Spectral spaces and affine realization

Written and self-checked by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. Original exposition, CC0. Independent AI review has not been completed.

The first lesson associated a space to a ring. We now solve the reverse problem: which spaces can arise? A spectral space is a sober, quasi-compact space with a basis of quasi-compact open sets closed under finite intersection. Sobriety includes the uniqueness of generic points, hence the separation condition \(T_0\). Hochster's theorem says that these conditions suffice, with no finiteness or countability assumption.

Read Spectra of rings, especially the radical criterion in Theorem 1.2 and the spectral-space proof in Solution 7.5, first. We use elementary polynomial rings, fraction fields and Zorn's lemma. Weighted valuations and the new topology used in the construction are developed here. The proof follows Hochster's mathematical construction in Prime ideal structure in commutative rings, §§2–7, with a fixed infinite ground field, explicit checks of the controlled-division step, and a direct recovery of every prime ideal. No functorial realization theorem is assumed.

1. Keeping track of both membership and nonmembership

Let \(\mathcal Q\) be the set of quasi-compact open subsets of a spectral space \(X\). It contains \(X,\varnothing\) and is closed under finite unions and intersections. Every quasi-compact open is a finite union of members of the given basis, so this follows even if the definition initially specified only a basis.

The patch topology is generated by all \(U\in\mathcal Q\) and their complements. Thus every such \(U\) is clopen in this finer topology.

Lemma 1.1. The patch topology is compact and Hausdorff.

Proof. Distinct points are distinguished by an original open set, hence by a member of \(\mathcal Q\); that set and its complement separate them in the patch topology.

For compactness, let \(\mathcal F\) be an ultrafilter of subsets of \(X\). Put \(\mathcal I=\{U\in\mathcal Q:U\notin\mathcal F\}\). It is closed under finite unions and passage to smaller members of \(\mathcal Q\). Define

The set \(C\) is nonempty: otherwise quasi-compactness of \(X\) would give a finite subcover from \(\mathcal I\), putting \(X\) in \(\mathcal I\). If \(U\in\mathcal Q\cap\mathcal F\), then \(U\cap C\ne\varnothing\). Indeed, a cover of \(U\) by members of \(\mathcal I\) would have a finite subcover, contradicting \(U\in\mathcal F\). Conversely, a member of \(\mathcal Q\) meeting \(C\) cannot belong to \(\mathcal I\).

Any two original opens meeting \(C\) therefore contain members \(U,V\) of \(\mathcal Q\cap\mathcal F\). Their intersection belongs to the filter and meets \(C\). Thus \(C\) is irreducible and closed. Let \(x\) be its unique generic point. An original open meets \(C=\overline{\{x\}}\) exactly when it contains \(x\). Consequently

Every patch neighborhood of \(x\) belongs to \(\mathcal F\), since a basic one is a finite intersection of sets \(U\) or \(X\setminus U\) having the corresponding membership. Hence every ultrafilter converges. This proves compactness: if a family of patch-closed sets has the finite-intersection property, extend the filter it generates to an ultrafilter by Zorn; a limit point lies in every closed member. The empty space is compact separately. \(\square\)

Write \(y\rightsquigarrow x\) when \(x\in\overline{\{y\}}\). Original opens contain \(y\) whenever they contain \(x\).

Lemma 1.2. A patch-closed subset \(E\) is closed in the original topology if and only if it contains every specialization of each of its points. A patch-clopen subset with this property has a quasi-compact original-open complement.

Proof. The forward implication is immediate. If \(x\) is in the original closure of \(E\), every finite intersection of quasi-compact open neighborhoods of \(x\) meets \(E\). Patch compactness gives a point \(y\in E\) in all these neighborhoods. Every original neighborhood of \(x\) contains such a basis member, so \(y\rightsquigarrow x\). Specialization stability puts \(x\) in \(E\). For the final assertion, the complement is original-open and patch-compact. Any original-open cover is also a patch-open cover, so has a finite subcover. \(\square\)

2. Functions whose nonzero sets are the desired opens

Assume \(X\ne\varnothing\), and fix \(k=\mathbb Q\). Introduce algebraically independent variables \(t_U\) for \(U\in\mathcal Q\), and let \(K=k(t_U:U\in\mathcal Q)\). Define a function

Let \(A_0\subset K^X\) be the \(k\)-algebra generated by these functions. For a function \(a\), put \(D_X(a)=\{x:a(x)\ne0\}\) and \(Z_X(a)=X\setminus D_X(a)\).

An element of \(A_0\) is a polynomial in finitely many \(e_U\)'s. Combine like monomials first. At a point, some monomials vanish; distinct surviving monomials remain distinct monomials in independent variables. They cannot cancel. Thus its nonzero set is the union of the finitely many intersections of opens corresponding to its nonconstant monomials, together with \(X\) if it has a nonzero constant term. In particular every \(D_X(a)\) is quasi-compact open, and the sets \(D_X(e_U)=U\) form a basis. Every function in \(A_0\) has finite image in \(K\).

Evaluation at a point has a domain as its image. Therefore its kernel is a prime ideal. The difficulty is that \(A_0\) can have additional primes. We will add functions to eliminate exactly those additional primes, preserving all the nonzero sets as quasi-compact opens.

3. Orders of vanishing along a specialization

For \(y\rightsquigarrow x\), the image \(A_0(y)\) is the polynomial ring generated by \(t_U\) with \(y\in U\). Give such a variable weight zero if \(x\in U\), and weight one otherwise. For a nonzero polynomial define \(v_{y,x}\) as the smallest weight of its surviving monomials. Minimal-weight homogeneous parts multiply without becoming zero, since their polynomial ring is a domain. Hence

whenever the expressions are nonzero. Extend to the fraction field of \(A_0(y)\) by \(v(f/g)=v(f)-v(g)\). The product identity makes this independent of the chosen fraction. The valuation may be trivial; this includes \(x=y\).

For \(a\in A_0\) with \(a(y)\ne0\),

The weight-zero monomials are precisely those surviving at \(x\), proving (1). For each fixed \(a\), the values in (1) are bounded above by its polynomial degree, uniformly over all relevant pairs.

We retain the following three properties as functions are added: their nonzero sets are quasi-compact open; (1) holds for the same valuations on their evaluation fraction fields; and each function has a uniform finite upper bound for the nonnegative valuations of its nonzero values. Every new value at \(y\) will lie in the original fraction field of \(A_0(y)\).

4. Division with control at the zero set

Suppose a ring \(A\subset K^X\) containing \(A_0\) has these three properties and all its functions have finite image. For \(a,b\in A\) with \(Z_X(b)\subseteq Z_X(a)\), define

Allow this function to be adjoined when, for every \(y\rightsquigarrow x\) with \(a(y)\ne0\),

The denominator is nonzero at \(y\) by the zero-set inclusion.

Lemma 4.1. Under (3), adjoining \(h\) preserves all three properties and finite images.

Proof. Consider \(r=\sum_{i=0}^m c_i h^i\) and the old function \(c=\sum_{i=0}^m c_i a^i b^{m-i}\). On \(D_X(b)\), \(c=b^m r\); on \(Z_X(b)\), \(r=c_0\). Thus

is patch-clopen.

Its complement is stable under specialization. Suppose \(r(y)=0\) and \(y\rightsquigarrow x\). If \(b(x)\ne0\), then \(b(y)\ne0\), so \(c(y)=0\) and therefore \(c(x)=0\), giving \(r(x)=0\). If \(b(x)=0=b(y)\), then \(c_0(y)=0\) implies \(c_0(x)=0\). In the remaining case \(b(y)\ne0\) and \(b(x)=0\), if \(a(y)=0\), again \(r(y)=c_0(y)=0\). Otherwise (3) gives \(v(h(y))>0\). Every nonconstant term \(c_i(y)h(y)^i\) then has positive valuation when nonzero. Their sum equals \(-c_0(y)\); hence either \(c_0(y)=0\), or its valuation is positive. Property (1) gives \(c_0(x)=0\) in either case. As \(h(x)=0\), we have \(r(x)=0\). Lemma 1.2 now makes (4) quasi-compact original-open.

We also verify (1) for \(r\). If \(r(y)\ne0\) and \(b(y)=0\), then \(b(x)=0\), and both values of \(r\) are those of \(c_0\). If \(b(y)\ne0\), all nonzero terms have nonnegative valuation by (3). When \(b(x)\ne0\), \(v(b(y))=0\), and \(c=b^mr\) gives \(v(r(y))=0\) exactly when \(c(x)\ne0\), or equivalently \(r(x)\ne0\). When \(b(x)=0\), the nonconstant terms have positive valuation, unless \(a(y)=0\), in which case they vanish. Thus \(v(r(y))=0\) exactly when \(c_0(x)\ne0\), again the required condition.

Finally, if \(b(y)\ne0\) and \(r(y)\ne0\), then \(c(y)\ne0\) and \(v(r(y))=v(c(y))-m v(b(y))\le N_c\), where \(N_c\) is an old uniform bound. If \(b(y)=0\), use \(N_{c_0}\). A uniform bound is therefore \(\max(N_c,N_{c_0})\). The values of \(h\) form a finite set, since the values of \(a,b\) do; a polynomial in finitely many finite-image functions also has finite image. \(\square\)

5. Making every zero-set containment algebraic

At each stage adjoin all functions (2) satisfying (3), and take the union over the stages \(A_0\subset A_1\subset A_2\subset\cdots\). This is a construction with sets: all rings are subrings of the fixed set \(K^X\). To justify adjoining many functions at once, each polynomial uses only finitely many of them. Adjoin those finitely many successively. Old evaluation fraction fields are unchanged, and the valuations of old functions are unchanged, so each old pair still satisfies (3). Lemma 4.1 applies successively. Its conclusions then pass to the union \(A\).

Proposition 5.1. In \(A\), if \(Z_X(b)\subseteq Z_X(a)\), then \(a\in\sqrt{(b)}\).

Proof. The two functions belong to some common stage. Let \(N\ge0\) bound all valuations of nonzero values of \(b\). The pair \((a^{N+1},b)\) satisfies (3): if \(b(x)\ne0\), its valuation is zero and the numerator has nonnegative valuation. If \(b(x)=0\), then \(a(x)=0\); for \(a(y)\ne0\), the integer valuation of \(a(y)\) is at least one. Thus \((N+1)v(a(y))>v(b(y))\). At the next stage the quotient extended by zero is present, and multiplying it by \(b\) gives \(a^{N+1}\) at every point. The identity holds also where \(b=0\), because \(a=0\) there. \(\square\)

Lemma 5.2. For finitely many \(b_i\in A\), their ideal contains a function \(c\) with \(D_X(c)=\bigcup_i D_X(b_i)\).

Proof. For two functions, choose \(\lambda\in\mathbb Q\) such that \(b_1(x)+\lambda b_2(x)\ne0\) whenever at least one summand function is nonzero. Where \(b_2\ne0\), failure forbids only the value \(-b_1(x)/b_2(x)\). There are finitely many forbidden values because both functions have finite image. The infinite field \(\mathbb Q\) therefore supplies a choice outside them. Where \(b_2=0\), no cancellation is possible. Take \(c=b_1+\lambda b_2\), and iterate. The empty list uses \(c=0\). \(\square\)

Combining the two results gives the full criterion

Indeed, the left side is \(Z_X(c)\), so Proposition 5.1 gives a power of \(a\) in \((c)\), which is contained in the indicated ideal.

6. Recovering every point and every prime

Theorem 6.1 (Hochster realization). Every spectral space is homeomorphic to the prime spectrum of a commutative ring with identity.

Proof. For nonempty \(X\), use \(A\) constructed above and define \(\phi(x)=\{a\in A:a(x)=0\}\). Evaluation has image in the field \(K\), so this is a prime ideal, and \(\phi^{-1}(D(a))=D_X(a)\). The map is continuous and injective: the \(e_U\)'s distinguish points because \(X\) is \(T_0\).

Let \(\mathfrak p\) be any prime of \(A\). The patch-closed sets

have the finite-intersection property. Otherwise, for finite lists, put \(a=\prod_j a_j\notin\mathfrak p\); the empty product is one. The empty intersection would say \(\bigcap_i Z_X(b_i)\subseteq Z_X(a)\). By (5), a power of \(a\) belongs to the ideal generated by the \(b_i\)'s and hence to \(\mathfrak p\), a contradiction. Patch compactness gives \(x\) in the intersection of all these sets. Exactly the members of \(\mathfrak p\) vanish at \(x\), so \(\phi(x)=\mathfrak p\). Thus \(\phi\) is onto.

The image of \(U=D_X(e_U)\) is now precisely \(D(e_U)\). Since these \(U\)'s form a basis, \(\phi\) is an open bijection and a homeomorphism. If \(X=\varnothing\), use the zero ring, whose prime spectrum is empty. No other case was excluded, and the construction allowed an arbitrary set of quasi-compact opens. \(\square\)

Together with Solution 7.5 of Spectra of rings, this proves the full characterization in both directions. It does not assert a canonical ring or a structure sheaf determined solely by the topology.

7. Examples and exercises

A one-point space is realized by \(\mathbb Q\), though the general construction is much larger. The two-point space with opens \(\varnothing,\{\eta\},\{\eta,s\}\) is realized by \(\mathbb Q[T]_{(T)}\): its primes are \((0)\) and \((T)\). In the construction, the variable associated to \(\{\eta\}\) has valuation one along \(\eta\rightsquigarrow s\). This is the mechanism that prevents an illicit inverse from destroying the closed point.

Exercise 7.1 (easy). Explain why a quasi-compact Hausdorff space with a basis of clopens is spectral. Realize a finite discrete space by a simpler ring than the general construction.

Solution 7.1. In a Hausdorff space a nonempty irreducible closed subset cannot contain two distinct points: disjoint neighborhoods would both meet it. It is therefore a singleton, giving sobriety. Closed subsets of a compact Hausdorff space are compact, so the clopen basis is quasi-compact and stable under intersections. A discrete space with \(n\) points is \(\operatorname{Spec}(\mathbb Q^n)\): its primes are the kernels of the coordinate maps, as follows from the orthogonal idempotents. For \(n=0\), use the zero ring.

Exercise 7.2 (medium). In the two-point example, show why adjoining the function which is \(t^{-1}\) at \(\eta\) and zero at \(s\) is forbidden. Determine the nonzero set of one minus its product with \(e_{\{\eta\}}\).

Solution 7.2. The putative inverse has valuation \(-1\) along \(\eta\rightsquigarrow s\), violating nonnegativity in (1). Its product with \(e_{\{\eta\}}\) is one at \(\eta\) and zero at \(s\). One minus that product has nonzero set \(\{s\}\), which is not open. Thus uncontrolled division really breaks the required topology.

Exercise 7.3 (hard). Let \(X\) be an infinite set with the discrete topology. Explain exactly which hypothesis fails and why no ring can have this space as its spectrum.

Solution 7.3. Each singleton is quasi-compact open, intersections remain quasi-compact, and the space is sober, since its irreducible closed subsets are singletons. The cover by all singletons has no finite subcover, so \(X\) is not quasi-compact. Every ring spectrum is quasi-compact by Proposition 2.3 of Spectra of rings. A homeomorphism would preserve this property, so no realization exists. This distinguishes unrestricted cardinality in the theorem from the omission of a required topological hypothesis.

Sources and proof scope

Melvin Hochster, Prime ideal structure in commutative rings, Transactions of the American Mathematical Society 142 (1969), 43–60, original article, DOI, §§2–7, supplies the realization method. The proof here uses its mathematical ideas with independently written exposition; it restricts the auxiliary ground field to \(\mathbb Q\), which places no restriction on the realized space, replaces finite prime avoidance by finite-value cancellation, and supplies the valuation-preservation checks. No source prose or images are reproduced.

The spectral-space definition and geometric context are also recorded in the AI Integrated Stacks Project topology chapter, with credit to the Stacks Project Authors and Hochster. That introductory passage alone is not a proof of the converse. Every step of the converse needed here is proved in this lesson; its exact internal forward provider is Spectra of rings, Solution 7.5.