--- title: "Lecture 21 - Discrete Valuation Rings and Nakayama's Lemma" stable_id: br-ak-2025-2026-l21 language: en upstream_title: "Kurs:Algebraische Kurven (Osnabrück 2025-2026)/Vorlesung 21" upstream_pageid: 165910 upstream_revid: 1112312 upstream_timestamp: "2026-08-21T09:27:05Z" upstream_mediawiki_sha1: 05c51f6e29f6ec12aef400195396ca517924b094 source_url: "https://de.wikiversity.org/w/index.php?oldid=1112312" authority_manifest: authority/wikiversity/unit-21/UNIT_AUTHORITY_MANIFEST.json authority_manifest_sha256: d85444ddfc66c8e77d52db3f3abc0a186e5dd598789edaaf890b3c09cf00f923 lecture_xml_sha256: 65c9dc086c930a46e53957102e7f742d3ea04fac707dae3595ac3983b54f75b8 lecture_expanded_tex_sha256: 0a6fc74c8d01069d327fe25c5203bf4587b4564c8409ad57f625c3ac16ceb62f license: "CC BY-SA 4.0 for translated course text; Unit 21 contains no substantive reader media" translation_status: complete source_semantic_entities: 13 edition_bridges: 3 source_corrections: 2 --- # Lecture 21: Discrete Valuation Rings and Nakayama's Lemma {#br-ak-2025-2026-l21} ## Discrete valuation rings {#br-ak-2025-2026-l21-s01} We now continue the local study of algebraic curves. In what follows, we shall obtain various characterisations of when a point on a curve is nonsingular—or smooth. Associated with a point $P$ on a curve is the local ring at $P$, obtained by localising the affine coordinate ring of the curve at the maximal ideal corresponding to $P$. If $$ P=(a,b)\in C=V(F)\subseteq\mathbb A_K^2, $$ this local ring can be described in two ways: $$ K[X,Y]_{(X-a,Y-b)}/(F) \cong (K[X,Y]/(F))_{\mathfrak m}, $$ where $\mathfrak m$ is the maximal ideal now regarded in the quotient ring. This ring captures the essential algebraic properties of the point. The first important concept is that of a discrete valuation ring. ### Definition: discrete valuation ring {#br-ak-2025-2026-l21-def-01} A *discrete valuation ring* is a principal ideal domain $R$ having exactly one prime element up to associates. A generator of the maximal ideal of a discrete valuation ring is also called a *local uniformiser*. ### Lemma: first properties of discrete valuation rings {#br-ak-2025-2026-l21-lem-01} A discrete valuation ring is a local Noetherian principal ideal domain with exactly two prime ideals, namely $$ (0)\subset\mathfrak m, $$ where $\mathfrak m$ is its maximal ideal. #### Proof {#br-ak-2025-2026-l21-lem-01-proof} A discrete valuation ring is not a field. In a principal ideal domain that is not a field, every maximal ideal is generated by a prime element, and prime generators of distinct maximal ideals cannot be associates. Since a discrete valuation ring has only one prime element up to associates, it has only one maximal ideal. Likewise, every nonzero prime ideal in a principal ideal domain is generated by a prime element. Hence its only nonzero prime ideal is $\mathfrak m$; together with the zero ideal, this gives exactly two prime ideals. ### Example: localisation of $K[X]$ at $(X)$ {#br-ak-2025-2026-l21-exa-01} Let $K$ be a field, $K[X]$ the polynomial ring, and $$ R=K[X]_{(X)} $$ the localisation at the maximal ideal $$ \mathfrak m=(X). $$ Then $R$ is a discrete valuation ring. Its two prime ideals are $$ (0)\subset(X). $$ The ring $R$ is a principal ideal domain because $K[X]$ is a principal ideal domain. Since there is only one maximal ideal, there is only one prime element up to associates, namely $X$. ### Example: localisation of $\mathbb Z$ at $(p)$ {#br-ak-2025-2026-l21-exa-02} Let $p$ be a prime number and $$ R=\mathbb Z_{(p)} $$ the localisation at the maximal ideal $$ \mathfrak m=(p). $$ Then $R$ is a discrete valuation ring. Its two prime ideals are $$ (0)\subset(p). $$ The ring $R$ is a principal ideal domain because $\mathbb Z$ is a principal ideal domain. Since there is only one maximal ideal, there is only one prime element up to associates, namely $p$. ### Definition: order in a discrete valuation ring {#br-ak-2025-2026-l21-def-02} Let $R$ be a discrete valuation ring with prime element $p$. For every nonzero element $f\in R$, there are an $n\in\mathbb N$ and a unit $u\in R^\times$ such that $$ f=up^n. $$ The number $n$ is called the *order* of $f$, written $$ \operatorname{ord}(f)=n. $$ Thus the order is simply the exponent of the unique prime element, up to associates, in the prime factorisation of $f$. ### Lemma: properties of the order {#br-ak-2025-2026-l21-lem-02} Let $R$ be a discrete valuation ring with maximal ideal $$ \mathfrak m=(p). $$ The order map $$ \begin{aligned} R\setminus\{0\}&\longrightarrow\mathbb N,\\ f&\longmapsto\operatorname{ord}(f) \end{aligned} $$ has the following properties. For $f,g\in R\setminus\{0\}$, 1. we have $$ \operatorname{ord}(fg) =\operatorname{ord}(f)+\operatorname{ord}(g); $$ 2. if $f+g\ne0$, then $$ \operatorname{ord}(f+g) \geq \min\{\operatorname{ord}(f),\operatorname{ord}(g)\}; $$ 3. $f\in\mathfrak m$ if and only if $$ \operatorname{ord}(f)\geq1; $$ 4. $f\in R^\times$ if and only if $$ \operatorname{ord}(f)=0. $$ **Proof.** See Exercise 21.2. **Edition note — source correction.** The source defines $\operatorname{ord}$ only on $R\setminus\{0\}$ but prints the inequality for sums without excluding $f+g=0$. This edition adds the condition $f+g\ne0$, so that every occurrence of $\operatorname{ord}$ lies within its domain. We shall now prove an important characterisation of discrete valuation rings. In particular, it shows that a normal Noetherian local integral domain with exactly two prime ideals is already a discrete valuation ring. First we need a lemma. ### Lemma 21.7: nilpotence of the maximal ideal {#br-ak-2025-2026-l21-lem-03} Let $S$ be a Noetherian local commutative ring. Suppose that its maximal ideal $\mathfrak m$ is its only prime ideal. Then there is an $n\in\mathbb N$ such that $$ \mathfrak m^n=0. $$ #### Proof {#br-ak-2025-2026-l21-lem-03-proof} First we claim that every element of $S$ is either a unit or nilpotent. Take a nonunit $f\in S$. Since $S$ is local, $$ f\in\mathfrak m. $$ Suppose that $f$ is not nilpotent. > **Edition bridge 21.A — the prime ideal lemma.** If an element $f$ of a > commutative ring is not nilpotent, there is a prime ideal $\mathfrak p$ > that does not contain $f$. Indeed, the multiplicative set > $T=\{1,f,f^2,\ldots\}$ does not contain zero. By Zorn's lemma, choose an > ideal $\mathfrak p$ maximal among ideals disjoint from $T$. If > $ab\in\mathfrak p$ but $a,b\notin\mathfrak p$, then both > $\mathfrak p+(a)$ and $\mathfrak p+(b)$ meet $T$. Multiplying one element > from each intersection gives an element of $T\cap\mathfrak p$, a > contradiction. Thus $\mathfrak p$ is prime and, since > $\mathfrak p\cap T=\varnothing$, we have $f\notin\mathfrak p$. > The source cites this fact using the numbering of another course; this > edition supplies its statement and proof to make the argument self-contained. This prime ideal $\mathfrak p$ differs from $\mathfrak m$, since $f\in\mathfrak m$ but $f\notin\mathfrak p$. This contradicts the assumption that $\mathfrak m$ is the only prime ideal. Thus every element of the maximal ideal is nilpotent. Since $S$ is Noetherian, the ideal $\mathfrak m$ has a finite generating system $$ \mathfrak m=(f_1,\ldots,f_k). $$ If $\mathfrak m=0$, take $n=1$. Hence we may now assume $k\geq1$. Choose $m\in\mathbb N$ such that $$ f_i^m=0 \qquad\text{for all }i=1,\ldots,k, $$ and set $n=km$. Every element of $\mathfrak m^n$ is a linear combination of products of the form $$ \left(\sum_{i=1}^k a_{i1}f_i\right) \left(\sum_{i=1}^k a_{i2}f_i\right) \cdots \left(\sum_{i=1}^k a_{in}f_i\right). $$ Upon expansion, each term is a monomial $$ f_1^{r_1}\cdots f_k^{r_k} \qquad\text{with}\qquad \sum_{i=1}^k r_i=n. $$ For at least one index $i$, $$ r_i\geq\frac nk=m. $$ Since $f_i^m=0$, every such monomial is zero. Hence $\mathfrak m^n=0$. ### Theorem: characterisation of discrete valuation rings {#br-ak-2025-2026-l21-thm-01} Let $R$ be a Noetherian local integral domain having exactly two prime ideals $$ (0)\subset\mathfrak m. $$ The following statements are equivalent. 1. $R$ is a discrete valuation ring. 2. $R$ is a principal ideal domain. 3. $R$ is a unique factorisation domain. 4. $R$ is normal. 5. The maximal ideal $\mathfrak m$ is principal. #### Proof {#br-ak-2025-2026-l21-thm-01-proof} The implication $(1)\Rightarrow(2)$ follows directly from Definition 21.1. The implication $(2)\Rightarrow(3)$ follows from Theorem 9.3 in Commutative Algebra. The implication $(3)\Rightarrow(4)$ follows from Theorem 20.2. To prove $(4)\Rightarrow(5)$, take $$ f\in\mathfrak m, \qquad f\ne0. $$ > **Edition bridge 21.B — application to $R/(f)$.** Set $S=R/(f)$. > The ring $S$ remains Noetherian and local, with maximal ideal > $\widetilde{\mathfrak m}=\mathfrak m/(f)$. Prime ideals of $S$ correspond > to prime ideals of $R$ containing $(f)$. Since the only prime ideals of > $R$ are $(0)$ and $\mathfrak m$, and $f\ne0$, the only prime ideal > containing $(f)$ is $\mathfrak m$. Thus $\widetilde{\mathfrak m}$ is the > only prime ideal of $S$. Lemma 21.7 gives > $\widetilde{\mathfrak m}^{\,n}=0$ for some $n$, which, pulled back to $R$, > means exactly that $\mathfrak m^n\subseteq(f)$. Choose $n$ minimal such that $$ \mathfrak m^n\subseteq(f) \qquad\text{and}\qquad \mathfrak m^{n-1}\nsubseteq(f). $$ Choose $$ g\in\mathfrak m^{n-1}\setminus(f) $$ and consider $$ h:=\frac fg\in Q(R). $$ Here $g\ne0$. Its inverse, $$ h^{-1}=\frac gf, $$ does not belong to $R$, since otherwise $g\in(f)$. Because $R$ is normal, $h^{-1}$ is not integral over $R$ either. By the module criterion for integrality in Lemma 19.9, applied in particular to the maximal ideal $\mathfrak m\subset R$, we have $$ h^{-1}\mathfrak m\nsubseteq\mathfrak m. $$ On the other hand, by the choice of $g$, $$ h^{-1}\mathfrak m =\frac gf\mathfrak m \subseteq\frac{\mathfrak m^n}{f} \subseteq R. $$ Thus $h^{-1}\mathfrak m$ is an ideal of $R$ not contained in the maximal ideal. Hence $$ h^{-1}\mathfrak m=R. $$ This equality gives $h\in\mathfrak m$. Moreover, for every $x\in\mathfrak m$, we have $h^{-1}x\in R$, so $$ x=h(h^{-1}x)\in(h). $$ Therefore $$ (h)=\mathfrak m, $$ and the maximal ideal is principal. Now prove $(5)\Rightarrow(1)$. Suppose $$ \mathfrak m=(\pi). $$ The element $\pi$ is prime and is the only prime element up to associates. Take a nonzero nonunit $f\in R$. Then $$ f\in\mathfrak m, $$ so $f=\pi g_1$. The element $g_1$ is either a unit or again belongs to $\mathfrak m$. In the latter case, $$ g_1=\pi g_2 \qquad\text{and}\qquad f=\pi^2g_2. $$ We claim that this process terminates, giving $$ f=\pi^ku $$ for some $k\in\mathbb N$ and unit $u$. Otherwise, for arbitrarily large $n$, we could write $$ f=\pi^ng_n. $$ Applying Lemma 21.7 to $R/(f)$ as in Bridge 21.B, there is an $m\in\mathbb N$ such that $$ (\pi^m)=\mathfrak m^m\subseteq(f). $$ If $n\geq m+1$, then for some $a,b\in R$ we obtain $$ \pi^m=af=a\pi^{m+1}b, $$ and, since $R$ is an integral domain, cancelling $\pi^m$ gives the contradiction $$ 1=ab\pi. $$ Thus every nonzero nonunit is a product of a power of $\pi$ and a unit. In particular, $R$ is a unique factorisation domain. Since $R$ is Noetherian, every ideal can be written as $$ \mathfrak a=(f_1,\ldots,f_s). $$ The zero ideal is already principal. If $\mathfrak a\ne0$, remove zero generators and write each remaining generator as $$ f_i=\pi^{n_i}u_i $$ with $u_i$ a unit. If $$ n=\min_i n_i, $$ then $$ \mathfrak a=(\pi^n). $$ Thus $R$ is a principal ideal domain with exactly one prime element up to associates; by definition, it is a discrete valuation ring. ## Nakayama's lemma {#br-ak-2025-2026-l21-s02} Within the class of Noetherian local integral domains having exactly two prime ideals $(0)\subset\mathfrak m$—and hence not fields—the preceding theorem shows that $R$ is a discrete valuation ring if and only if the maximal ideal $\mathfrak m$ is generated by one element. It is therefore natural to ask, more generally, how many generators are needed for the maximal ideal of the local ring at a point on an algebraic curve. This leads to the embedding dimension, which we have already encountered for monomial curves. It is also the dimension of the vector space over $R/\mathfrak m$ $$ \mathfrak m/\mathfrak m^2. $$ To explain this connection, we need some preparation, in particular Nakayama's lemma. **Edition note — correction to the source's scope.** The source's transition sentence states the equivalence between “discrete valuation ring” and “principal maximal ideal” without repeating the hypotheses. This edition explicitly retains the scope of the theorem: Noetherian local integral domains with exactly two prime ideals. The following construction is used in Nakayama's lemma. Let $V$ be an $R$-module, $U\subseteq V$ a submodule, and $I\subseteq R$ an ideal. The notation $IU$ denotes the $R$-submodule of $V$ generated by all elements $$ fv, \qquad f\in I, \qquad v\in U. $$ This is also a submodule of $U$. If $U$ itself is an ideal—that is, an $R$-submodule of $R$—the construction agrees with the product of ideals. The quotient module $V/IV$ is naturally not only an $R$-module but also an $(R/I)$-module. If $I$ is maximal, this quotient module is even a vector space over the residue field $R/I$. ### Nakayama's lemma {#br-ak-2025-2026-l21-lem-04} Let $(R,\mathfrak m)$ be a local ring and $V$ a finitely generated $R$-module. If $$ \mathfrak mV=V, $$ then $$ V=0. $$ #### Proof {#br-ak-2025-2026-l21-lem-04-proof} Let $v_1,\ldots,v_n$ be a generating system for $V$. Since $v_i\in\mathfrak mV$, for each $v_i$ there is a representation $$ v_i=a_{i1}v_1+\cdots+a_{in}v_n, \qquad a_{ij}\in\mathfrak m. $$ Thus, for every $i$, $$ (1-a_{ii})v_i =a_{i1}v_1+\cdots+a_{i,i-1}v_{i-1} +a_{i,i+1}v_{i+1}+\cdots+a_{in}v_n. $$ Since $a_{ii}\in\mathfrak m$, the coefficient $1-a_{ii}$ is a unit in the local ring $R$. We can solve for $v_i$, so $v_i$ is redundant in the generating system. Removing generators one by one eventually leaves no generators. Hence $V$ is the zero module. > **Edition bridge 21.C — minimal generators corollary.** Let > $(R,\mathfrak m)$ be a local ring, $k=R/\mathfrak m$ its residue field, > and $V$ a finitely generated $R$-module. Elements $v_1,\ldots,v_r$ > generate $V$ if and only if their classes generate the $k$-vector space > $V/\mathfrak mV$. The forward implication is immediate. Conversely, set > $N=Rv_1+\cdots+Rv_r$. The hypothesis on the classes gives > $V=N+\mathfrak mV$, so $V/N=\mathfrak m(V/N)$. Nakayama's lemma gives > $V/N=0$, hence $V=N$. Consequently the minimal number of generators of > $V$ is $\dim_k(V/\mathfrak mV)$. If $\mathfrak m$ is finitely generated, > in particular in the Noetherian setting above, then > $\mu_R(\mathfrak m)=\dim_k(\mathfrak m/\mathfrak m^2)$, the embedding > dimension. This corollary supplies the tool needed to assess the number > of ideal generators in Exercises 21.25–21.26, without providing > solutions to those exercises. --- **Edition provenance.** Translation and reader production: OpenAI Codex gpt-5.6-sol, Ultra. Sources, authors, and component licences are retained as stated in the metadata and the edition's rights files.