\mathfrak p\) be the largest element of \(U\). Since \(U\) is open there is \(f\) with \(\mathfrak p\in D(f)\subseteq U\). Every \(\mathfrak q\in U\) satisfies \(\mathfrak q\subseteq\mathfrak p\), hence \(f\notin\mathfrak q\). So \(U\subseteq D(f)\). \(\square\)

Corollary 3.5. Let \(f\in M\) be not nilpotent. If a family of open subsets of \(\operatorname{Spec}M\) covers \(D(f)\), then one member of the family contains \(D(f)\).

Proof. One member \(W\) contains the point \(\mathfrak p_f\). By Proposition 3.3(b) it contains \(\Lambda(\mathfrak p_f)=D(f)\). \(\square\)

The set \(\Lambda(\mathfrak p)\) need not be open (Example 9.2), and in condition (ii) a largest element cannot be weakened to a unique maximal element:

Example 3.6 (one maximal point, no largest point). Let \(Q\) be the set of all sequences \(\gamma=(\gamma_1,\gamma_2,\dots)\) of integers with only finitely many non-zero terms, such that \(\gamma=0\) or the first non-zero term of \(\gamma\) is positive. \(Q\) is closed under addition, and \(Q\cap(-Q)=\{0\}\). Let \(e_n\in Q\) be the sequence with \(n\)-th term \(1\) and all other terms \(0\). Let \(M=\mathbb F_1[Q\times\mathbb N]\), and put \(f_n=\chi^{(e_n,0)}\) and \(t=\chi^{(0,1)}\). Then:

• \(f_{n+1}\) divides \(f_n\), because \(e_n-e_{n+1}\in Q\);
• \(f_n\) divides no power of \(f_{n+1}\), because the first non-zero term of \(ke_{n+1}-e_n\) is \(-1\);
• the divisors of \(t^k\) are the \(t^j\) with \(j\le k\), because \(\gamma\in Q\) and \(-\gamma\in Q\) force \(\gamma=0\);
• \(t\) divides no power of \(f_n\), because the second coordinate of \((ke_n,0)-(0,1)\) is negative.

By Proposition 3.2(e) the first two facts give \(D(f_1)\subsetneq D(f_2)\subsetneq\cdots\). Let \(\mathfrak w_n=\mathfrak p_{f_n}\) and \(\mathfrak z=\mathfrak p_t\) be the largest points of \(D(f_n)\) and \(D(t)\). By Theorem 3.4 the \(\mathfrak w_n\) form a strictly increasing chain. By the third fact \(\mathfrak z=\{\chi^{(\gamma,j)}:\gamma\neq0\}\cup\{0\}\), so \(f_n\in\mathfrak z\) for all \(n\). By the fourth fact \(t\in\mathfrak w_n\) for all \(n\). Now consider the open set

The point \(\mathfrak z\) is maximal in \(U\): a point \(\mathfrak q\in U\) with \(\mathfrak q\supsetneq\mathfrak z\) is not in \(D(t)\), so \(f_n\notin\mathfrak q\) for some \(n\); but \(f_n\in\mathfrak z\subseteq\mathfrak q\). It is the only maximal point: a maximal point \(\mathfrak q\) of \(U\) that lies in \(D(t)\) satisfies \(\mathfrak q\subseteq\mathfrak z\), so \(\mathfrak q=\mathfrak z\); and a point of \(D(f_n)\) lies in \(\mathfrak w_n\subsetneq\mathfrak w_{n+1}\in U\), so it is not maximal. But \(\mathfrak z\) is not the largest point of \(U\), because \(t\in\mathfrak w_1\) and \(t\notin\mathfrak z\). So \(U\) has exactly one maximal point and no largest point. By Theorem 3.4, \(U\) is not of the form \(D(f)\). It is not even homeomorphic to the spectrum of a monoid: by Proposition 3.3 the spectrum of a non-zero monoid has a point that lies in the closure of every point, and the closure of a point \(\mathfrak q\) in \(U\) is \(V(\mathfrak q)\cap U\).

Reference: [Cortiñas–Haesemeyer–Walker–Weibel 2015, Lemma 2.4] states that an open subset of a monoid scheme with a unique maximal point is affine; this fails for the set \(U\) above, so Theorem 3.4 asks for a largest point. For finitely generated monoids the two conditions agree (Corollary 7.3).

4. Localization

Let \(S\) be a multiplicative subset of \(M\). If \(0\notin S\), Theorem 2.8 gives the face \(S^{\mathrm{sat}}\) and the prime ideal \(\mathfrak p_S\). If \(0\in S\) we put \(S^{\mathrm{sat}}=M\). In both cases \(S^{\mathrm{sat}}\) is the set of elements that divide an element of \(S\).

Proposition 4.1 (localization). Let \(S\) be a multiplicative subset of \(M\). On \(M\times S\) define \((a,s)\sim(a',s')\) if \(tas'=ta's\) for some \(t\in S\).

(a) This is an equivalence relation. Write \(a/s\) for the class of \((a,s)\) and \(S^{-1}M\) for the set of classes. The product \((a/s)(b/u)=ab/su\) is well defined and makes \(S^{-1}M\) a monoid with zero \(0/1\) and one \(1/1\). The map \(\iota:M\to S^{-1}M\), \(a\mapsto a/1\), is a morphism, and \(\iota(s)\) is a unit for every \(s\in S\).

(b) If \(\varphi:M\to N\) is a morphism with \(\varphi(S)\subseteq N^\times\), there is exactly one morphism \(\varphi':S^{-1}M\to N\) with \(\varphi'\circ\iota=\varphi\). It is given by \(\varphi'(a/s)=\varphi(a)\varphi(s)^{-1}\).

(c) \(\iota(a)=\iota(b)\) if and only if \(ta=tb\) for some \(t\in S\). The monoid \(S^{-1}M\) is zero if and only if \(0\in S\).

(d) \(a/s\) is a unit of \(S^{-1}M\) if and only if \(a\in S^{\mathrm{sat}}\).

Proof. (a) The relation is reflexive and symmetric. If \(tas'=ta's\) and \(ua's''=ua''s'\) with \(t,u\in S\), then

and \(tus'\in S\). So the relation is transitive. If \(tas'=ta's\), then \(t(ab)(s'u)=t(a'b)(su)\), so the product does not depend on the representative of the first factor; by commutativity the same holds for the second. The monoid axioms are inherited from \(M\). The map \(\iota\) is a morphism by the definition of the product, and \((s/1)(1/s)=s/s=1/1\).

(b) Since \(a/s=\iota(a)\iota(s)^{-1}\), the morphism \(\varphi'\) is unique. It is well defined: from \(tas'=ta's\) we get \(\varphi(t)\varphi(a)\varphi(s')=\varphi(t)\varphi(a')\varphi(s)\), and \(\varphi(t)\), \(\varphi(s)\), \(\varphi(s')\) are units. It is multiplicative and preserves \(0\) and \(1\).

(c) The first statement is the definition of the relation for \(s=s'=1\). The monoid \(S^{-1}M\) is zero if and only if \(1/1=0/1\), that is, \(t=t\cdot0=0\) for some \(t\in S\).

(d) If \(ab=s'\in S\), then \((a/s)(bs/s')=abs/ss'=1/1\). Conversely, if \((a/s)(b/u)=1/1\), then \(t\,ab=t\,su\) for some \(t\in S\), so \(a\cdot tb\in S\) and \(a\in S^{\mathrm{sat}}\). \(\square\)

For \(f\in M\) we write \(M_f=S^{-1}M\) with \(S=\{f^n:n\ge0\}\). For a prime ideal \(\mathfrak p\) we write \(M_{\mathfrak p}=S^{-1}M\) with \(S=M\setminus\mathfrak p\), which is a face and so a multiplicative set. If \(S\subseteq T\) are multiplicative subsets, (b) gives a morphism \(S^{-1}M\to T^{-1}M\), \(a/s\mapsto a/s\). In particular, for prime ideals \(\mathfrak q\subseteq\mathfrak p\) there is a morphism

These morphisms satisfy \(\rho_{\mathfrak p,\mathfrak p}=\mathrm{id}\) and \(\rho_{\mathfrak q,\mathfrak r}\circ\rho_{\mathfrak p,\mathfrak q}=\rho_{\mathfrak p,\mathfrak r}\) for \(\mathfrak r\subseteq\mathfrak q\subseteq\mathfrak p\).

Proposition 4.2 (prime ideals of a localization). Let \(S\) be a multiplicative subset of \(M\). The map \(\iota^*:\operatorname{Spec}S^{-1}M\to\operatorname{Spec}M\) is a homeomorphism onto the subspace \(\{\mathfrak p\in\operatorname{Spec}M:\mathfrak p\cap S=\emptyset\}\). Its inverse sends \(\mathfrak p\) to \(S^{-1}\mathfrak p:=\{a/s:a\in\mathfrak p,\ s\in S\}\).

Proof. Let \(\mathfrak Q\) be a prime ideal of \(S^{-1}M\). Then \(\iota^{-1}(\mathfrak Q)\) is prime by Lemma 2.6, and it is disjoint from \(S\), because \(\iota(S)\) consists of units. Now let \(\mathfrak p\) be a prime ideal of \(M\) with \(\mathfrak p\cap S=\emptyset\). We claim that \(a/s\in S^{-1}\mathfrak p\) if and only if \(a\in\mathfrak p\). Indeed, if \(a/s=a'/s'\) with \(a'\in\mathfrak p\), then \(tas'=ta's\in\mathfrak p\) for some \(t\in S\); since \(ts'\in S\) is not in \(\mathfrak p\), we get \(a\in\mathfrak p\). The claim shows that \(S^{-1}\mathfrak p\) is an ideal, that \(1/1\notin S^{-1}\mathfrak p\), that \((a/s)(b/u)\in S^{-1}\mathfrak p\) implies \(a\in\mathfrak p\) or \(b\in\mathfrak p\), and that \(\iota^{-1}(S^{-1}\mathfrak p)=\mathfrak p\). So \(S^{-1}\mathfrak p\) is a prime ideal that maps to \(\mathfrak p\). Conversely, for a prime ideal \(\mathfrak Q\) of \(S^{-1}M\) we have \(a/s\in\mathfrak Q\) if and only if \(a/1\in\mathfrak Q\), because \(s/1\) is a unit. So \(\mathfrak Q=S^{-1}(\iota^{-1}(\mathfrak Q))\). Hence the two maps are inverse bijections. The map \(\iota^*\) is continuous. It sends \(D(a/s)=D(a/1)\) onto the intersection of \(D(a)\) with the image, so it is a homeomorphism onto its image. \(\square\)

Theorem 4.3 (every localization is local at a prime). Let \(S\) be a multiplicative subset of \(M\) with \(0\notin S\).

(a) The canonical morphism \(S^{-1}M\to(S^{\mathrm{sat}})^{-1}M=M_{\mathfrak p_S}\) is an isomorphism.

(b) The maximal ideal of \(S^{-1}M\) is \(S^{-1}\mathfrak p_S\), and its preimage in \(M\) is \(\mathfrak p_S\).

(c) If \(f\in M\) is not nilpotent, then \(M_f\cong M_{\mathfrak p_f}\). If \(\mathfrak p\) is a prime ideal, then the maximal ideal of \(M_{\mathfrak p}\) is \(\mathfrak pM_{\mathfrak p}:=\{a/s:a\in\mathfrak p,\ s\notin\mathfrak p\}\), its preimage in \(M\) is \(\mathfrak p\), and the units of \(M_{\mathfrak p}\) are the fractions \(a/s\) with \(a,s\notin\mathfrak p\).

Proof. (a) By Proposition 4.1(d) the elements of \(S^{\mathrm{sat}}\) become units in \(S^{-1}M\). So Proposition 4.1(b) gives morphisms \(S^{-1}M\to(S^{\mathrm{sat}})^{-1}M\) and \((S^{\mathrm{sat}})^{-1}M\to S^{-1}M\) that are compatible with the maps from \(M\). Both composites are compatible with the maps from \(M\), so they are identities by the uniqueness in Proposition 4.1(b). (b) The monoid \(S^{-1}M\) is not zero. By Proposition 4.1(d) its non-units are the fractions \(a/s\) with \(a\notin S^{\mathrm{sat}}\), that is, with \(a\in\mathfrak p_S\). By the claim in the proof of Proposition 4.2 these form the set \(S^{-1}\mathfrak p_S\), whose preimage in \(M\) is \(\mathfrak p_S\). (c) For \(S=\{f^n\}\) we have \(\mathfrak p_S=\mathfrak p_f\) (Corollary 2.9). For \(S=M\setminus\mathfrak p\) we have \(S^{\mathrm{sat}}=S\), because \(S\) is a face, and \(\mathfrak p_S=\mathfrak p\). \(\square\)

Reference: [Cortiñas–Haesemeyer–Walker–Weibel 2015, Lemma 1.1]; [Chu–Lorscheid–Santhanam 2012, Corollary 2.5].

For a ring \(R\) the localization \(R_f\) is rarely a local ring. For a monoid, \(M_f\) and \(M_{\mathfrak p_f}\) are the same thing.

Examples 4.4.

(a) If \(M\neq0\), the map \(\iota:M\to M_{\mathfrak m_M}\) is an isomorphism: the set \(M\setminus\mathfrak m_M\) consists of units already, so the identity of \(M\) has the universal property of Proposition 4.1(b).

(b) \(\mathbb F_1[x]_x=\mathbb F_1[x^{\pm1}]\), and \(\mathbb F_1[x_1,\dots,x_n]_{x_1\cdots x_n}=\mathbb F_1[x_1^{\pm1},\dots,x_n^{\pm1}]\).

(c) Let \(M=\{0,1\}\cup\{x^a:a\ge1\}\cup\{y^b:b\ge1\}\), with \(x^ax^{a'}=x^{a+a'}\), \(y^by^{b'}=y^{b+b'}\) and \(x^ay^b=0\). In \(M_x\) we have \(y/1=0/1\), because \(x\cdot y=x\cdot0\). So \(M_x=\{0\}\cup\{x^n:n\in\mathbb Z\}\cong\mathbb F_1[x^{\pm1}]\), and \(\iota:M\to M_x\) is not injective.

5. The structure sheaf

Throughout this section \(X=\operatorname{Spec}M\). An element of \(M_{\mathfrak p}\) is called a germ at \(\mathfrak p\).

Definition 5.1. For an open subset \(U\subseteq X\) let \(\mathcal O(U)=\mathcal O_M(U)\) be the set of all families \(s=(s_{\mathfrak p})_{\mathfrak p\in U}\) with \(s_{\mathfrak p}\in M_{\mathfrak p}\) and

For open sets \(V\subseteq U\) the restriction \(\mathcal O(U)\to\mathcal O(V)\) forgets the components outside \(V\).

The set \(\mathcal O(U)\) is a submonoid of the product monoid \(\prod_{\mathfrak p\in U}M_{\mathfrak p}\), because the maps \(\rho_{\mathfrak p,\mathfrak q}\) are morphisms. The restriction maps are morphisms. The monoid \(\mathcal O(\emptyset)\) has one element.

Theorem 5.2.

(a) \(\mathcal O\) is a sheaf of monoids on \(X\).

(b) If an open set \(U\) has a largest point \(\mathfrak p\), then \(\mathcal O(U)\to M_{\mathfrak p}\), \(s\mapsto s_{\mathfrak p}\), is an isomorphism. In particular \(\mathcal O(D(f))\cong M_f\) for every \(f\in M\), and \(\mathcal O(X)\cong M\).

(c) For every \(\mathfrak p\in X\) the map from the stalk \(\mathcal O_{\mathfrak p}\) to \(M_{\mathfrak p}\) induced by \(s\mapsto s_{\mathfrak p}\) is an isomorphism.

Proof. (a) Let \(U=\bigcup_iU_i\) be an open cover. Two elements of \(\mathcal O(U)\) with the same restrictions to all \(U_i\) have the same components, so they are equal. Let \(s^i\in\mathcal O(U_i)\) be sections that agree on all intersections \(U_i\cap U_j\). Define \(s_{\mathfrak p}=s^i_{\mathfrak p}\) for any \(i\) with \(\mathfrak p\in U_i\); this does not depend on \(i\). If \(\mathfrak q\subseteq\mathfrak p\) are in \(U\) and \(\mathfrak p\in U_i\), then \(\mathfrak q\in U_i\) by Proposition 3.3(b), so \(\rho_{\mathfrak p,\mathfrak q}(s_{\mathfrak p})=\rho_{\mathfrak p,\mathfrak q}(s^i_{\mathfrak p}) =s^i_{\mathfrak q}=s_{\mathfrak q}\). So \(s\in\mathcal O(U)\), and \(s\) restricts to \(s^i\) on \(U_i\).

(b) For \(x\in M_{\mathfrak p}\) the family \((\rho_{\mathfrak p,\mathfrak q}(x))_{\mathfrak q\in U}\) lies in \(\mathcal O(U)\), because \(\rho_{\mathfrak q,\mathfrak r}\circ\rho_{\mathfrak p,\mathfrak q}=\rho_{\mathfrak p,\mathfrak r}\), and its component at \(\mathfrak p\) is \(x\). Every \(s\in\mathcal O(U)\) satisfies \(s_{\mathfrak q}=\rho_{\mathfrak p,\mathfrak q}(s_{\mathfrak p})\) for all \(\mathfrak q\in U\), so \(s\) is the family attached to \(x=s_{\mathfrak p}\). Hence \(s\mapsto s_{\mathfrak p}\) is bijective. If \(f\) is not nilpotent, \(D(f)\) has the largest point \(\mathfrak p_f\) (Theorem 3.4) and \(M_{\mathfrak p_f}\cong M_f\) (Theorem 4.3). If \(f\) is nilpotent, then \(D(f)=\emptyset\) and \(M_f=0\). If \(M\neq0\), then \(X\) has the largest point \(\mathfrak m_M\), and \(M_{\mathfrak m_M}=M\) (Example 4.4(a)).

(c) The maps \(s\mapsto s_{\mathfrak p}\) are compatible with restriction, so they induce a morphism \(\mathcal O_{\mathfrak p}\to M_{\mathfrak p}\). It is surjective: an element of \(M_{\mathfrak p}\) is \(a/u\) with \(u\notin\mathfrak p\), and the family of the fractions \(a/u\in M_{\mathfrak q}\), \(\mathfrak q\in D(u)\), is a section over \(D(u)\) with value \(a/u\) at \(\mathfrak p\). It is injective: let \(s\in\mathcal O(U)\) and \(s'\in\mathcal O(U')\) with \(\mathfrak p\in U\cap U'\) and \(s_{\mathfrak p}=s'_{\mathfrak p}\). Choose \(f\notin\mathfrak p\) with \(D(f)\subseteq U\cap U'\), and let \(\mathfrak z=\mathfrak p_f\) be the largest point of \(D(f)\). Since \(M_f\cong M_{\mathfrak z}\), we can write \(s_{\mathfrak z}=a/f^n\) and \(s'_{\mathfrak z}=a'/f^n\) with \(a,a'\in M\) and the same \(n\). Then \(s_{\mathfrak q}=a/f^n\) and \(s'_{\mathfrak q}=a'/f^n\) in \(M_{\mathfrak q}\) for all \(\mathfrak q\in D(f)\). For \(\mathfrak q=\mathfrak p\) this gives \(t\,a\,f^n=t\,a'\,f^n\) for some \(t\notin\mathfrak p\). If \(\mathfrak q\in D(tf)\), then \(t\notin\mathfrak q\), so \(a/f^n=a'/f^n\) in \(M_{\mathfrak q}\). So \(s\) and \(s'\) agree on the neighbourhood \(D(tf)\) of \(\mathfrak p\). \(\square\)

For the spectrum of a ring, the structure sheaf is built in [Stacks, Tag 01HR], and the proof that \(D(f)\) has the sections \(R_f\) needs an argument with partitions of unity. For a monoid, part (b) is a formality, because \(D(f)\) has a largest point. The next proposition shows that Definition 5.1 agrees with the definition by local fractions used in [Deitmar 2005, Section 2.1] and [Cortiñas–Haesemeyer–Walker–Weibel 2015, Section 1].

Proposition 5.3 (sections are locally fractions). Let \(U\subseteq X\) be open and let \(s=(s_{\mathfrak p})_{\mathfrak p\in U}\) be a family with \(s_{\mathfrak p}\in M_{\mathfrak p}\). Then \(s\in\mathcal O(U)\) if and only if every point of \(U\) has an open neighbourhood \(V\subseteq U\) and elements \(a,u\in M\) such that \(u\notin\mathfrak q\) and \(s_{\mathfrak q}=a/u\) in \(M_{\mathfrak q}\) for all \(\mathfrak q\in V\).

Proof. Suppose the local condition holds and let \(\mathfrak q\subseteq\mathfrak p\) be in \(U\). Choose \(V\), \(a\), \(u\) for the point \(\mathfrak p\). Then \(\mathfrak q\in V\) by Proposition 3.3(b), so \(s_{\mathfrak q}=a/u=\rho_{\mathfrak p,\mathfrak q}(a/u)=\rho_{\mathfrak p,\mathfrak q}(s_{\mathfrak p})\). Conversely let \(s\in\mathcal O(U)\) and \(\mathfrak p\in U\). Choose \(g\) with \(\mathfrak p\in D(g)\subseteq U\), and let \(\mathfrak z=\mathfrak p_g\). Write \(s_{\mathfrak z}=a/g^n\), using \(M_g\cong M_{\mathfrak z}\). For \(\mathfrak q\in D(g)\) we have \(\mathfrak q\subseteq\mathfrak z\), hence \(s_{\mathfrak q}=\rho_{\mathfrak z,\mathfrak q}(s_{\mathfrak z})=a/g^n\), and \(g^n\notin\mathfrak q\). So \(V=D(g)\) and \(u=g^n\) work. \(\square\)

Reference: [Deitmar 2005, Proposition 2.1] proves the statements on stalks and global sections of Theorem 5.2 for the sheaf of local fractions. [Cortiñas–Haesemeyer–Walker–Weibel 2015, Remark 2.12] states the description of sections as compatible families.

Remark 5.4 (morphisms). Let \(\varphi:M\to N\) be a morphism, \(\mathfrak Q\in\operatorname{Spec}N\) and \(\mathfrak p=\varphi^{-1}(\mathfrak Q)\). By Proposition 4.1(b), \(\varphi\) induces a morphism \(\varphi_{\mathfrak Q}:M_{\mathfrak p}\to N_{\mathfrak Q}\), \(a/s\mapsto\varphi(a)/\varphi(s)\). It is local: by Theorem 4.3(c), \(\varphi(a)/\varphi(s)\) is a unit if and only if \(\varphi(a)\notin\mathfrak Q\), that is, \(a\notin\mathfrak p\). For an open set \(U\subseteq\operatorname{Spec}M\) the rule \((s_{\mathfrak p})\mapsto(\varphi_{\mathfrak Q}(s_{\varphi^*\mathfrak Q}))_{\mathfrak Q}\) is a morphism \(\mathcal O_M(U)\to\mathcal O_N((\varphi^*)^{-1}(U))\), compatible with restrictions. For \(U=\operatorname{Spec}M\) it is \(\varphi\) itself. The lesson Monoid schemes takes this up: it shows that \(M\mapsto(\operatorname{Spec}M,\mathcal O_M)\) is a contravariant equivalence from monoids to affine monoid schemes.

Example 5.5 (the punctured plane). Let \(M=\mathbb F_1[x,y]\). A monomial lies in a face if and only if each variable that occurs in it does. So \(M\) has four faces and four prime ideals: \(\{0\}\), \(xM\), \(yM\) and \(\mathfrak m=xM\cup yM\). Since \(M\) is a submonoid of the group with zero \(G_0=\mathbb F_1[x^{\pm1},y^{\pm1}]\), every localization \(S^{-1}M\) with \(0\notin S\) embeds in \(G_0\) by \(a/s\mapsto as^{-1}\), and the maps \(\rho\) become inclusions. We get

Let \(U=X\setminus\{\mathfrak m\}=D(x)\cup D(y)=\{\{0\},xM,yM\}\). A section over \(U\) is a triple of germs at the three points with the same image in \(G_0\). So

The restriction \(\mathcal O(X)\to\mathcal O(U)\) is an isomorphism, although \(U\neq X\). The set \(U\) has two maximal points and no largest point, so it is not of the form \(D(f)\). Exercise 4 shows that the restriction to the punctured spectrum is not always an isomorphism.

Example 5.6 (a disconnected open set). Let \(M\) be the monoid of Example 4.4(c). Its faces are \(\{1\}\), \(\{1\}\cup\{x^a\}\) and \(\{1\}\cup\{y^b\}\); the set of all non-zero elements is not a face, because \(xy=0\). So \(X\) has three points: \(\mathfrak p_x=\{0\}\cup\{x^a\}\), \(\mathfrak p_y=\{0\}\cup\{y^b\}\) and \(\mathfrak m=\mathfrak p_x\cup\mathfrak p_y\). We have \(D(x)=\{\mathfrak p_y\}\), \(D(y)=\{\mathfrak p_x\}\) and \(D(x)\cap D(y)=D(xy)=\emptyset\). The open set \(U=D(x)\cup D(y)\) is a discrete space with two points, and

The space \(X\) itself is connected, and it has two minimal points.

6. Quotients: congruences and ideals

Definition 6.1. A congruence on \(M\) is an equivalence relation \(\sim\) on \(M\) such that \(a\sim b\) implies \(ac\sim bc\) for all \(c\in M\).

Proposition 6.2.

(a) Let \(\sim\) be a congruence on \(M\). The set \(M/{\sim}\) of classes \([a]\) is a monoid with \([a][b]=[ab]\), and \(\pi:M\to M/{\sim}\), \(a\mapsto[a]\), is a surjective morphism.

(b) Let \(\varphi:M\to N\) be a morphism. The relation \(a\sim_\varphi b\iff\varphi(a)=\varphi(b)\) is a congruence, called the kernel congruence of \(\varphi\), and \(\varphi\) induces an isomorphism of \(M/{\sim_\varphi}\) onto the submonoid \(\varphi(M)\) of \(N\).

(c) Let \(\sim\) be a congruence and \(\varphi:M\to N\) a morphism such that \(a\sim b\) implies \(\varphi(a)=\varphi(b)\). Then there is exactly one morphism \(\bar\varphi:M/{\sim}\to N\) with \(\bar\varphi\circ\pi=\varphi\).

(d) The intersection of a non-empty family of congruences, as subsets of \(M\times M\), is a congruence. So every subset \(R\subseteq M\times M\) lies in a smallest congruence, the congruence generated by \(R\).

Proof. (a) If \(a\sim a'\) and \(b\sim b'\), then \(ab\sim a'b\sim a'b'\). So the product is well defined, and the monoid axioms pass to the classes. (b) If \(\varphi(a)=\varphi(b)\) then \(\varphi(ac)=\varphi(bc)\). The map \([a]\mapsto\varphi(a)\) is a well defined injective morphism with image \(\varphi(M)\). (c) Put \(\bar\varphi([a])=\varphi(a)\). (d) The conditions of Definition 6.1 pass to intersections, and \(M\times M\) is a congruence containing \(R\). \(\square\)

So congruences on \(M\) describe the surjective morphisms out of \(M\), as ideals do for a ring.

Definition 6.3. Let \(I\) be an ideal of \(M\). Define \(a\sim_Ib\) if \(a=b\) or \(a,b\in I\). This is a congruence. The monoid \(M/I:=M/{\sim_I}\) is the quotient of \(M\) by the ideal \(I\). As a set it is \((M\setminus I)\sqcup\{0\}\); the product of \(a,b\in M\setminus I\) is \(ab\) if \(ab\notin I\) and \(0\) otherwise.

For example \(\mathbb F_1[x]/\langle x^n\rangle=\{0,1,x,\dots,x^{n-1}\}\) with \(x^n=0\); for \(n=2\) this is the monoid \(\{0,1,\varepsilon\}\) of Example 1.3(f). The monoid of Example 4.4(c) is \(\mathbb F_1[x,y]/\langle xy\rangle\). For every non-zero monoid, \(M/\mathfrak m_M=(M^\times)_0\).

Proposition 6.4. Let \(\sim\) be a congruence on \(M\) and let \(I_\sim=[0]\) be the class of \(0\).

(a) \(I_\sim\) is an ideal, and \(a\sim_{I_\sim}b\) implies \(a\sim b\).

(b) \(\sim\) is of the form \(\sim_I\) for an ideal \(I\) if and only if every class other than \(I_\sim\) has exactly one element. In that case \(I=I_\sim\).

(c) A morphism \(\varphi:M\to N\) factors through \(\pi:M\to M/I\) if and only if \(\varphi(I)=\{0\}\), and the factorization is unique.

Proof. (a) If \(a\sim0\) then \(ab\sim0b=0\). Two elements of \(I_\sim\) are both equivalent to \(0\). (b) The classes of \(\sim_I\) are \(I\) and the one-element sets \(\{a\}\), \(a\notin I\). (c) If \(\varphi(I)=\{0\}\), then \(a\sim_Ib\) implies \(\varphi(a)=\varphi(b)\), and Proposition 6.2(c) applies. Conversely \(\pi(I)=\{0\}\). \(\square\)

Proposition 6.5 (the spectrum of a quotient). Let \(\sim\) be a congruence on \(M\) and \(\pi:M\to M/{\sim}\).

(a) The map \(\pi^*:\operatorname{Spec}(M/{\sim})\to\operatorname{Spec}M\) is a homeomorphism onto its image. The image is the set of prime ideals of \(M\) that are unions of classes of \(\sim\).

(b) If \(\sim\) is \(\sim_I\) for an ideal \(I\), the image is the closed set \(V(I)\). So \(\operatorname{Spec}(M/I)\cong V(I)\).

(c) In general the image is not closed.

Proof. (a) For a prime ideal \(\mathfrak Q\) of \(M/{\sim}\) the set \(\pi^{-1}(\mathfrak Q)\) is a union of classes, and \(\mathfrak Q=\pi(\pi^{-1}(\mathfrak Q))\), because \(\pi\) is surjective. So \(\pi^*\) is injective. Let \(\mathfrak p\) be a prime ideal of \(M\) that is a union of classes. Then \(\pi(\mathfrak p)\) is an ideal of \(M/{\sim}\), and \(\pi^{-1}(\pi(\mathfrak p))=\mathfrak p\). So \([1]\notin\pi(\mathfrak p)\), and \([a][b]\in\pi(\mathfrak p)\) implies \(ab\in\mathfrak p\), hence \(a\in\mathfrak p\) or \(b\in\mathfrak p\). So \(\pi(\mathfrak p)\) is a prime ideal with \(\pi^*(\pi(\mathfrak p))=\mathfrak p\). Finally \(\pi^*\) is continuous and maps \(D([a])\) onto the intersection of \(D(a)\) with the image. (b) A prime ideal \(\mathfrak p\) is a union of classes of \(\sim_I\) if and only if \(I\subseteq\mathfrak p\) or \(I\cap\mathfrak p=\emptyset\). The second case does not occur, because \(0\in I\cap\mathfrak p\). (c) See Example 6.6. \(\square\)

Example 6.6 (the diagonal). Let \(\Delta:\mathbb F_1[x,y]\to\mathbb F_1[t]\) be the morphism with \(x\mapsto t\) and \(y\mapsto t\). It is surjective. Its kernel congruence identifies \(x^ay^b\) and \(x^{a'}y^{b'}\) when \(a+b=a'+b'\), and the class of \(0\) is \(\{0\}\). The monoid \(\mathbb F_1[t]\) has two faces, \(\{1\}\) and the set of all powers of \(t\), so it has the two prime ideals \(\langle t\rangle\) and \(\{0\}\). Their preimages under \(\Delta\) are \(\mathfrak m=xM\cup yM\) and \(\{0\}\), with \(M=\mathbb F_1[x,y]\). So the image of \(\Delta^*\) in the four-point space of Example 5.5 is \(\{\{0\},\mathfrak m\}\). It is not closed, because the closure of the point \(\{0\}\) is the whole space. After base change to \(\mathbb Z\) (Section 8) the map \(\Delta\) becomes the surjection \(\mathbb Z[x,y]\to\mathbb Z[t]\) with kernel \((x-y)\), whose spectrum is the closed diagonal of the affine plane. Reference: [Cortiñas–Haesemeyer–Walker–Weibel 2015, Remark 2.7.1].

Remark 6.7 (how quotients differ from the ring case).

1. In a ring, congruences and ideals are the same thing: \(a\equiv b\) if and only if \(a-b\in I\). A monoid has no subtraction, and a congruence is not determined by the class of \(0\). The kernel congruence of \(\Delta\) and the kernel congruence of \(\mathbb F_1[x]\to\mathbb F_1\), \(x\mapsto1\), both have \(I_\sim=\{0\}\), and neither morphism is injective. A morphism is injective if and only if its kernel congruence is equality; \(\varphi^{-1}(0)=\{0\}\) is not enough. [Deitmar 2013, Introduction] puts it this way: for a ring all congruences come from ideals, and a monoid has far more congruences than ideals.
2. The quotients by ideals are the special congruences of Proposition 6.4(b). The spectrum of such a quotient is a closed subset of \(\operatorname{Spec}M\) (Proposition 6.5). This property does not single them out: the morphism \(\mathbb F_{1^2}\to\mathbb F_1\) that sends \(-1\) to \(1\) is a quotient by a congruence that does not come from an ideal, and both spectra are one point.
3. The sum of two ideals of a ring corresponds to the union of two ideals of a monoid (Lemma 2.2). The quotient \(M/(I\cup J)\) is also the quotient of \(M/I\) by the image of \(J\): both are obtained from \(M\) by collapsing \(I\cup J\) to \(0\).
4. If \(\mathfrak p\) is a prime ideal, then \(M/\mathfrak p\) is without zero divisors, because \(M\setminus\mathfrak p\) is closed under multiplication. But a monoid without zero divisors need not embed in a group with zero (Section 7), so \(M/\mathfrak p\) is a weaker analogue of an integral domain.
5. After base change to \(\mathbb Z\), a quotient by an ideal becomes the quotient of \(\mathbb Z[M]\) by an ideal spanned by elements of \(M\), and a quotient by a congruence becomes the quotient by an ideal spanned by differences \(a-b\) of elements of \(M\) (Proposition 8.4).

Exercise 3 lists all quotients of \(\mathbb F_1[x]\).

Remark 6.8 (prime congruences and dimension). The space \(\operatorname{Spec}M\) has few points, and two things are lost. It does not see closed conditions: the image of the diagonal in Example 6.6 is not closed. And it does not see dimension: a group with zero \(G_0\) has the single prime ideal \(\{0\}\), so its spectrum is a point. This holds also for \(G=\mathbb Z^n\), although the corresponding ring over \(\mathbb C\) (Section 8) is the ring of Laurent polynomials in \(n\) variables, of dimension \(n\). [Jarra 2023a, Introduction] makes this point for the \(n\)-torus over any group with zero.

A second space, built from congruences, repairs both defects. A congruence \(\sim\) on \(M\) is prime if \(M/{\sim}\) is cancellative (Definition 7.4), that is, if \(1\not\sim0\) and \(ab\sim ac\) implies \(a\sim0\) or \(b\sim c\). The congruence space \(\operatorname{Cong}M\) is the set of prime congruences on \(M\), with the topology generated by the sets \(U_{a,b}=\{\sim\ :\ a\not\sim b\}\) for \(a,b\in M\) [Lorscheid–Ray 2024, Definitions 2.3 and 2.9]. The same space, for a class of objects that contains monoids and rings, is the starting point of [Deitmar 2013, Section 2.1]. Three facts are elementary.

1. If \(\sim\) is prime, then \(I_\sim\) is a prime ideal: \(1\notin I_\sim\), and \(ab\sim0=a0\) with \(a\not\sim0\) gives \(b\sim0\). The map \(\operatorname{Cong}M\to\operatorname{Spec}M\), \(\sim\ \mapsto I_\sim\), is continuous, because the preimage of \(D(a)\) is \(U_{a,0}\). It is surjective, because the kernel congruence of \(\chi_{\mathfrak p}:M\to\mathbb F_1\) (Proposition 2.7) is prime and has \(I_\sim=\mathfrak p\).
2. Let \(\varphi:M\to N\) be a morphism and \(\approx\) a prime congruence on \(N\). The kernel congruence of \(M\to N/{\approx}\) is prime, because its quotient is isomorphic to a submonoid of \(N/{\approx}\) (Proposition 6.2), and a submonoid of a cancellative monoid is cancellative. This defines a map \(\operatorname{Cong}N\to\operatorname{Cong}M\). It is continuous, because the preimage of \(U_{a,b}\) is \(U_{\varphi(a),\varphi(b)}\). For the diagonal \(\Delta\) of Example 6.6 its image is closed. Indeed, the kernel congruence of \(\Delta\) is generated by the pair \((x,y)\): replace \(y\) by \(x\) in a monomial, one factor at a time. So by Proposition 6.2 the image consists of the prime congruences on \(\mathbb F_1[x,y]\) with \(x\sim y\), and this set is the complement of \(U_{x,y}\). Reference: [Lorscheid–Ray 2024, Example 2.15].
3. On a group with zero \(G_0\) the prime congruences correspond to the subgroups \(H\) of \(G\), by: \(a\sim_Hb\) if \(a=b=0\), or \(a,b\neq0\) and \(ab^{-1}\in H\). Indeed, if \(\sim\) is prime and \(g\sim0\) for some \(g\in G\), then \(1=g^{-1}g\sim0\). So no element of \(G\) is equivalent to \(0\), and on \(G\) the relation is the coset relation of the subgroup \(H=\{g:g\sim1\}\). Conversely \(G_0/{\sim_H}=(G/H)_0\) is cancellative. So \(\operatorname{Cong}((\mathbb Z^n)_0)\) is the set of subgroups of \(\mathbb Z^n\). It contains chains \(0\subset H_1\subset\dots\subset H_n\) with \(H_i\) of rank \(i\), while the spectrum is a point.

For an arbitrary monoid, the fibre of \(\operatorname{Cong}M\to\operatorname{Spec}M\) over \(\mathfrak p\) is the congruence space of the group with zero \(M_{\mathfrak p}/\mathfrak pM_{\mathfrak p}\) [Lorscheid–Ray 2024, Proposition 2.18]. A space with the expected dimension is obtained by restricting the congruences and enlarging the base. [Jarra 2023a, Definitions 3.1 and 5.2] calls a monoid a domain if it is isomorphic to a submonoid of \((K,\cdot)\) for a field \(K\), calls a congruence strong if its quotient is a domain, and calls a group with zero \(F\) algebraically closed if for every \(\alpha\in F^\times\) and \(n\ge1\) the equation \(x^n=\alpha\) has exactly \(n\) solutions in \(F\). An example is the group of all complex roots of unity, with zero. For such an \(F\) and a fan \(\Sigma\), [Jarra 2023a, Theorem A] states: the space of those strong congruences on the toric monoid scheme of \(\Sigma\) over \(F\) for which \(F\) maps injectively to the quotient is catenary, and its Krull dimension is that of the complex toric variety of \(\Sigma\). For the \(n\)-torus over \(F\) these congruences are the ones generated by relations \(X^{z_i}\sim\lambda_i\), \(i=1,\dots,c\), where \(z_1,\dots,z_c\) is part of a basis of \(\mathbb Z^n\) and \(\lambda_i\in F^\times\) [Jarra 2023a, Proposition 6.1]. The proof of Theorem A in that work (Theorems 5.13 and 5.15 of the third arXiv version) uses [Jarra 2023a, Lemma 5.12]: if \(G\) is a torsion-free abelian group of finite rank and \(H_1,H_2\) are subgroups with \(H_1\cap H_2=\{0\}\) such that \(G/H_1\) and \(G/H_2\) are torsion-free, then \(G/(H_1\oplus H_2)\) is torsion-free. This lemma is false: for \(G=\mathbb Z^2\), \(H_1=\mathbb Z(1,0)\) and \(H_2=\mathbb Z(1,2)\) both quotients \(G/H_i\) are infinite cyclic, but \(G/(H_1\oplus H_2)\cong\mathbb Z/2\mathbb Z\). So the proof of Theorem A in that version is incomplete; the example does not refute Theorem A. Theorem A and Proposition 6.1 are proved in Monoid schemes, Section 6.6 (Theorem 6.19 and Corollary 6.20), with a different argument; this lesson does not use them.

7. Finiteness and regularity conditions

Finitely generated monoids

Definition 7.1. A subset \(\Gamma\subseteq M\) generates \(M\) if every non-zero element of \(M\) is a product of finitely many elements of \(\Gamma\); the empty product is \(1\). The monoid \(M\) is finitely generated if a finite set generates it. By Proposition 1.4 this holds if and only if there is a surjective morphism \(\mathbb F_1[x_1,\dots,x_n]\to M\) for some \(n\).

Proposition 7.2. Let \(\Gamma\) generate \(M\).

(a) For every prime ideal \(\mathfrak p\) we have \(\mathfrak p=\langle\Gamma\cap\mathfrak p\rangle\), and \(M\setminus\mathfrak p\) is the set of all products of elements of \(\Gamma\setminus\mathfrak p\). The map \(\mathfrak p\mapsto\Gamma\cap\mathfrak p\) is injective. If \(\Gamma\) is finite, \(M\) has at most \(2^{|\Gamma|}\) prime ideals.

(b) Let \(\Gamma\) be finite, let \(\mathfrak p\) be a prime ideal, and let \(f\) be the product of the elements of \(\Gamma\setminus\mathfrak p\). Then \(D(f)=\Lambda(\mathfrak p)\), \(\mathfrak p=\mathfrak p_f\) and \(M_{\mathfrak p}\cong M_f\).

(c) If \(\Gamma\) is finite, every localization \(S^{-1}M\) is isomorphic, compatibly with the maps from \(M\), to \(M_f\) for some \(f\in M\), and it is finitely generated.

Proof. (a) Let \(a\in\mathfrak p\), \(a\neq0\), and write \(a=\gamma_1\cdots\gamma_k\) with \(\gamma_i\in\Gamma\). Here \(k\ge1\), because \(1\notin\mathfrak p\). Since \(\mathfrak p\) is prime, some \(\gamma_i\) lies in \(\mathfrak p\), and \(a\in\gamma_iM\). So \(\mathfrak p\subseteq\langle\Gamma\cap\mathfrak p\rangle\), and the other inclusion is clear. If \(a\notin\mathfrak p\), then \(a\neq0\) and \(a=\gamma_1\cdots\gamma_k\) with no \(\gamma_i\) in \(\mathfrak p\), because \(\mathfrak p\) is an ideal. Conversely a product of elements outside \(\mathfrak p\) is outside \(\mathfrak p\). The remaining statements follow.

(b) We have \(f\notin\mathfrak p\). If \(\mathfrak q\in D(f)\), then no element of \(\Gamma\setminus\mathfrak p\) lies in \(\mathfrak q\), since each of them divides \(f\). So \(\Gamma\cap\mathfrak q\subseteq\Gamma\cap\mathfrak p\), and \(\mathfrak q=\langle\Gamma\cap\mathfrak q\rangle\subseteq\mathfrak p\) by (a). Conversely \(\mathfrak q\subseteq\mathfrak p\) implies \(f\notin\mathfrak q\). So \(D(f)=\Lambda(\mathfrak p)\). Theorem 3.4 gives \(\mathfrak p=\mathfrak p_f\), and Theorem 4.3(c) gives \(M_f\cong M_{\mathfrak p}\).

(c) If \(0\in S\), then \(S^{-1}M=0=M_f\) for \(f=0\). Otherwise \(S^{-1}M\cong M_{\mathfrak p_S}\) by Theorem 4.3(a), and \(M_{\mathfrak p_S}\cong M_f\) by (b). The monoid \(M_f\) is generated by the images of the elements of \(\Gamma\) and by \(1/f\). \(\square\)

Reference: [Chu–Lorscheid–Santhanam 2012, Lemma 2.9 and Proposition 2.10]; [Cortiñas–Haesemeyer–Walker–Weibel 2015, Lemma 1.5].

Corollary 7.3 (finite spectra). Let \(M\) be finitely generated and \(X=\operatorname{Spec}M\). Then \(X\) is finite. A subset of \(X\) is open if and only if it is stable under generization. Every set \(\Lambda(\mathfrak p)\) is open. A non-empty open set is of the form \(D(f)\) if and only if it has exactly one maximal element.

Proof. \(X\) is finite by Proposition 7.2(a), and \(\Lambda(\mathfrak p)\) is open by Proposition 7.2(b). Open sets are stable under generization (Proposition 3.3). A subset \(U\) that is stable under generization is the union of the open sets \(\Lambda(\mathfrak p)\), \(\mathfrak p\in U\). In a finite partially ordered set every element lies below a maximal one; so a subset with exactly one maximal element has a largest element, and Theorem 3.4 applies. \(\square\)

So the spectrum of a finitely generated monoid is a finite partially ordered set, with the topology whose open sets are the subsets closed under passing to smaller elements. Example 3.6 shows that the last statement of the corollary needs finite generation.

Cancellative monoids and the group completion

Definition 7.4. A monoid \(M\) is cancellative, or integral, if \(M\neq0\) and

Both words are in use: [Chu–Lorscheid–Santhanam 2012, Section 2.1] says integral, and [Cortiñas–Haesemeyer–Walker–Weibel 2015, Section 1] says cancellative. For a commutative monoid \(A\) without zero, [Deitmar 2008, Introduction] calls \(A\) integral if \(ab=ac\) implies \(b=c\); this holds if and only if \(A_0\) is cancellative in the sense of Definition 7.4. A cancellative monoid is without zero divisors: \(ab=0=a0\) with \(a\neq0\) gives \(b=0\). The converse fails: in \(\{0,e,1\}\) we have \(e\cdot e=e\cdot1\).

Proposition 7.5 (group completion). Let \(M\) be a monoid without zero divisors, and let \(M^{\mathrm{gp}}=M_{\{0\}}\) be the localization at the prime ideal \(\{0\}\), that is, at \(S=M\setminus\{0\}\).

(a) \(M^{\mathrm{gp}}=G_0\), where \(G=(M^{\mathrm{gp}})^\times\) is the abelian group of the fractions \(a/s\) with \(a,s\neq0\).

(b) Let \(H\) be an abelian group and \(\varphi:M\to H_0\) a morphism with \(\varphi^{-1}(0)=\{0\}\). Then there is exactly one morphism \(\varphi':M^{\mathrm{gp}}\to H_0\) with \(\varphi'\circ\iota=\varphi\), and it restricts to a group homomorphism \(G\to H\).

(c) \(\iota:M\to M^{\mathrm{gp}}\) is injective if and only if \(M\) is cancellative. Hence a monoid is cancellative if and only if it is isomorphic to a submonoid of \(G_0\) for some abelian group \(G\).

Proof. (a) By the definition of the relation in Proposition 4.1, \(a/s=0/1\) if and only if \(ta=0\) for some \(t\neq0\), that is, \(a=0\). For \(a\neq0\) the fraction \(a/s\) has the inverse \(s/a\). (b) We have \(\varphi(S)\subseteq H=(H_0)^\times\), so Proposition 4.1(b) applies, and a morphism maps units to units. (c) By Proposition 4.1(c), \(\iota(a)=\iota(b)\) if and only if \(ta=tb\) for some \(t\neq0\). If \(M\) is a submonoid of \(G_0\), then \(M\neq0\), and \(ab=ac\) with \(a\neq0\) gives \(b=c\) after multiplication by \(a^{-1}\) in \(G_0\). \(\square\)

The monoid \(M^{\mathrm{gp}}\) is the group completion of \(M\). For a cancellative monoid we regard \(M\) as a submonoid of \(M^{\mathrm{gp}}\). All localizations \(S^{-1}M\) with \(0\notin S\) are then submonoids of \(M^{\mathrm{gp}}\) as well, by \(a/s\mapsto as^{-1}\). More generally, let \(M\) be a submonoid of a group with zero \(H_0\), and let \(0\notin S\). Then \(a/s\mapsto as^{-1}\) is an injective morphism \(S^{-1}M\to H_0\): it exists by Proposition 4.1(b), and \(as^{-1}=a's'^{-1}\) gives \(as'=a's\), hence \(a/s=a'/s'\). In particular \(M^{\mathrm{gp}}\) is the submonoid of \(H_0\) that consists of \(0\) and the fractions \(ab^{-1}\) with \(a,b\in M\setminus\{0\}\). Example 5.5 used this for \(M=\mathbb F_1[x,y]\).

Definition 7.6. A monoid \(M\) is torsion-free if \(a^n=b^n\) for some \(n\ge1\) implies \(a=b\).

Proposition 7.7. Let \(M\) be cancellative, with \(M^{\mathrm{gp}}=G_0\). Then \(M\) is torsion-free if and only if the group \(G\) is torsion-free.

Proof. Let \(G\) be torsion-free and \(a^n=b^n\) in \(M\). If \(a=0\), then \(b^n=0\), so \(b=0\). Otherwise \(b\neq0\), and \((a/b)^n=1\) in \(G\), so \(a/b=1\) and \(a=b\). Conversely let \(M\) be torsion-free and \(g=a/b\in G\) with \(g^n=1\). Then \(a^n=b^n\) in \(M\), so \(a=b\) and \(g=1\). \(\square\)

Reference: [Cortiñas–Haesemeyer–Walker–Weibel 2015, Section 1] defines torsion-free monoids and notes the first implication.

The monoid \(\mathbb F_{1^n}\) is cancellative, and it is torsion-free only for \(n=1\). The monoid \(\{0,e,1\}\) is torsion-free and not cancellative. A torsion-free monoid has no nilpotent element other than \(0\), since \(a^n=0=0^n\) gives \(a=0\).

Normal monoids

Definition 7.8. Let \(M\) be cancellative. The normalization of \(M\) is

\(M\) is normal if \(M_{\mathrm{nor}}=M\): every \(\alpha\in M^{\mathrm{gp}}\) with a power in \(M\) lies in \(M\).

Proposition 7.9. Let \(M\) be cancellative.

(a) \(M_{\mathrm{nor}}\) is a submonoid of \(M^{\mathrm{gp}}\) that contains \(M\).

(b) If \(N\) is a submonoid with \(M\subseteq N\subseteq M^{\mathrm{gp}}\), then \(N\) is cancellative and the inclusion induces an isomorphism \(N^{\mathrm{gp}}\cong M^{\mathrm{gp}}\).

(c) \(M_{\mathrm{nor}}\) is normal, and it is contained in every normal monoid \(N\) with \(M\subseteq N\subseteq M^{\mathrm{gp}}\).

Proof. (a) If \(\alpha^n\in M\) and \(\beta^m\in M\), then \((\alpha\beta)^{nm}=(\alpha^n)^m(\beta^m)^n\in M\). (b) \(N\) is cancellative by Proposition 7.5(c). By Proposition 7.5(b) the inclusion induces a morphism \(N^{\mathrm{gp}}\to M^{\mathrm{gp}}\), \(a/b\mapsto ab^{-1}\). It is injective, because \(ab^{-1}=a'b'^{-1}\) gives \(ab'=a'b\). It is surjective, because every non-zero element of \(M^{\mathrm{gp}}\) is a fraction of elements of \(M\subseteq N\). (c) By (b), \(N\) is normal if and only if every \(\alpha\in M^{\mathrm{gp}}\) with a power in \(N\) lies in \(N\). If \(\alpha^n\in M_{\mathrm{nor}}\), then \((\alpha^n)^m\in M\) for some \(m\), so \(\alpha\in M_{\mathrm{nor}}\). If \(N\) is normal and \(\alpha\in M_{\mathrm{nor}}\), then \(\alpha^n\in M\subseteq N\), so \(\alpha\in N\). \(\square\)

Passing to the normalization does not change the spectrum. This holds in a more general form, without any cancellation.

Theorem 7.10 (extensions by roots). Let \(M\) be a submonoid of a monoid \(N\) such that every \(b\in N\) has a power \(b^n\in M\) with \(n\ge1\). Then

is a homeomorphism. Its inverse sends \(\mathfrak p\) to \(\mathfrak p'=\{b\in N:b^n\in\mathfrak p\ \text{for some}\ n\ge1\}\). In particular \(\operatorname{Spec}M_{\mathrm{nor}}\to\operatorname{Spec}M\) is a homeomorphism for every cancellative monoid \(M\).

Proof. The map is \(j^*\) for the inclusion \(j:M\to N\), so it is continuous. Let \(\mathfrak p\) be a prime ideal of \(M\). We show that \(\mathfrak p'\) is a prime ideal of \(N\). It contains \(0\). Let \(b\in\mathfrak p'\) with \(b^n\in\mathfrak p\), and let \(c\in N\) with \(c^m\in M\). Then \((bc)^{nm}=(b^n)^m(c^m)^n\in\mathfrak p\), so \(bc\in\mathfrak p'\). Since \(1\notin\mathfrak p\), we have \(1\notin\mathfrak p'\). Let \(b,c\in N\) with \(bc\in\mathfrak p'\), say \((bc)^n\in\mathfrak p\). Choose \(k\ge1\) with \(b^k\in M\) and \(c^k\in M\). Then \((b^k)^n(c^k)^n=((bc)^n)^k\in\mathfrak p\), and both factors lie in \(M\). So \((b^k)^n\in\mathfrak p\) or \((c^k)^n\in\mathfrak p\), that is, \(b\in\mathfrak p'\) or \(c\in\mathfrak p'\). Next, \(\mathfrak p'\cap M=\mathfrak p\): for \(b\in M\), \(b^n\in\mathfrak p\) if and only if \(b\in\mathfrak p\). Finally let \(\mathfrak Q\) be a prime ideal of \(N\) and \(\mathfrak p=\mathfrak Q\cap M\). If \(b\in\mathfrak Q\), choose \(n\) with \(b^n\in M\); then \(b^n\in\mathfrak Q\cap M=\mathfrak p\), so \(b\in\mathfrak p'\). If \(b\in\mathfrak p'\), then \(b^n\in\mathfrak p\subseteq\mathfrak Q\), so \(b\in\mathfrak Q\). Hence \(\mathfrak Q=\mathfrak p'\), and the two maps are inverse bijections. The map is open: for \(b\in N\) with \(b^n\in M\) we have \(D(b)=D(b^n)\) in \(\operatorname{Spec}N\), and its image is the set \(D(b^n)\) of \(\operatorname{Spec}M\). \(\square\)

Reference: [Cortiñas–Haesemeyer–Walker–Weibel 2015, Remark 1.6.1] treats the case \(N=M_{\mathrm{nor}}\).

Example 7.11 (the cusp monoid). Let \(M=\{0\}\cup\{t^n:n=0\ \text{or}\ n\ge2\}\), a submonoid of \(\mathbb F_1[t]\). It is generated by \(t^2\) and \(t^3\). It is cancellative, with \(M^{\mathrm{gp}}=\mathbb F_1[t^{\pm1}]\), since \(t=t^3/t^2\); so it is torsion-free. It is not normal: \(t\) lies in \(M^{\mathrm{gp}}\) and \(t^2\in M\), but \(t\notin M\). A power \(t^{kn}\) with \(n\ge1\) lies in \(M\) only if \(k\ge0\), so \(M_{\mathrm{nor}}=\mathbb F_1[t]\). By Theorem 7.10, \(\operatorname{Spec}M\) has two points, like \(\operatorname{Spec}\mathbb F_1[t]\).

8. The monoid ring and the map from Spec Z[M] to Spec M

The monoid ring

Definition 8.1. Let \(k\) be a ring and \(M\) a monoid. The monoid algebra \(k[M]\) is the free \(k\)-module with basis \(M\setminus\{0\}\), with the \(k\)-bilinear multiplication that extends the product of \(M\); a product of basis elements that is \(0\) in \(M\) is \(0\) in \(k[M]\). It is a commutative \(k\)-algebra. The map \(M\to k[M]\) that sends \(a\neq0\) to the basis element \(a\) and \(0\) to \(0\) is a morphism of monoids \(M\to(k[M],\cdot)\). If \(k\neq0\) it is injective, and we regard \(M\) as a subset of \(k[M]\). For an ideal \(I\) of \(M\) we write \(k[I]\) for the \(k\)-span of \(I\setminus\{0\}\); it is an ideal of \(k[M]\). The ring \(\mathbb Z[M]\) is the base change of \(M\) to the integers.

In other words, \(k[M]\) is the usual monoid algebra of \(M\) modulo the ideal spanned by the zero of \(M\).

Examples 8.2. \(k[\mathbb F_1[P]]\) is the usual monoid algebra \(k[P]\) of the additive monoid \(P\). In particular \(k[\mathbb F_1[x_1,\dots,x_n]]=k[x_1,\dots,x_n]\), \(k[G_0]\) is the group algebra \(k[G]\), and \(\mathbb Z[\mathbb F_1]=\mathbb Z\). Further \(\mathbb Z[\mathbb F_{1^n}]=\mathbb Z[x]/(x^n-1)\) and \(\mathbb Z[\{0,1,\varepsilon\}]=\mathbb Z[\varepsilon]/(\varepsilon^2)\). The ring \(\mathbb Z[\{0,e,1\}]\) has the basis \(1,e\) with \(e^2=e\), so it is isomorphic to \(\mathbb Z\times\mathbb Z\) by \(1\mapsto(1,1)\), \(e\mapsto(0,1)\).

Proposition 8.3 (adjunction). Let \(R\) be a \(k\)-algebra. Restriction to \(M\) is a bijection

For \(k=\mathbb Z\): ring homomorphisms \(\mathbb Z[M]\to R\) are the same as monoid morphisms \(M\to(R,\cdot)\).

Proof. A homomorphism of \(k\)-algebras restricts to a morphism of monoids, and it is determined by its values on the basis. A morphism \(\psi:M\to(R,\cdot)\) extends \(k\)-linearly from the basis. The extension is multiplicative, because \(\psi(ab)=\psi(a)\psi(b)\) also when \(ab=0\). \(\square\)

Reference: [Deitmar 2005, Theorem 1.1], for monoids without zero.

Proposition 8.4 (localization and quotients).

(a) For a multiplicative subset \(S\subseteq M\) there is a canonical isomorphism of \(k\)-algebras \(S^{-1}k[M]\cong k[S^{-1}M]\).

(b) For a congruence \(\sim\) on \(M\) let \(J_\sim\subseteq k[M]\) be the \(k\)-span of the elements \(a-b\) with \(a\sim b\). Then \(J_\sim\) is an ideal and \(k[M/{\sim}]\cong k[M]/J_\sim\). For an ideal \(I\) of \(M\) this gives \(k[M/I]\cong k[M]/k[I]\).

Proof. (a) The homomorphism \(k[M]\to k[S^{-1}M]\) induced by \(\iota\) sends \(S\) to units, so it factors through a homomorphism \(u:S^{-1}k[M]\to k[S^{-1}M]\) [Stacks, Tag 00CP]. The morphism of monoids \(M\to(S^{-1}k[M],\cdot)\) sends \(S\) to units. By Proposition 4.1(b) it extends to \(S^{-1}M\), and by Proposition 8.3 to a homomorphism \(v:k[S^{-1}M]\to S^{-1}k[M]\). Both \(u\circ v\) and \(v\circ u\) fix the images of \(M\) and the inverses of the elements of \(S\). These generate the two algebras, so \(u\) and \(v\) are inverse isomorphisms.

(b) If \(a\sim b\) and \(c\in M\), then \(c(a-b)=ca-cb\) with \(ca\sim cb\); so \(J_\sim\) is an ideal. The surjective homomorphism \(k[M]\to k[M/{\sim}]\) induced by \(\pi\) vanishes on \(J_\sim\). Define a \(k\)-linear map \(\sigma:k[M/{\sim}]\to k[M]/J_\sim\) on the basis by \([a]\mapsto a+J_\sim\) for \([a]\neq[0]\). It is well defined, because \(a\sim a'\) gives \(a-a'\in J_\sim\). The composite \(k[M]/J_\sim\to k[M/{\sim}]\to k[M]/J_\sim\) is the identity: it fixes \(a+J_\sim\) if \(a\not\sim0\), and if \(a\sim0\) then \(a=a-0\in J_\sim\). So \(k[M]/J_\sim\to k[M/{\sim}]\) is injective, hence an isomorphism. For \(\sim\,=\,\sim_I\) the span of the differences \(a-b\) with \(a,b\in I\) is \(k[I]\). \(\square\)

Reference: [Chu–Lorscheid–Santhanam 2012, Lemma 2.7] and [Cortiñas–Haesemeyer–Walker–Weibel 2015, Lemma 5.5] for (a).

So ideals of \(M\) correspond to ideals of \(k[M]\) spanned by monomials, and congruences to ideals spanned by binomials and monomials. For the diagonal of Example 6.6, \(J_\sim\) is the ideal generated by \(x-y\).

Theorem 8.5 (properties of \(M\) and of \(k[M]\)). Let \(k\neq0\) be a ring and \(M\) a monoid.

(a) \(M\) is finitely generated if and only if \(k[M]\) is a finitely generated \(k\)-algebra.

(b) \(k[M]\) is an integral domain if and only if \(k\) is an integral domain and \(M\) is cancellative and torsion-free.

(c) If \(k[M]\) is a normal domain, then \(M\) is normal and torsion-free.

Proof. (a) If \(\Gamma\) generates \(M\), it generates \(k[M]\) as a \(k\)-algebra. Conversely let \(g_1,\dots,g_r\) generate \(k[M]\). Each \(g_j\) is a \(k\)-linear combination of finitely many elements of \(M\). Let \(\Gamma\) be the finite set of all elements of \(M\) that occur, and let \(M'\subseteq M\) be the submonoid generated by \(\Gamma\). The \(k\)-span of \(M'\setminus\{0\}\) is a subalgebra that contains all \(g_j\), so it is \(k[M]\). Since \(M\setminus\{0\}\) is a basis and \(k\neq0\), this forces \(M'=M\).

(b) Let \(k[M]\) be a domain. Then \(M\neq0\), and \(k\) is a domain, being a subring of \(k[M]\). If \(ab=ac\) in \(M\) with \(a\neq0\), then \(a(b-c)=0\) in \(k[M]\), so \(b=c\). Hence \(M\) is cancellative. Suppose that \(M\) is not torsion-free. Choose \(a\neq b\) in \(M\) such that \(a^n=b^n\) for some \(n\ge1\), and let \(n\) be the smallest such exponent for this pair. Then \(n\ge2\). Neither \(a\) nor \(b\) is \(0\): if \(b=0\) then \(a^n=0\) and \(a=0\), because \(M\) is without zero divisors. In \(k[M]\) we have

The \(c_i\) are non-zero elements of \(M\). They are pairwise different: if \(c_i=c_j\) with \(i<j\), cancelling \(a^{n-1-j}b^i\) gives \(a^{j-i}=b^{j-i}\) with \(1\le j-i<n\), against the choice of \(n\). So \(c_0+\dots+c_{n-1}\) is a sum of \(n\) different basis elements, and it is not zero. Also \(a-b\neq0\). This contradicts the assumption that \(k[M]\) is a domain.

Conversely let \(k\) be a domain and \(M\) cancellative and torsion-free. Then \(M\subseteq G_0=M^{\mathrm{gp}}\) with \(G\) torsion-free (Proposition 7.7), and \(k[M]\) is a subring of the group algebra \(k[G]\). Let \(x,y\in k[G]\) be non-zero. The finitely many elements of \(G\) that occur in \(x\) and \(y\) generate a subgroup \(G'\). It is finitely generated and torsion-free, hence \(G'\cong\mathbb Z^r\) [Stacks, Tag 0ASV]. So \(x\) and \(y\) lie in \(k[G']\cong k[t_1^{\pm1},\dots,t_r^{\pm1}]\), which is a localization of a polynomial ring over a domain. Hence \(xy\neq0\).

(c) Recall that a domain is normal if it is integrally closed in its fraction field [Stacks, Tag 0309]. By (b), \(M\) is cancellative and torsion-free. Let \(G_0=M^{\mathrm{gp}}\). By Proposition 8.4(a), \(k[G]=k[M^{\mathrm{gp}}]\) is a localization of \(k[M]\), so it lies in the fraction field \(K\) of \(k[M]\). Let \(\alpha\in G\) with \(\alpha^n\in M\). Then \(\alpha\in K\) is a root of the monic polynomial \(X^n-\alpha^n\) over \(k[M]\), so \(\alpha\in k[M]\). Now \(\alpha\) is an element of the basis \(G\) of \(k[G]\), and \(k[M]\) is the \(k\)-span of the basis elements in \(M\setminus\{0\}\). Since \(k\neq0\), this gives \(\alpha\in M\). \(\square\)

Reference: [Chu–Lorscheid–Santhanam 2012, Lemma 2.11] for (a). [Cortiñas–Haesemeyer–Walker–Weibel 2015, Lemma 5.10] proves that \(k[M]\) is a domain when \(k\) is a domain and \(M\) is cancellative with torsion-free group completion. A converse of (c) for finitely generated \(M\) is quoted in Section 12, for coefficient rings that contain a field and for \(k=\mathbb Z\).

Together with Proposition 8.4(b), part (b) decides when a prime ideal of \(M\) spans a prime ideal of \(k[M]\): for \(\mathfrak p\in\operatorname{Spec}M\), the ideal \(k[\mathfrak p]\) is prime if and only if \(k\) is a domain and \(M/\mathfrak p\) is cancellative and torsion-free. For \(M=\{0,e,1\}\) and \(\mathfrak p=\{0\}\) it is not: the zero ideal of \(\mathbb Z\times\mathbb Z\) is not prime.

Corollary 8.6. Let \(M\) be finitely generated. Then every ideal of \(M\) is generated by finitely many elements, and every ascending chain of ideals of \(M\) becomes stationary.

Proof. The ring \(\mathbb Z[M]\) is finitely generated, hence Noetherian [Stacks, Tag 00FN]. Let \(I\) be an ideal of \(M\). The ideal \(\mathbb Z[I]\) is generated by finitely many elements \(g_1,\dots,g_r\). Each \(g_j\) is a \(\mathbb Z\)-linear combination of finitely many elements of \(I\). Let \(T\subseteq I\) be the finite set of all elements that occur, and \(J=\langle T\rangle\subseteq I\). Then \(\mathbb Z[J]\) is an ideal of \(\mathbb Z[M]\) that contains all \(g_j\), so \(\mathbb Z[J]=\mathbb Z[I]\) and \(J=I\). The union of an ascending chain of ideals is an ideal (Lemma 2.2). It is finitely generated, and its generators lie in one member of the chain. \(\square\)

The converse fails: if \(G\) is an abelian group that is not finitely generated, the only ideals of \(G_0\) are \(\{0\}\) and \(G_0\), but \(G_0\) is not finitely generated.

Examples 8.7 (three warnings).

(a) Torsion. \(\operatorname{Spec}\mathbb F_{1^n}\) is one point, but \(\mathbb Z[\mathbb F_{1^n}]=\mathbb Z[x]/(x^n-1)\) is not a domain for \(n\ge2\).

(b) The cusp. For the monoid \(M\) of Example 7.11, \(\mathbb Z[M]=\mathbb Z[t^2,t^3]\subseteq\mathbb Z[t]\). It is a domain and it is not normal, in agreement with Theorem 8.5: \(t\) lies in the fraction field, \(t^2\) lies in \(\mathbb Z[M]\), and \(t\) does not.

(c) Nilpotents from non-cancellation. Let \(M=\{0,1,a,b,c_2,c_3,\dots\}\), with zero \(0\) and unit \(1\), where \(a\) and \(b\) have degree \(1\), \(c_n\) has degree \(n\), and the product of two elements of degrees \(i,j\ge1\) is \(c_{i+j}\). This is a monoid without zero divisors and without nilpotent elements other than \(0\). But \(a^2=ab=b^2=c_2\), so in \(\mathbb Z[M]\) we get \((a-b)^2=c_2-2c_2+c_2=0\). The ring \(\mathbb Z[M]\) is not reduced, and \(M\) is neither cancellative nor torsion-free.

Prime ideals of rings and of monoids

Proposition 8.8. Let \(R\) be a ring. Every prime ideal of the ring \(R\) is a prime ideal of the monoid \((R,\cdot)\), and the inclusion \(\operatorname{Spec}R\to\operatorname{Spec}(R,\cdot)\) is a homeomorphism onto its image.

Proof. A prime ideal \(P\) of the ring contains \(0\), is stable under multiplication by \(R\), is different from \(R\), and \(ab\in P\) implies \(a\in P\) or \(b\in P\). For \(f\in R\) the set \(D(f)\) of \(\operatorname{Spec}(R,\cdot)\) meets \(\operatorname{Spec}R\) in the basic open set \(D(f)\) of the Zariski topology of the ring [Stacks, Tag 00DY]. So the subspace topology and the Zariski topology have the same basis. \(\square\)

Reference: [Cortiñas–Haesemeyer–Walker–Weibel 2015, Observation 5.2.1].

Example 8.9 (the integers). The monoid \((\mathbb Z,\cdot)\) is generated by \(-1\) and the prime numbers. By Proposition 7.2(a) a prime ideal \(\mathfrak p\) of \((\mathbb Z,\cdot)\) is determined by the set \(\Sigma\) of prime numbers it contains, and

Conversely, for every set \(\Sigma\) of prime numbers, \(\mathfrak p_\Sigma\) is a prime ideal of \((\mathbb Z,\cdot)\): it is an ideal, it does not contain \(1\), and a prime number that divides \(ab\) divides \(a\) or \(b\). So \(\operatorname{Spec}(\mathbb Z,\cdot)\) is in bijection with the set of all sets of prime numbers, ordered by inclusion. The prime ideals of the ring \(\mathbb Z\) are the \(\mathfrak p_\Sigma\) with \(\Sigma\) empty or a single prime. The others, such as \(2\mathbb Z\cup3\mathbb Z\), are prime ideals of the monoid only; compare Lemma 2.10.

The map from Spec Z[M] to Spec M

Definition 8.10. For a monoid \(M\) let

It is the composite of the inclusion of Proposition 8.8 for \(R=\mathbb Z[M]\) and the map induced by the morphism \(M\to(\mathbb Z[M],\cdot)\). So it is continuous. We have \(\beta_M^{-1}(D(f))=D(f)\) and \(\beta_M^{-1}(V(I))=V(\mathbb Z[I])\) for \(f\in M\) and ideals \(I\) of \(M\). For a morphism \(\varphi:M\to N\), with induced ring homomorphism \(\mathbb Z[\varphi]\), the maps satisfy \(\beta_M\circ\operatorname{Spec}(\mathbb Z[\varphi])=\varphi^*\circ\beta_N\).

Reference: [Chu–Lorscheid–Santhanam 2012, Theorem 3.2].

Definition 8.11. For a prime ideal \(\mathfrak p\) of \(M\) put

By Theorem 4.3(c), \(\mathfrak pM_{\mathfrak p}\) is the set of non-units of \(M_{\mathfrak p}\), so \(\kappa(\mathfrak p)=(G_{\mathfrak p})_0\) is a group with zero. We call it the residue monoid of \(M\) at \(\mathfrak p\). It plays the role of the residue field of a ring at a prime ideal, and [Lorscheid–Ray 2024, Section 2.6] calls it the residue field. If \(M\) is without zero divisors, then \(\kappa(\{0\})=M^{\mathrm{gp}}\).

Lemma 8.12 (morphisms to a group with zero). Let \(H\) be an abelian group.

(a) Let \(\varphi:M\to H_0\) be a morphism. Then \(\mathfrak p=\varphi^{-1}(0)\) is a prime ideal, and there is exactly one group homomorphism \(\chi:G_{\mathfrak p}\to H\) with \(\varphi(a)=\chi(a/1)\) for all \(a\notin\mathfrak p\).

(b) Conversely, for a prime ideal \(\mathfrak p\) and a group homomorphism \(\chi:G_{\mathfrak p}\to H\), the map \(\varphi\) with \(\varphi(a)=\chi(a/1)\) for \(a\notin\mathfrak p\) and \(\varphi(a)=0\) for \(a\in\mathfrak p\) is a morphism \(M\to H_0\) with \(\varphi^{-1}(0)=\mathfrak p\).

So \(\operatorname{Hom}(M,H_0)\) is the disjoint union of the sets \(\operatorname{Hom}_{\mathrm{groups}}(G_{\mathfrak p},H)\), \(\mathfrak p\in\operatorname{Spec}M\).

Proof. (a) \(\{0\}\) is a prime ideal of \(H_0\), so \(\mathfrak p\) is prime by Lemma 2.6. We have \(\varphi(M\setminus\mathfrak p)\subseteq H=(H_0)^\times\). By Proposition 4.1(b), \(\varphi\) extends to a morphism \(\varphi':M_{\mathfrak p}\to H_0\), which maps units to units. Let \(\chi\) be its restriction to \(G_{\mathfrak p}\). For \(a\notin\mathfrak p\) we get \(\varphi(a)=\varphi'(a/1)=\chi(a/1)\). The homomorphism \(\chi\) is unique, because \(G_{\mathfrak p}\) consists of the elements \((a/1)(s/1)^{-1}\) with \(a,s\notin\mathfrak p\). (b) \(\varphi(1)=1\) and \(\varphi(0)=0\). If \(a\) or \(b\) lies in \(\mathfrak p\), then \(ab\in\mathfrak p\) and \(\varphi(ab)=0=\varphi(a)\varphi(b)\). Otherwise \(ab\notin\mathfrak p\), and \(\varphi(ab)=\chi(ab/1)=\varphi(a)\varphi(b)\). The two constructions are inverse to each other. \(\square\)

For \(H=\{1\}\) this is Proposition 2.7. For \(H=\mu_n\) it describes the morphisms \(M\to\mathbb F_{1^n}\). For \(H=k^\times\), \(k\) a field, it describes the morphisms \(M\to(k,\cdot)\), that is, by Proposition 8.3, the ring homomorphisms \(\mathbb Z[M]\to k\).

Theorem 8.13 (the fibres of \(\beta\)). Let \(M\) be a monoid.

(a) For \(\mathfrak p\in\operatorname{Spec}M\), the ring homomorphism \(\mathbb Z[M]\to\mathbb Z[\kappa(\mathfrak p)]=\mathbb Z[G_{\mathfrak p}]\) induces a homeomorphism of \(\operatorname{Spec}\mathbb Z[G_{\mathfrak p}]\) onto the fibre \(\beta_M^{-1}(\mathfrak p)\), with its subspace topology.

(b) \(\beta_M\) is surjective.

(c) For every field \(k\), the set of ring homomorphisms \(\mathbb Z[M]\to k\) is the disjoint union of the sets \(\operatorname{Hom}_{\mathrm{groups}}(G_{\mathfrak p},k^\times)\), \(\mathfrak p\in\operatorname{Spec}M\). The homomorphisms that belong to \(\mathfrak p\) are those whose kernel \(P\) satisfies \(\beta_M(P)=\mathfrak p\).

Proof. (a) Let \(S=M\setminus\mathfrak p\) and \(R=\mathbb Z[M]\). By Proposition 8.4, \(\mathbb Z[M_{\mathfrak p}]=S^{-1}R\) and \(\mathbb Z[\kappa(\mathfrak p)]=S^{-1}R/\mathbb Z[\mathfrak pM_{\mathfrak p}]\). The ideal \(\mathbb Z[\mathfrak pM_{\mathfrak p}]\) is spanned by the fractions \(a/s=(a/1)(1/s)\) with \(a\in\mathfrak p\), so it is the ideal of \(S^{-1}R\) generated by the image of \(\mathfrak p\). By the description of the spectrum of a localization and of a quotient of a ring [Stacks, Tags 00E3 and 00E5], the map \(\operatorname{Spec}\mathbb Z[\kappa(\mathfrak p)]\to\operatorname{Spec}R\) is a homeomorphism onto the set of prime ideals \(P\) of \(R\) with \(P\cap S=\emptyset\) and \(P\supseteq\mathfrak p\). These two conditions say \(P\cap M=\mathfrak p\). (b) The ring \(\mathbb Z[G_{\mathfrak p}]\) has the homomorphism to \(\mathbb Z\) that sends every element of \(G_{\mathfrak p}\) to \(1\). Its kernel is a prime ideal, so the fibre is not empty. (c) Combine Proposition 8.3 and Lemma 8.12 for \(H=k^\times\). If \(\Phi:\mathbb Z[M]\to k\) extends \(\varphi:M\to(k,\cdot)\), then \(\ker\Phi\cap M=\varphi^{-1}(0)\). \(\square\)

So \(\operatorname{Spec}\mathbb Z[M]\) is, as a set, the disjoint union of the spectra of the group rings \(\mathbb Z[G_{\mathfrak p}]\), indexed by the points of \(\operatorname{Spec}M\). When \(G_{\mathfrak p}\) is free of rank \(r\), the ring \(\mathbb Z[G_{\mathfrak p}]\) is a ring of Laurent polynomials in \(r\) variables, and the fibre is a torus of dimension \(r\) over \(\mathbb Z\). The space \(\operatorname{Spec}M\) records how these tori fit together. Note that \(\operatorname{Spec}M\) is connected while \(\operatorname{Spec}\mathbb Z[M]\) need not be: for \(M=\{0,e,1\}\) the spectrum has two points, \(\{0\}\subset\{0,e\}\), and \(\operatorname{Spec}\mathbb Z[M]=\operatorname{Spec}(\mathbb Z\times\mathbb Z)\) is the disjoint union of the two fibres.

Corollary 8.14 (counting points). Let \(M\) be finitely generated.

(a) For every prime ideal \(\mathfrak p\), the group \(G_{\mathfrak p}\) is finitely generated. So \(G_{\mathfrak p}\cong\mathbb Z^{r(\mathfrak p)}\times T_{\mathfrak p}\) with \(T_{\mathfrak p}\) finite.

(b) For every finite abelian group \(H\),

(c) Suppose that all groups \(G_{\mathfrak p}\) are torsion-free, and put \(N(q)=\sum_{\mathfrak p}(q-1)^{r(\mathfrak p)}\). Then \(\mathbb Z[M]\) has \(N(q)\) homomorphisms to the finite field \(\mathbb F_q\), for every prime power \(q\), and \(M\) has \(N(n+1)\) morphisms to \(\mathbb F_{1^n}\), for every \(n\ge1\).

Proof. (a) \(M_{\mathfrak p}\) is generated by a finite set \(\Gamma'\) (Proposition 7.2(c)). If a product of elements of \(\Gamma'\) is a unit, every factor is a unit. So \(G_{\mathfrak p}\) is generated by the units in \(\Gamma'\), and the structure theorem for finitely generated abelian groups applies [Stacks, Tag 0ASV]. (b) \(\operatorname{Hom}(\mathbb Z^r\times T,H)=H^r\times\operatorname{Hom}(T,H)\); now use Lemma 8.12 and the finiteness of \(\operatorname{Spec}M\). (c) Take \(H=\mathbb F_q^\times\), of order \(q-1\), and use Proposition 8.3; then take \(H=\mu_n\), of order \(n\). \(\square\)

Reference: [Deitmar 2005, Section 6] observes the relation between the two counts of (c) in examples. [Connes–Consani 2010, Proposition 3.18 and Theorem 4.10] prove the decomposition of Lemma 8.12 and part (c) for schemes built from monoids. [Deitmar 2006, Theorem 1] treats unit groups with torsion: the count over \(\mathbb F_q\) is a fixed polynomial in \(q\) for all \(q\) such that \(q-1\) is prime to a number \(e\). By (b) one can take the polynomial \(\sum_{\mathfrak p}(q-1)^{r(\mathfrak p)}\) and for \(e\) the product of the orders of the groups \(T_{\mathfrak p}\).

Part (c) gives a count over \(\mathbb F_q\) that is a polynomial in \(q\) and has a meaning "over \(\mathbb F_{1^n}\)" at \(q=n+1\); compare Counting over finite fields and the limit q → 1. The lesson Monoid schemes extends the count to spaces glued from spectra.

9. Examples: free monoids and the monoids of cones

Free monoids

Example 9.1 (the free monoid on \(n\) generators). Let \(M=\mathbb F_1[x_1,\dots,x_n]\). For a subset \(J\subseteq\{1,\dots,n\}\) let \(\mathfrak p_J=\langle x_j:j\in J\rangle\). It consists of \(0\) and the monomials divisible by some \(x_j\) with \(j\in J\). Its complement is the set of monomials in the \(x_i\) with \(i\notin J\), which is a face. By Proposition 7.2(a) these \(2^n\) ideals are all the prime ideals of \(M\), and \(\mathfrak p_J\subseteq\mathfrak p_K\) if and only if \(J\subseteq K\). So \(\operatorname{Spec}M\) is the set of subsets of \(\{1,\dots,n\}\), and by Corollary 7.3 a family of subsets is open when it contains, with a subset, all smaller subsets. For every \(J\),

The ring \(\mathbb Z[M]\) is \(\mathbb Z[x_1,\dots,x_n]\), and \(\beta_M\) sends a prime ideal \(P\) of this ring to \(\mathfrak p_J\) with \(J=\{j:x_j\in P\}\). The fibre over \(\mathfrak p_J\) is the set of prime ideals that contain \(x_j\) for \(j\in J\) and no \(x_i\) with \(i\notin J\). By Theorem 8.13 it is the spectrum of the Laurent polynomial ring in the variables \(x_i\), \(i\notin J\). So affine \(n\)-space over \(\mathbb Z\) is cut into \(2^n\) tori, one for each point of \(\operatorname{Spec}M\). The count of Corollary 8.14 is

and \(M\) has \(N(m+1)=(m+1)^n\) morphisms to \(\mathbb F_{1^m}\), as Proposition 1.4 shows directly.

Example 9.2 (infinitely many generators). Let \(I\) be an infinite set and \(M\) the free monoid on generators \(x_i\), \(i\in I\): its non-zero elements are the monomials, each involving finitely many \(x_i\). As in Example 9.1 the prime ideals are the \(\mathfrak p_J=\langle x_j:j\in J\rangle\) for arbitrary subsets \(J\subseteq I\). We claim:

If \(I\setminus J\) is finite, let \(f\) be the product of the \(x_i\) with \(i\notin J\). Then \(D(f)\) consists of the \(\mathfrak p_K\) with \(K\subseteq J\), so \(D(f)=\Lambda(\mathfrak p_J)\). Conversely let \(\Lambda(\mathfrak p_J)\) be open. By Theorem 3.4 it equals \(D(f)\) for some \(f\), which is a monomial in finitely many variables. If \(I\setminus J\) were infinite, there would be \(i\in I\setminus J\) such that \(x_i\) does not occur in \(f\). Then \(\mathfrak p_{\{i\}}\in D(f)\), but \(\mathfrak p_{\{i\}}\not\subseteq\mathfrak p_J\). This proves the claim. Reference: [Cortiñas–Haesemeyer–Walker–Weibel 2015, Example 1.4] treats finite \(J\).

When \(I\setminus J\) is infinite, the stalk at \(\mathfrak p_J\) is not reached on any open set: for every open neighbourhood \(U\) of \(\mathfrak p_J\) the map \(\mathcal O(U)\to M_{\mathfrak p_J}\) is not surjective. Indeed, choose \(f\) with \(\mathfrak p_J\in D(f)\subseteq U\) and \(i\in I\setminus J\) such that \(x_i\) does not occur in \(f\). A section over \(U\) with germ \(1/x_i\) at \(\mathfrak p_J\) would restrict to an element of \(\mathcal O(D(f))=M_f\) with image \(1/x_i\) in \(M_{\mathfrak p_J}\). But \(M_f\to M_{\mathfrak p_J}\) is injective, since \(M\) is cancellative, and \(1/x_i\notin M_f\). For finitely generated monoids this cannot happen, by Proposition 7.2(b).

The monoids of cones

Definition 9.3. A lattice is a free abelian group \(L\) of finite rank \(n\). We write \(L_{\mathbb R}=L\otimes\mathbb R\cong\mathbb R^n\) and regard \(L\) as a subgroup of \(L_{\mathbb R}\). A rational polyhedral cone in \(L_{\mathbb R}\) is a set of the form

with \(v_1,\dots,v_r\in L\). Then \(C\cap L\) is a submonoid of the additive group \(L\), and \(\mathbb F_1[C\cap L]\) is the monoid of the cone \(C\). Its elements are \(0\) and the \(\chi^m\) with \(m\in C\cap L\).

Lemma 9.4 (Gordan's lemma). Let \(C=\operatorname{cone}(v_1,\dots,v_r)\) with \(v_i\in L\), and let \(K=\{\sum t_iv_i:0\le t_i\le1\}\). Then \(K\cap L\) is finite and generates the additive monoid \(C\cap L\). So \(\mathbb F_1[C\cap L]\) is finitely generated.

Proof. \(K\) is bounded and \(L\cong\mathbb Z^n\), so \(K\cap L\) is finite. Let \(m=\sum t_iv_i\in C\cap L\) with \(t_i\ge0\). Then \(m=\sum\lfloor t_i\rfloor v_i+m'\) with \(m'=\sum(t_i-\lfloor t_i\rfloor)v_i\in K\), and \(m'=m-\sum\lfloor t_i\rfloor v_i\in L\). The \(v_i\) lie in \(K\cap L\) too. \(\square\)

Lemma 9.5 (rational coefficients). Let \(w_1,\dots,w_s\in L\) and \(m\in L\). If \(m=\sum_jt_jw_j\) with real numbers \(t_j>0\), then there are integers \(N\ge1\) and \(n_j\ge1\) with \(Nm=\sum_jn_jw_j\).

Proof. Choose a basis of \(L\). The set \(W=\{t\in\mathbb R^s:\sum t_jw_j=m\}\) is the solution set of a system of linear equations with integer coefficients. Gaussian elimination over \(\mathbb Q\) shows that \(W=t^0+\mathbb Rz_1+\dots+\mathbb Rz_d\) with \(t^0,z_1,\dots,z_d\in\mathbb Q^s\). So the points of \(W\) with rational coordinates are dense in \(W\). The points of \(W\) with all coordinates positive form a non-empty open subset of \(W\), so one of them is rational. Multiply by a common denominator. \(\square\)

Proposition 9.6. The monoid of a rational polyhedral cone is finitely generated, cancellative, torsion-free and normal.

Proof. Let \(M=\mathbb F_1[C\cap L]\). It is finitely generated by Lemma 9.4. It is a submonoid of the group with zero \(\mathbb F_1[L]\), so it is cancellative, and \(M^{\mathrm{gp}}=\mathbb F_1[G]\) for the subgroup \(G\) of \(L\) generated by \(C\cap L\) (Proposition 7.5). \(G\) is torsion-free, so \(M\) is torsion-free (Proposition 7.7). Let \(\alpha\in G\) with \(n\alpha\in C\cap L\) for some \(n\ge1\). Then \(\alpha=\frac1n(n\alpha)\in C\) and \(\alpha\in L\). So \(M\) is normal. \(\square\)

Theorem 9.7 (the monoids of cones). For a monoid \(M\) the following are equivalent.

(i) \(M\cong\mathbb F_1[C\cap L]\) for a lattice \(L\) and a rational polyhedral cone \(C\subseteq L_{\mathbb R}\).

(ii) \(M\) is finitely generated, cancellative, torsion-free and normal.

Proof. (i)⇒(ii) is Proposition 9.6. Let \(M\) satisfy (ii), let \(a_1,\dots,a_r\) be non-zero generators and \(M^{\mathrm{gp}}=G_0\). The group \(G\) is generated by the \(a_i\) and torsion-free (Proposition 7.7), hence free of finite rank [Stacks, Tag 0ASV]. Put \(L=G\), written additively, and \(C=\operatorname{cone}(a_1,\dots,a_r)\). Then \(M\setminus\{0\}\subseteq C\cap L\). Let \(b\in C\cap L\), \(b\neq0\). Write \(b=\sum t_ia_i\) with \(t_i\ge0\), and drop the terms with \(t_i=0\). By Lemma 9.5 some multiple \(Nb\), \(N\ge1\), is a sum of the \(a_i\) with positive integer coefficients, so \(Nb\in M\). Since \(M\) is normal, \(b\in M\). Hence \(M=\mathbb F_1[C\cap L]\). \(\square\)

Reference: the argument for (ii)⇒(i) is part of the proof of [Cortiñas–Haesemeyer–Walker–Weibel 2015, Theorem 4.4].

The cusp monoid of Example 7.11 satisfies all conditions of (ii) except normality. It is not the monoid of a cone.

Definition 9.8. A face of a rational polyhedral cone \(C\) is a subset \(F\subseteq C\) that contains \(0\), is closed under addition and under multiplication by non-negative real numbers, and satisfies

For example \(C\) is a face. If \(u:L_{\mathbb R}\to\mathbb R\) is a linear form with \(u\ge0\) on \(C\), then \(C\cap\ker u\) is a face: \(u(x)+u(y)=0\) and \(u(x),u(y)\ge0\) force \(u(x)=u(y)=0\). The dimension of a face \(F\) is the dimension of the linear subspace \(F-F=\{x-y:x,y\in F\}\) that it spans.

Lemma 9.9. Let \(C=\operatorname{cone}(v_1,\dots,v_r)\) and let \(F\) be a face of \(C\). Then \(F=\operatorname{cone}(v_i:v_i\in F)\). In particular \(C\) has at most \(2^r\) faces, and every face is a rational polyhedral cone.

Proof. Let \(x\in F\) and write \(x=\sum t_iv_i\) with \(t_i\ge0\). If \(t_i>0\), then \(x\) is the sum of \(t_iv_i\in C\) and \(\sum_{k\neq i}t_kv_k\in C\), so \(t_iv_i\in F\) and \(v_i\in F\). Hence \(x\in\operatorname{cone}(v_i:v_i\in F)\). The other inclusion is clear. \(\square\)

Theorem 9.10 (prime ideals and faces). Let \(C=\operatorname{cone}(v_1,\dots,v_r)\) be a rational polyhedral cone in \(L_{\mathbb R}\), \(P=C\cap L\) and \(M=\mathbb F_1[P]\).

(a) The map \(F\mapsto\mathfrak p_F:=\{\chi^m:m\in P\setminus F\}\cup\{0\}\) is an inclusion-reversing bijection from the set of faces of \(C\) to \(\operatorname{Spec}M\).

(b) \(M/\mathfrak p_F\cong\mathbb F_1[F\cap L]\).

(c) \(M_{\mathfrak p_F}\cong\mathbb F_1[(C-F)\cap L]\), where \(C-F=\{x-y:x\in C,\ y\in F\}\) is a rational polyhedral cone.

(d) \(G_{\mathfrak p_F}\cong(F-F)\cap L\), and this group is free abelian of rank \(\dim F\).

Proof. Call a subset \(Q\subseteq P\) a face of \(P\) if \(0\in Q\) and, for \(m,m'\in P\), \(m+m'\in Q\) holds exactly when \(m,m'\in Q\). The faces of the monoid \(M\) (Definition 2.3) are the sets \(\{\chi^m:m\in Q\}\) for faces \(Q\) of \(P\). So by Lemma 2.4, (a) says that \(F\mapsto F\cap L\) is a bijection from the faces of \(C\) to the faces of \(P\).

If \(F\) is a face of \(C\), then \(F\cap L\) is a face of \(P\). By Lemma 9.9, \(F\) is the cone generated by the \(v_i\in F\), which lie in \(F\cap L\). So \(F=\operatorname{cone}(F\cap L)\), and the map is injective.

Let \(Q\) be a face of \(P\), and let \(F=\operatorname{cone}(Q)\) be the set of all finite sums \(\sum t_jq_j\) with \(q_j\in Q\) and \(t_j\ge0\). We show that \(F\cap L=Q\) and that \(F\) is a face of \(C\).

\(F\cap L=Q\). Let \(m\in F\cap L\), \(m\neq0\), and write \(m=\sum t_jq_j\) with \(q_j\in Q\) and \(t_j>0\). By Lemma 9.5, \(Nm=\sum n_jq_j\in Q\) for some integers \(N,n_j\ge1\). Now \(m\in P\) and \((N-1)m\in P\), and their sum \(Nm\) lies in \(Q\). Since \(Q\) is a face of \(P\), \(m\in Q\).

\(F\) is a face. \(F\) contains \(0\) and is closed under addition and under multiplication by non-negative numbers. Let \(x,y\in C\) with \(x+y\in F\). Write \(x=\sum a_iv_i\), \(y=\sum b_iv_i\) with \(a_i,b_i\ge0\), and \(x+y=\sum_jt_jq_j\) with \(q_j\in Q\), \(t_j>0\). Let \(I=\{i:a_i+b_i>0\}\). Then

with positive coefficients. By Lemma 9.5, applied to \(m=0\) and the vectors \(v_i\) (\(i\in I\)) and \(-q_j\), there are positive integers \(n_i,k_j\) with \(\sum_{i\in I}n_iv_i=\sum_jk_jq_j\in Q\). For each \(i_0\in I\) this element of \(Q\) is the sum of \(v_{i_0}\in P\) and another element of \(P\). Since \(Q\) is a face of \(P\), \(v_{i_0}\in Q\). Hence \(x=\sum_{i\in I}a_iv_i\in F\), and in the same way \(y\in F\).

This proves (a); the bijection reverses inclusions by construction.

(b) \(M/\mathfrak p_F\) consists of \(0\) and the \(\chi^m\) with \(m\in F\cap L\), and the product of two such elements is never in \(\mathfrak p_F\).

(c) Let \(w_1,\dots,w_s\in F\cap L\) generate the cone \(F\) (Lemma 9.9). Then \(C-F=\operatorname{cone}(v_1,\dots,v_r,-w_1,\dots,-w_s)\). Since \(M\) is cancellative, \(M_{\mathfrak p_F}\) is the submonoid of \(\mathbb F_1[L]\) of all \(\chi^{m-g}\) with \(m\in P\), \(g\in F\cap L\), together with \(0\). These exponents lie in \((C-F)\cap L\). Conversely let \(z\in(C-F)\cap L\), \(z=x-y\) with \(x\in C\) and \(y=\sum t_jw_j\in F\), \(t_j\ge0\). Put \(g=\sum\lceil t_j\rceil w_j\in F\cap L\). Then \(g-y\in F\subseteq C\), so \(z+g=x+(g-y)\in C\cap L=P\), and \(z=(z+g)-g\).

(d) By (c), \(G_{\mathfrak p_F}\) is the group of lattice points of \((C-F)\cap-(C-F)\). We show that this set is \(F-F\). Clearly \(F-F\subseteq C-F\), and \(F-F\) is stable under \(z\mapsto-z\). Conversely let \(z=x-y\) and \(-z=x'-y'\) with \(x,x'\in C\) and \(y,y'\in F\). Then \(x+x'=y+y'\in F\), so \(x,x'\in F\) and \(z\in F-F\). The group \(\Lambda=(F-F)\cap L\) is finitely generated (Corollary 8.14(a)) and torsion-free, hence free [Stacks, Tag 0ASV]. It contains the \(w_j\), which span \(F-F\), so its rank is at least \(\dim F\). Elements of \(L\cong\mathbb Z^n\) that are linearly independent over \(\mathbb Z\) are linearly independent over \(\mathbb R\), because the rank of an integer matrix is the same over \(\mathbb Q\) and over \(\mathbb R\). So the rank of \(\Lambda\) is at most \(\dim F\). \(\square\)

Reference: [Jarra 2023a, Lemma 5.9] states the bijection of (a) for faces cut out by linear forms.

The smallest face of \(C\) is \(C\cap(-C)\). It is a face: it is a linear subspace contained in \(C\), and if \(x,y\in C\) and \(x+y\in C\cap(-C)\), then \(-x=y-(x+y)\in C\), so \(x\in C\cap(-C)\), and likewise \(y\). Every face \(F\) contains it: for \(z\in C\cap(-C)\) both \(z\) and \(-z\) lie in \(C\), and their sum \(0\) lies in \(F\), so \(z\in F\). This face corresponds to the maximal ideal of \(M\), and the face \(C\) corresponds to the ideal \(\{0\}\).

Corollary 9.11. Let \(M=\mathbb F_1[C\cap L]\) be the monoid of a rational polyhedral cone. Then for every prime power \(q\) and every \(m\ge1\),

where \(F\) runs over the faces of \(C\).

Proof. Theorem 9.10(a),(d) and Corollary 8.14(c). \(\square\)

Example 9.12 (the quadric cone). Let \(L=\mathbb Z^2\) and \(C=\operatorname{cone}((1,0),(1,2))=\{(a,b):0\le b\le2a\}\). The lattice points of the set \(K\) of Lemma 9.4 are \((0,0)\), \((1,0)\), \((1,1)\), \((1,2)\) and \((2,2)=(1,0)+(1,2)\). So \(M=\mathbb F_1[C\cap L]\) is generated by

By Lemma 9.9 every face of \(C\) is generated by a subset of \(\{(1,0),(1,2)\}\). All four subsets give faces: \(\{0\}\), the rays \(\rho_1=\mathbb R_{\ge0}(1,0)\) and \(\rho_2=\mathbb R_{\ge0}(1,2)\), which are cut out by the linear forms \(b\) and \(2a-b\), and \(C\). So \(M\) has four prime ideals,

As an ordered set, \(\operatorname{Spec}M\) is the same as \(\operatorname{Spec}\mathbb F_1[x_1,x_2]\). The monoids differ: \(M\) needs three generators, because none of \(x,y,z\) is a product of two non-units. The localization at \(\mathfrak p_{\rho_1}\) is \(M_x\), the monoid of the half-plane \(C-\rho_1=\{(a,b):b\ge0\}\), and \(G_{\mathfrak p_{\rho_1}}=\mathbb Z(1,0)\) has rank \(1=\dim\rho_1\).

The ring \(\mathbb Z[M]\) is isomorphic to \(\mathbb Z[X,Y,Z]/(XZ-Y^2)\). Indeed, this quotient is spanned by the monomials \(X^iZ^k\) and \(X^iYZ^k\). The homomorphism \(X\mapsto x\), \(Y\mapsto y\), \(Z\mapsto z\) sends them to \(\chi^{(i+k,2k)}\) and \(\chi^{(i+k+1,2k+1)}\). These are pairwise different, and they are all the basis elements of \(\mathbb Z[M]\). So the homomorphism is bijective. Corollary 9.11 gives

This agrees with a direct count of the solutions of \(XZ=Y^2\) in \(\mathbb F_q^3\): there are \(2q-1\) solutions with \(Y=0\) and \((q-1)^2\) solutions with \(Y\neq0\).

10. Monoids without zero and the convention of [Deitmar 2005]

In [Deitmar 2005] a monoid is a commutative monoid without zero: a set \(A\) with an associative and commutative multiplication and a unit. The symbol \(\mathbb F_1\) denotes the trivial monoid \(\{1\}\) there. An ideal is a subset \(\mathfrak a\subseteq A\) with \(\mathfrak aA\subseteq\mathfrak a\); the empty set is an ideal. A prime ideal is an ideal \(\mathfrak p\neq A\) whose complement is closed under multiplication; the empty set is a prime ideal. The closed sets of the spectrum are the sets \(V(\mathfrak a)\). Localization is defined as in Proposition 4.1, the structure sheaf is defined by local fractions as in Proposition 5.3, and the base change to \(\mathbb Z\) is the monoid ring \(\mathbb Z[A]\). The next proposition translates all of this into the language of this lesson.

Proposition 10.1. Let \(A\) be a commutative monoid without zero and \(M=A_0\).

(a) \(\mathfrak a\mapsto\mathfrak a\cup\{0\}\) is an inclusion-preserving bijection from the ideals of \(A\) in the sense of [Deitmar 2005] to the ideals of \(M\). Prime ideals correspond to prime ideals, and the empty set corresponds to \(\{0\}\).

(b) This bijection is a homeomorphism from the spectrum of \(A\) in the sense of [Deitmar 2005] to \(\operatorname{Spec}M\).

(c) For a multiplicative subset \(S\subseteq A\) we have \(S^{-1}M=(S^{-1}A)_0\).

(d) Let \(U\subseteq\operatorname{Spec}M\) be open and non-empty. A section of \(\mathcal O_M\) over \(U\) is either the family with all components \(0\), or a family of non-zero germs \(s_{\mathfrak p}\in A_{\mathfrak p}\). The families of the second kind are exactly the sections over \(U\) of the structure sheaf of [Deitmar 2005].

(e) \(\mathbb Z[M]\) is the monoid ring \(\mathbb Z[A]\).

(f) Let \(B\) be another commutative monoid without zero. The morphisms \(A_0\to B_0\) that are induced by morphisms \(A\to B\) are exactly the morphisms \(\varphi\) with \(\varphi^{-1}(0)=\{0\}\).

Proof. (a) If \(\mathfrak aA\subseteq\mathfrak a\), then \(\mathfrak a\cup\{0\}\) is an ideal of \(M\). Conversely, if \(I\) is an ideal of \(M\), then \((I\setminus\{0\})A\subseteq I\setminus\{0\}\), because a product of non-zero elements of \(M\) is not zero. The conditions for a prime ideal agree on non-zero elements, and a product \(ab\) in \(M\) is \(0\) only if \(a=0\) or \(b=0\). (b) The closed sets correspond. (c) The relation of Proposition 4.1 on \(M\times S\) restricts to the same relation on \(A\times S\), and \((0,s)\sim(a,s')\) holds only for \(a=0\). (d) The point \(\{0\}\) is prime and lies in \(U\) (Proposition 3.3(b)). The map \(\rho_{\mathfrak p,\{0\}}:M_{\mathfrak p}\to M_{\{0\}}=M^{\mathrm{gp}}\) sends only \(0\) to \(0\). So for \(s\in\mathcal O_M(U)\): if one component of \(s\) is \(0\), then \(s_{\{0\}}=0\), and then all components are \(0\). If all components are non-zero, Proposition 5.3 describes \(s\) by local fractions \(a/u\), in which necessarily \(a,u\in A\). This is the condition of [Deitmar 2005, Section 2.1]. (e) Both rings have the basis \(A\). (f) A morphism \(A\to B\) extends by \(0\mapsto0\) (Proposition 1.6), and the extension maps \(A\) into \(B\). Conversely \(\varphi^{-1}(0)=\{0\}\) means \(\varphi(A)\subseteq B\). \(\square\)

So nothing is lost by adjoining a zero. Three things change.

1. More monoids. By Proposition 1.6 the monoids \(A_0\) are exactly the monoids without zero divisors. Monoids with zero divisors are needed for closed subsets: for an ideal \(I\) the closed set \(V(I)\) is the spectrum of \(M/I\) (Proposition 6.5), and \(M/I\) is without zero divisors only if \(I\) is prime. Without zero, \(V(I)\) is in general not a spectrum at all. Every spectrum \(\operatorname{Spec}A_0\) has the smallest point \(\{0\}\), but the closed set \(V(xy)\) in \(\operatorname{Spec}\mathbb F_1[x,y]\) has two minimal points (Example 5.6).
2. More morphisms and more points. By part (f), \(\operatorname{Hom}(A_0,B_0)\) is larger than the set of morphisms \(A\to B\). For example \(\operatorname{Hom}(M,\mathbb F_1)=\operatorname{Spec}M\) (Proposition 2.7), whereas a monoid without zero has exactly one morphism to the trivial monoid. Accordingly [Deitmar 2005, Proposition 2.4] finds that the \(\mathbb F_1\)-points of a scheme are its connected components, while with zero the \(\mathbb F_1\)-points are all the points [Connes–Consani 2010b, Theorem 3.1].
3. Spectra with several minimal points. In \(\operatorname{Spec}A_0\) the point \(\{0\}\) lies in every non-empty open set, as noted in [Deitmar 2005, Section 1.2] for the empty prime ideal. The spectrum of a monoid with zero divisors has no such point in general.

The works [Connes–Consani 2010b], [Chu–Lorscheid–Santhanam 2012] and [Cortiñas–Haesemeyer–Walker–Weibel 2015] use monoids with zero, under the names pointed monoids or \(\mathcal M_0\). So does [Deitmar 2013, Section 1.1]. This course follows them: a monoid has a zero, and \(\mathbb F_1=\{0,1\}\).

11. Exercises

Exercise 1 (groups with zero). Let \(M\neq0\) be a monoid. Show that the following are equivalent: (i) every non-zero element of \(M\) is a unit; (ii) \(\{0\}\) and \(M\) are the only ideals of \(M\); (iii) \(\operatorname{Spec}M\) has one point and \(M\) has no nilpotent element other than \(0\). Give a monoid with a one-point spectrum that is not a group with zero.

Solution. (i)⇒(ii): a proper ideal contains no unit, so it is \(\{0\}\). (ii)⇒(i): for \(a\neq0\) the ideal \(aM\) is not \(\{0\}\), so \(aM=M\) and \(a\) is a unit. (i)⇒(iii): every prime ideal lies in \(\mathfrak m_M=\{0\}\), so \(\{0\}\) is the only prime ideal; a unit is not nilpotent. (iii)⇒(i): the only prime ideal is \(\mathfrak m_M\). By Corollary 2.9(c) the set of nilpotent elements is the intersection of all prime ideals, that is, \(\mathfrak m_M\). So \(\mathfrak m_M=\{0\}\). The monoid \(\{0,1,\varepsilon\}\) with \(\varepsilon^2=0\) has the single prime ideal \(\{0,\varepsilon\}\), and \(\varepsilon\) is not a unit.

Exercise 2 (products). Let \(M\) and \(N\) be non-zero monoids.

(a) Show that the faces of \(M\times N\) are the sets \(F_1\times F_2\), where \(F_1\) is a face of \(M\) or \(F_1=M\), \(F_2\) is a face of \(N\) or \(F_2=N\), and not both \(F_1=M\) and \(F_2=N\).

(b) Describe \(\operatorname{Spec}(\mathbb F_1\times\mathbb F_1)\).

(c) Show that \(\mathbb Z[\mathbb F_1\times\mathbb F_1]\cong\mathbb Z\times\mathbb Z\times\mathbb Z\), and describe \(\beta\).

Solution. (a) Let \(F\) be a face of \(M\times N\). Put \(F_1=\{a:(a,1)\in F\}\) and \(F_2=\{b:(1,b)\in F\}\). Since \((a,b)=(a,1)(1,b)\), we have \((a,b)\in F\) if and only if \(a\in F_1\) and \(b\in F_2\). So \(F=F_1\times F_2\). The set \(F_1\) contains \(1\), and \(aa'\in F_1\) if and only if \(a,a'\in F_1\). If \(0\notin F_1\), then \(F_1\) is a face of \(M\). If \(0\in F_1\), then \(0=0\cdot a\in F_1\) gives \(a\in F_1\) for all \(a\), so \(F_1=M\). The same holds for \(F_2\). Not both are improper, because \((0,0)\notin F\). Conversely every such product contains \((1,1)\), does not contain \((0,0)\), and satisfies the condition of Definition 2.3.

So \(\operatorname{Spec}(M\times N)\) has \(mn+m+n\) points if \(\operatorname{Spec}M\) has \(m\) points and \(\operatorname{Spec}N\) has \(n\) points. It contains the disjoint open subsets \(D((1,0))\) and \(D((0,1))\), which are copies of \(\operatorname{Spec}M\) and \(\operatorname{Spec}N\) by Proposition 4.2, because \((M\times N)_{(1,0)}\cong M\). But it has \(mn\) further points, it has a single closed point, and it is connected (Proposition 3.3). For rings, the spectrum of a product is the disjoint union [Stacks, Tag 00ED].

(b) Write \(e_1=(1,0)\), \(e_2=(0,1)\). The only face of \(\mathbb F_1\) is \(\{1\}\). By (a) the faces are \(\{1\}\), \(\{1,e_1\}\) and \(\{1,e_2\}\), and the prime ideals are \(\mathfrak m=\{0,e_1,e_2\}\), \(\mathfrak p_1=\{0,e_2\}\) and \(\mathfrak p_2=\{0,e_1\}\). The ideal \(\{0\}\) is not prime, because \(e_1e_2=0\). The open sets are \(\emptyset\), \(\{\mathfrak p_1\}=D(e_1)\), \(\{\mathfrak p_2\}=D(e_2)\), \(\{\mathfrak p_1,\mathfrak p_2\}\) and the whole space.

(c) The ring has the basis \(1,e_1,e_2\) with \(e_i^2=e_i\) and \(e_1e_2=0\). Put \(e_3=1-e_1-e_2\). Then \(e_3^2=1-2e_1-2e_2+e_1+e_2=e_3\) and \(e_3e_1=e_3e_2=0\). So \(e_1,e_2,e_3\) are orthogonal idempotents with sum \(1\), and \(x\mapsto(xe_1,xe_2,xe_3)\) is an isomorphism onto \(\mathbb Ze_1\times\mathbb Ze_2\times\mathbb Ze_3\cong\mathbb Z^3\). The spectrum is the disjoint union of three copies of \(\operatorname{Spec}\mathbb Z\). On the first copy \(e_1=1\) and \(e_2=0\), so \(\beta\) maps it to \(\mathfrak p_1\). The second copy maps to \(\mathfrak p_2\). On the third copy \(e_1=e_2=0\), so it maps to \(\mathfrak m\). Each fibre is \(\operatorname{Spec}\mathbb Z\), in agreement with Theorem 8.13: all groups \(G_{\mathfrak p}\) are trivial. In particular \(\mathbb Z[M\times N]\) is not \(\mathbb Z[M]\times\mathbb Z[N]\).

Exercise 3 (the quotients of \(\mathbb F_1[x]\)).

(a) Show that every congruence on \(\mathbb F_1[x]\) is exactly one of the following: equality; the congruence \(\sim_I\) of the ideal \(I=\langle x^n\rangle\), \(n\ge0\); or, for integers \(r\ge0\) and \(m\ge1\), the congruence \(\sim_{r,m}\) with

in which \(0\) is equivalent only to itself.

(b) For each quotient determine the spectrum and the base change to \(\mathbb Z\).

(c) Which of the quotients are cancellative?

Solution. (a) Let \(\sim\) be a congruence. Case 1: \(x^n\sim0\) for some \(n\ge0\). Take \(n\) minimal. Then \(x^a\sim0\) for all \(a\ge n\). Suppose \(x^a\sim x^b\) with \(a<b\) and \(a<n\), and put \(d=b-a\). Multiplying by \(x^d\) repeatedly gives \(x^a\sim x^{a+d}\sim x^{a+2d}\sim\cdots\), and \(x^{a+kd}\sim0\) for large \(k\). So \(x^a\sim0\), against the choice of \(n\). Hence the classes of \(1,x,\dots,x^{n-1}\) have one element each, and \(\sim\) is \(\sim_I\) for \(I=\langle x^n\rangle\). Case 2: no power of \(x\) is equivalent to \(0\), and \(\sim\) is not equality. Let \(r\) be the smallest exponent such that \(x^r\sim x^b\) for some \(b>r\), and let \(m\ge1\) be minimal with \(x^r\sim x^{r+m}\). Multiplying by powers of \(x\) gives \(x^a\sim x^{a+km}\) for all \(a\ge r\), \(k\ge0\). Conversely let \(x^a\sim x^b\) with \(a<b\). Then \(a\ge r\) by the choice of \(r\). Write \(b-a=km+j\) with \(0\le j<m\), and put \(a'=a+km\). Then \(x^{a'}\sim x^a\sim x^b=x^{a'+j}\). Choose \(k'\) with \(r+k'm\ge a'\) and multiply by \(x^{r+k'm-a'}\): this gives \(x^{r+k'm}\sim x^{r+k'm+j}\), hence \(x^r\sim x^{r+j}\). By the choice of \(m\), \(j=0\). So \(\sim\) is \(\sim_{r,m}\). The listed congruences are pairwise different.

(b) \(\mathbb F_1[x]\) has the prime ideals \(\{0\}\) and \(\langle x\rangle\), and base change \(\mathbb Z[x]\). The quotient \(\mathbb F_1[x]/\langle x^n\rangle\) is the zero monoid for \(n=0\). For \(n\ge1\) the element \(x\) is nilpotent, the only prime ideal is \(\langle x\rangle\), and the base change is \(\mathbb Z[x]/(x^n)\). The quotient \(C_{r,m}=\mathbb F_1[x]/{\sim_{r,m}}\) has the \(r+m\) non-zero elements \(1,x,\dots,x^{r+m-1}\), with \(x^{r+m}=x^r\). For \(r=0\) it is \(\mathbb F_{1^m}\), with one prime ideal. For \(r\ge1\) it is without zero divisors, \(x\) is neither a unit nor nilpotent, and a face that contains a power \(x^k\), \(k\ge1\), contains \(x\) and so all non-zero elements. So there are two prime ideals, \(\{0\}\) and the maximal ideal. By Proposition 8.4(b) the base change is \(\mathbb Z[x]/(x^{r+m}-x^r)\), because every difference \(x^{a+km}-x^a\) with \(a\ge r\) is a multiple of \(x^{r+m}-x^r\). The fibres of \(\beta\) over the two points are \(V(x^m-1)\setminus V(x)\) and \(V(x)\): the spectra of \(\mathbb Z[x]/(x^m-1)\) and of \(\mathbb Z\). This agrees with Theorem 8.13, since the localization of \(C_{r,m}\) at \(\{0\}\) is \(\mathbb F_{1^m}\).

(c) \(\mathbb F_1[x]\) is cancellative. \(\mathbb F_1[x]/\langle x^n\rangle\) is cancellative only for \(n=1\), where it is \(\mathbb F_1\): for \(n\ge2\) we have \(x\cdot x^{n-1}=x\cdot0\). \(C_{r,m}\) is cancellative only for \(r=0\): for \(r\ge1\) we have \(x^r\cdot x^m=x^r\cdot1\) and \(x^m\neq1\). So the congruences on \(\mathbb F_1[x]\) with a cancellative quotient, that is, the prime congruences of Remark 6.8, are: equality, \(x\sim0\), and \(x^m\sim1\) for \(m\ge1\). This agrees with fact 3 of Remark 6.8 and the fibre description quoted there: the group \(G_{\mathfrak p}\) is \(\mathbb Z\) at \(\mathfrak p=\{0\}\), with subgroups \(m\mathbb Z\), and it is trivial at \(\mathfrak p=\langle x\rangle\).

Reference: [Lorscheid–Ray 2024, Example 2.10] lists the prime congruences on \(\mathbb F_1[t]\) as equality, \(\langle(t,0)\rangle\) and \(\langle(t,t^k)\rangle\), \(k\ge0\); this fails for \(k\ge2\), because the quotient by \(t\sim t^k\) is \(C_{1,k-1}\), which is not cancellative, so the list above has \(\langle(t^m,1)\rangle\) instead. [Lorscheid–Ray 2024, Example 2.15] describes the prime congruences on free monoids by subgroups, in agreement with the list above.

Exercise 4 (sections over the punctured spectrum). Let \(M\) be cancellative.

(a) Show that for every non-empty open set \(U\subseteq\operatorname{Spec}M\), \(\mathcal O(U)\cong\bigcap_{\mathfrak p\in U}M_{\mathfrak p}\), the intersection being taken in \(M^{\mathrm{gp}}\).

(b) Let \(P=\mathbb N^2\setminus\{(1,0)\}\) and \(M=\mathbb F_1[P]\), with \(x^ay^b=\chi^{(a,b)}\). Show that \(M\) is finitely generated with four prime ideals, and that for \(U=\operatorname{Spec}M\setminus\{\mathfrak m_M\}\) the restriction \(\mathcal O(\operatorname{Spec}M)\to\mathcal O(U)\) is injective and not surjective. Compare with Example 5.5.

Solution. (a) \(\{0\}\) is a prime ideal, and it lies in \(U\) by Proposition 3.3(b). The maps \(\rho_{\mathfrak p,\{0\}}:M_{\mathfrak p}\to M^{\mathrm{gp}}\) are injective, because \(M\) is cancellative. A section \(s\in\mathcal O(U)\) is determined by \(g=s_{\{0\}}\in M^{\mathrm{gp}}\), since \(\rho_{\mathfrak p,\{0\}}(s_{\mathfrak p})=g\) for all \(\mathfrak p\in U\); and \(g\) lies in every \(M_{\mathfrak p}\), \(\mathfrak p\in U\). Conversely, for \(g\) in the intersection let \(s_{\mathfrak p}\) be the element of \(M_{\mathfrak p}\) with image \(g\). For \(\mathfrak q\subseteq\mathfrak p\) in \(U\) the elements \(\rho_{\mathfrak p,\mathfrak q}(s_{\mathfrak p})\) and \(s_{\mathfrak q}\) have the same image \(g\), so they are equal.

(b) \(P\) is closed under addition: if \(u+v=(1,0)\) with \(u,v\in\mathbb N^2\), then \(u\) or \(v\) is \((1,0)\). \(M\) is generated by \(y\), \(x^2\), \(x^3\) and \(xy\): every integer \(a\ge2\) is of the form \(2i+3j\), and \((1,b)=(1,1)+(0,b-1)\) for \(b\ge1\). \(M\) is a submonoid of \(\mathbb F_1[\mathbb Z^2]\), so it is cancellative, and \(M^{\mathrm{gp}}=\mathbb F_1[\mathbb Z^2]\), because \(x=x^3/x^2\). A face that contains \(x^2\) or \(x^3\) contains both, because \((x^2)^3=(x^3)^2\). A face that contains \(xy\) contains \(x^2y^2=x^2\cdot y\cdot y\), hence \(x^2\) and \(y\). A face that contains \(x^2\) and \(y\) contains \(x^2y^2=(xy)^2\), hence \(xy\). So there are four faces: \(\{1\}\), the powers of \(y\), the elements \(x^a\) with \(a\neq1\), and the set of all non-zero elements. Let \(\mathfrak p_1\) be the complement of the powers of \(y\), and \(\mathfrak p_2\) the complement of the set of the \(x^a\). Then \(U=\{\{0\},\mathfrak p_1,\mathfrak p_2\}\), and

because \(x=xy/y\) and \(x=x^3/x^2\). By (a), \(\mathcal O(U)=M_{\mathfrak p_1}\cap M_{\mathfrak p_2} =\mathbb F_1[x,y]\). It contains \(x\), which is not in \(M=\mathcal O(\operatorname{Spec}M)\). In Example 5.5 the restriction was bijective. Here \(\mathcal O(U)\) is the normalization of \(M\).

Exercise 5 (the cone over a square). Let \(M\) be the quotient of \(\mathbb F_1[x,y,z,w]\) by the congruence generated by the pair \((xw,yz)\). Let \(L=\mathbb Z^3\) and \(C=\operatorname{cone}(v_1,v_2,v_3,v_4)\) with

(a) Show that \(x\mapsto\chi^{v_1}\), \(y\mapsto\chi^{v_2}\), \(z\mapsto\chi^{v_3}\), \(w\mapsto\chi^{v_4}\) induces an isomorphism \(M\cong\mathbb F_1[C\cap L]\).

(b) Show that \(M\) has ten prime ideals.

(c) Count the solutions of \(xw=yz\) in \(\mathbb F_q^4\) in two ways.

Solution. (a) Let \(\varphi:\mathbb F_1[x,y,z,w]\to\mathbb F_1[C\cap L]\) be the morphism of Proposition 1.4. Since \(v_1+v_4=v_2+v_3\), we have \(\varphi(xw)=\varphi(yz)\). So the kernel congruence of \(\varphi\) contains the generated congruence \(\sim\), and \(\varphi\) induces \(\bar\varphi:M\to\mathbb F_1[C\cap L]\) (Proposition 6.2). A point \(\sum t_iv_i\) of \(C\) is \((t_1+t_3,\ t_2+t_4,\ t_3+t_4)\). So

for a point of the right-hand side take \(t_3=\min(p,r)\), \(t_4=r-t_3\), \(t_1=p-t_3\), \(t_2=q-t_4\); these numbers are non-negative, and they are integers if \(p,q,r\) are. Hence the \(v_i\) generate the monoid \(C\cap L\), and \(\bar\varphi\) is surjective. For injectivity, use \(xw\sim yz\) to replace a factor \(xw\) by \(yz\) as long as possible. So every monomial is equivalent to one of the form \(x^ay^bz^c\) or \(y^bz^cw^d\). Their images are \(\chi^{(a+c,\,b,\,c)}\) and \(\chi^{(c,\,b+d,\,c+d)}\). In the first family the third coordinate is at most the first, with equality only for \(a=0\). In the second family the third coordinate is at least the first, with equality only for \(d=0\). So \(\varphi\) is injective on these monomials. No monomial is equivalent to \(0\), because \(\varphi\) sends monomials to non-zero elements. Hence \(\bar\varphi\) is injective.

(b) By Theorem 9.10 we count the faces of \(C\). By Lemma 9.9 a face \(F\) is the cone generated by the set \(E\) of the \(v_i\) it contains. If \(v_1,v_4\in F\), then \(v_2+v_3=v_1+v_4\in F\), so \(v_2,v_3\in F\); and conversely. This excludes six of the sixteen subsets \(E\): those that contain \(\{v_1,v_4\}\) but not \(\{v_2,v_3\}\), or the reverse. The other ten are faces, of the form \(C\cap\ker u\) with \(u\ge0\) on \(C\) (Definition 9.8). In the coordinates \((p,q,r)\): \(u=0\) gives \(C\); the forms \(p\), \(q\), \(r\) and \(p+q-r\) give the cones generated by \(\{v_2,v_4\}\), \(\{v_1,v_3\}\), \(\{v_1,v_2\}\) and \(\{v_3,v_4\}\); the forms \(q+r\), \(p+r\), \(p+2q-r\) and \(2p+q-r\) give the rays through \(v_1\), \(v_2\), \(v_3\) and \(v_4\); and \(p+q\) gives \(\{0\}\). So \(C\) has one face of dimension \(0\), four of dimension \(1\), four of dimension \(2\) and one of dimension \(3\).

(c) By (a), Proposition 1.4, Proposition 6.2(c) and Proposition 8.3, the ring homomorphisms \(\mathbb Z[M]\to\mathbb F_q\) are the quadruples \((x,y,z,w)\in\mathbb F_q^4\) with \(xw=yz\). By Corollary 9.11 and (b) their number is

Directly: the solutions are the \(2\times2\) matrices over \(\mathbb F_q\) with determinant \(0\). There are \((q^2-1)(q^2-q)\) invertible matrices, since the first column is any non-zero vector and the second any vector outside the line spanned by the first. So the number of solutions is \(q^4-(q^2-1)(q^2-q)=q^3+q^2-q\).

12. What this lesson does not prove

• The structure theorem for finitely generated abelian groups, in the form: a finitely generated abelian group is the product of a free group of finite rank and a finite group. This is the case of the ring \(\mathbb Z\) of the structure theorem for finite modules over a principal ideal domain [Stacks, Tag 0ASV]. It is used in Theorem 8.5(b), Corollary 8.14, Theorem 9.7 and Theorem 9.10(d).
• The Hilbert basis theorem: a finitely generated ring is Noetherian [Stacks, Tag 00FN]. It is used in Corollary 8.6.
• The universal property of the localization of a ring [Stacks, Tag 00CP], and the description of the spectrum of a localization and of a quotient of a ring [Stacks, Tags 00E3 and 00E5]. They are used in Proposition 8.4 and Theorem 8.13.
• A converse of Theorem 8.5(c). Let \(M\) be finitely generated, normal and torsion-free, and let \(k\) be an integrally closed domain that contains a field. Then \(k[M]\) is an integrally closed domain. This is the affine case of [Cortiñas–Haesemeyer–Walker–Weibel 2015, Proposition 6.1(1)], where it is deduced from the normality of toric varieties. The ring \(\mathbb Z\) contains no field, but the statement for \(k=\mathbb Z\) follows from the statement for \(k=\mathbb Q\). Let \(M^{\mathrm{gp}}=G_0\). Then \(\mathbb Z[M]=\mathbb Q[M]\cap\mathbb Z[G]\) inside \(\mathbb Q[G]\), and all these rings have the same fraction field. The group \(G\) is free of finite rank, as in the proof of Theorem 9.7, so \(\mathbb Z[G]\) is a ring of Laurent polynomials over \(\mathbb Z\). It is a normal domain, because a principal ideal domain is normal, a polynomial ring over a normal domain is normal, and a localization of a normal domain is normal [Stacks, Tags 00GZ, 00H1 and 00GY]. So an element of the fraction field that is integral over \(\mathbb Z[M]\) lies in \(\mathbb Q[M]\) and in \(\mathbb Z[G]\), hence in \(\mathbb Z[M]\).
• The statement that \(M\mapsto(\operatorname{Spec}M,\mathcal O_M)\) is a contravariant equivalence between monoids and affine monoid schemes ([Deitmar 2005, Proposition 2.2] for monoids without zero). It belongs to the lesson Monoid schemes and is not used here.
• The description of the fibres of the congruence space quoted in Remark 6.8 [Lorscheid–Ray 2024, Proposition 2.18]. The results of [Jarra 2023a] quoted there are proved in Monoid schemes, Section 6.6.

References

• [Deitmar 2005] A. Deitmar, Schemes over F_1, arXiv:math/0404185.
• [Deitmar 2006] A. Deitmar, Remarks on zeta functions and K-theory over F_1, arXiv:math/0605429.
• [Deitmar 2008] A. Deitmar, F_1-schemes and toric varieties, arXiv:math/0608179.
• [Deitmar 2013] A. Deitmar, Congruence schemes, International Journal of Mathematics 24 (2013), no. 2, arXiv:1102.4046.
• [Connes–Consani 2010] A. Connes, C. Consani, Schemes over F_1 and zeta functions, arXiv:0903.2024. Free at https://alainconnes.org/wp-content/uploads/schemesF1zeta.pdf
• [Connes–Consani 2010b] A. Connes, C. Consani, From monoids to hyperstructures: in search of an absolute arithmetic, arXiv:1006.4810. Free at https://alainconnes.org/wp-content/uploads/From-monoids-to-hyperstructures-2010.pdf
• [Chu–Lorscheid–Santhanam 2012] C. Chu, O. Lorscheid, R. Santhanam, Sheaves and K-theory for F_1-schemes, arXiv:1010.2896.
• [Cortiñas–Haesemeyer–Walker–Weibel 2015] G. Cortiñas, C. Haesemeyer, M. E. Walker, C. Weibel, Toric varieties, monoid schemes and cdh descent, arXiv:1106.1389.
• [Lorscheid–Ray 2024] O. Lorscheid, S. Ray, The topological shadow of F_1-geometry: congruence spaces, arXiv:2305.12801.
• [Jarra 2023a] M. Jarra, Strong congruence spaces and dimension in F_1-geometry, arXiv:2305.15953.
• [Stacks] The Stacks project, cited by tag. Each tag links to the same place in AI Integrated Stacks Project, the programme's edition of the Stacks project, which agrees with it modulo corrections made by GPT-6 Astra (OpenAI, Ultra setting) on suggestions of GPT-5.6 Sol (OpenAI, Ultra setting). Of the tags cited in this lesson, Tags 006S, 0078, 00AR, 00CM, 00DY, 00E3 and 0ASV carry such corrections: wording, notation and small precisions in statements and proofs. None of them changes the results used here.

Result numbers in the cited preprints are those of the arXiv versions.