Affine Scheme

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

The most basic example of a sheaf on a topological space is the sheaf of continuous functions, and the sheaf $\mathcal{O}_{\Spec A}$ we are about to define is analogous: the only difference is that we use regular functions in place of continuous ones.

Locally ringed space

Sheaves on topological spaces were already treated in [Topology] §Sheaves, but that definition is somewhat inadequate for describing the structure sheaf on $\Spec A$.

Definition 1 A pair $(X,\mathcal{O}_X)$ consisting of a topological space $X$ and a $\cRing$-valued sheaf $\mathcal{O}_X$ on it is called a ringed space. If for every point $x\in X$, the stalk $\mathcal{O}_{X,x}$ at $x$ is a local ring, then this pair $(X, \mathcal{O}_X)$ is called a locally ringed space.

Our claim is that we can define a suitable structure sheaf $\mathcal{O}_{\Spec A}$ on $\Spec A$ so that $(\Spec A, \mathcal{O}_{\Spec A})$ becomes a locally ringed space, and that this $\Spec$ construction enjoys the same functoriality as in [§Spectrums, ⁋Proposition 2] or [§Spectrums, ⁋Proposition 8]. To state this precisely, we first define morphisms between locally ringed spaces.

Definition 2 For two ringed spaces $(X, \mathcal{O}_X)$ and $(Y, \mathcal{O}_Y)$, a morphism between them is a pair consisting of a continuous map $\varphi:X \rightarrow Y$ and a morphism $\varphi^\sharp:\mathcal{O}_Y \rightarrow \varphi_\ast \mathcal{O}_X$ in $\Sh(Y;\cRing)$.

A morphism between two locally ringed spaces $(X, \mathcal{O}_X)$ and $(Y, \mathcal{O}_Y)$ is a morphism $(\varphi,\varphi^\sharp)$ of ringed spaces such that, for each $x\in X$, the induced map on stalks $\varphi_x^\sharp:\mathcal{O}_{Y,\varphi(x)} \rightarrow \mathcal{O}_{X,x}$ is a local homomorphism.

Algebraic functions on $\Spec A$

We now define $\mathcal{O}_{\Spec A}$. As mentioned at the outset, this is the sheaf of algebraic functions on $\Spec A$, and it is precisely the generalization of [[Algebraic Varieties] §Affine Varieties, ⁋Definition 14]](/en/math/algebraic_varieties/affine_varieties#def14).

Let us carry this discussion over to schemes. First, as with algebraic varieties, we regard elements of $A$ as functions $f$. Then the function value of $f$ at a point $\mathfrak{p}\in\Spec A$ is the image of $f$ under the canonical projection $\pi: A \rightarrow A/\mathfrak{p}$. In particular, $f$ vanishing at $\mathfrak{p}$ means

\[f\equiv 0\pmod{\mathfrak{p}}\iff f\in \mathfrak{p}\iff \mathfrak{p}\in Z(f)\]

Thus $Z(f)$ can be understood as the locus where $f=0$, and its complement, the principal open set $D(f)$, as the locus where $f\neq 0$.

From this perspective we can describe what the algebraic functions on $\Spec A$ are. Just as in [[Algebraic Varieties] §Affine Varieties, ⁋Definition 14]](/en/math/algebraic_varieties/affine_varieties#def14), they are defined to be functions that can be represented, in a suitable neighborhood of each point, as rational functions whose denominators do not vanish on that neighborhood. Note that this condition is local: a single fractional expression valid on the entire open set does not generally exist, and as we shall see below, this is guaranteed only on principal open sets.

Now suppose a principal open set $D(f)$ is given. If we consider only functions representable as a single rational function $g/h$ on all of $D(f)$, then the admissible denominators $h$ must satisfy $D(f)\subseteq D(h)$.

Lemma 3 For a fixed element $f\in A$, define

\[S(f)=\{h\in A\mid D(f)\subseteq D(h)\}\]

Then $S(f)$ is a multiplicative subset of $A$.

Proof

Since $D(1)=\Spec A$, it is obvious that $S(f)$ contains the empty product $1$. Now if $h_1,h_2\in S(f)$, then from the identity

\[D(h_1h_2)=\Spec A\setminus Z(h_1h_2)=\Spec A\setminus (Z(h_1)\cup Z(h_2))=(\Spec A\setminus Z(h_1))\cap (\Spec A\setminus Z(h_2))=D(h_1)\cap D(h_2)\]

we see that $D(f)\subseteq D(h_1)\cap D(h_2)=D(h_1h_2)$. This identity is merely a geometric interpretation of [[Algebraic Structures] §Field of Fractions, ⁋Proposition 9]](/en/math/algebraic_structures/field_of_fractions#prop9).

It is now intuitively clear, and indeed we shall define it so, that the collection of algebraic functions on the subset $D(f)$ of $\Spec A$ should be $S(f)^{-1}A$. Before doing so, we prove the following lemma.

Lemma 4 $D(f)\subseteq D(h)$ holds if and only if there exists $n\geq 1$ such that $f^n\in (h)$.

Proof

$D(f)\subseteq D(h)$ is equivalent to $Z(h)\subseteq Z(f)$, which by the third result of [§Spectrums, ⁋Lemma 6] is equivalent to $\sqrt{(f)}\subseteq \sqrt{(h)}$.

If $\sqrt{(f)}\subseteq \sqrt{(h)}$, then from $(f)\subseteq \sqrt{(f)}\subseteq \sqrt{(h)}$ we get $f\in \sqrt{(h)}$, so there exists $n\geq 1$ with $f^n\in (h)$. Conversely, if $f^n\in (h)$ for some $n\geq 1$, then $f\in \sqrt{(h)}$, whence $(f)\subseteq \sqrt{(h)}$, and therefore

\[\sqrt{(f)}\subseteq\sqrt{\sqrt{(h)}}=\sqrt{(h)}\]

Using this lemma, we can express $S(f)^{-1}A$ in a cleaner form.

Lemma 5 For any $f\in A$, there is an isomorphism

\[S(f)^{-1}A\cong S_f^{-1}A\]

Moreover, if $S(g)\subseteq S(f)$, then the following diagram

commutes.

Proof

Write the canonical morphisms as $\epsilon(f): A \rightarrow S(f)^{-1}A$ and $\epsilon_f:A \rightarrow S_f^{-1}A$. Since $D(f)=D(f^n)$ for any $n\geq 1$, we have $S_f\subseteq S(f)$, so the image of $S_f$ under $\epsilon(f)$ consists entirely of units in $S(f)^{-1}A$. Conversely, for any $h\in S(f)$, Lemma 4 gives $n\geq 1$ and $a\in A$ with $f^n=ah$, so

\[\frac{h}{1}\frac{a}{f^n}=1\qquad\text{in $S_f^{-1}A$}\]

and thus the image of $S(f)$ under $\epsilon_f$ also consists entirely of units. Therefore, by [[Commutative Algebra] §Localization, ⁋Proposition 6]](/en/math/commutative_algebra/localization#prop6), there exist unique maps

\[\overline{\epsilon_f}: S(f)^{-1}A \rightarrow S_f^{-1}A,\qquad \overline{\epsilon(f)}: S_f^{-1}A \rightarrow S(f)^{-1}A\]

satisfying $\overline{\epsilon_f}\circ\epsilon(f)=\epsilon_f$ and $\overline{\epsilon(f)}\circ\epsilon_f=\epsilon(f)$. Then the two composites $\overline{\epsilon(f)}\circ\overline{\epsilon_f}$ and $\overline{\epsilon_f}\circ\overline{\epsilon(f)}$ each extend $\epsilon(f)$ and $\epsilon_f$, so by the same uniqueness they are identity maps. That is, they are inverses of each other, yielding the claimed isomorphism.

Now suppose $S(g)\subseteq S(f)$. Then elements of $S(g)$ are sent to units by both $\epsilon(f)$ and $\epsilon_f$, and since $S_g\subseteq S(g)$, again by [[Commutative Algebra] §Localization, ⁋Proposition 6]](/en/math/commutative_algebra/localization#prop6) there exist unique maps

\[\widehat{\epsilon(f)}:S(g)^{-1}A \rightarrow S(f)^{-1}A,\qquad \widecheck{\epsilon_f}: S_g^{-1}A \rightarrow S_f^{-1}A\]

extending $\epsilon(f)$ and $\epsilon_f$ respectively, and these are the remaining two arrows of the claimed diagram. Now both composites $\widecheck{\epsilon_f}\circ\overline{\epsilon_g}$ and $\overline{\epsilon_f}\circ\widehat{\epsilon(f)}$ compose with $\epsilon(g)$ to give $\epsilon_f$, so by the same uniqueness they are equal.

Thus, it suffices to regard algebraic functions on $D(f)$ as elements of $S_f^{-1}A$. For convenience, in the previous post we agreed to denote $S_f^{-1}A$ by $A_f$.

Lemma 6 For the base $\{D(f)\}_{f\in A}$ of $\Spec A$, define for each $f_i\in A$

\[\mathcal{F}(D(f_i))=S(f_i)^{-1}A\cong A_{f_i}\]

Also, for each $f_i,f_j\in A$ with $D(f_i)\subseteq D(f_j)$, define the restriction map

\[\rho_{ji}: S(f_j)^{-1}(A) \rightarrow S(f_i)^{-1}(A)\]

to be the map obtained by applying [[Commutative Algebra] §Localization, ⁋Proposition 6]](/en/math/commutative_algebra/localization#prop6) to the canonical morphism $A\rightarrow S(f_i)^{-1}(A)$. Then these data satisfy the two conditions of [[Topology] §Sheaves, ⁋Proposition 8]](/en/math/topology/sheaves#prop8), and therefore determine uniquely a ($\cRing$-valued) sheaf $\mathcal{F}$ on $\Spec A$ extending this assignment.

Proof

That the $\rho_{ji}$ satisfy the conditions for restriction maps in [[Topology] §Presheaves, ⁋Definition 2]](/en/math/topology/presheaves#def2) is immediate from the universal property of [[Commutative Algebra] §Localization, ⁋Proposition 6]](/en/math/commutative_algebra/localization#prop6). Here, $\rho_{ji}: S(f_j)^{-1}(A) \rightarrow S(f_i)^{-1}(A)$ is, by Lemma 5, simply the map that regards an element of $S(f_j)^{-1}(A)$ written in the form

\[g/h,\qquad\text{where $h\in S(f_j)$}\tag{$\ast$}\]

as an element of $S(f_i)^{-1}(A)$ via the implication

\[h\in S(f_j)\iff D(f_j)\subseteq D(h)\implies D(f_i)\subseteq D(h)\iff h\in S(f_i)\]

We now verify the two conditions of [[Topology] §Sheaves, ⁋Proposition 8]](/en/math/topology/sheaves#prop8). For notational convenience, since $D(f)=\Spec A_f$, it suffices to consider only the case $f=1$ after replacing $A$ by $A_f$. Fix $f_i\in A$ with $\Spec A=\bigcup_{i\in I}D(f_i)$.

First, to verify the first condition, suppose an element $s\in A$ satisfies $s=0$ in $S(f_i)^{-1}A$ for all $i\in I$, and let us show that $s=0$ in $A$. By [§Spectrums, ⁋Lemma 12] we can choose $f_1,\ldots, f_n$ from among the $f_i$ such that $\Spec A=\bigcup_{i=1}^n D(f_i)$, and by assumption there exist $m_i\geq 1$ such that

\[f_i^{m_i}s=0\]

for all $i=1,\ldots, n$. On the other hand, from the calculation after [§Spectrums, ⁋Lemma 11] we have $D(f_i^{m_i})=D(f_i)$ for all $i$, so

\[\Spec A=\bigcup_{i=1}^n D(f_i^{m_i})\]

and hence there exist $a_i\in A$ with $1=\sum_{i=1}^n a_i f_i^{m_i}$. (See the proof of [§Spectrums, ⁋Lemma 12], or the proof of [[Commutative Algebra] §Integral Extension, ⁋Proposition 15]](/en/math/commutative_algebra/integral_extension#prop15).)

Therefore

\[s=1s=\left(\sum_{i=1}^n a_i f_i^{m_i}\right)s=\sum_{i=1}^n a_i (f_i^{m_i}s)=0\]

Now to verify the second condition, suppose for each $i$ there exists $s_i=a_i/f_i^{m_i}\in S(f_i)^{-1}A$ such that for each $i,j$

\[\frac{a_i}{f_i^{m_i}}=\frac{a_j}{f_j^{m_j}}\quad\text{ in }D(f_i)\cap D(f_j)=D(f_if_j)\]

Here, since $a/1=af_i/f_i$, we may choose $m_i\geq 1$ for all $i$. Since $D(f_i)=D(f_i^{m_i})$ and $D(f_j)=D(f_j^{m_j})$,

\[D(f_if_j)=D(f_i)\cap D(f_j)=D(f_i^{m_i})\cap D(f_j^{m_j})=D(f_i^{m_i}f_j^{m_j})\]

so there exists $N_{ij}$ such that

\[(f_i^{m_i}f_j^{m_j})^{N_{ij}}(a_if_j^{m_j}-a_jf_i^{m_i})=0\]

Let $N=\max_{i,j}\{N_{ij}\}$ and obtain

\[(f_i^{m_i}f_j^{m_j})^N(a_if_j^{m_j}-a_jf_i^{m_i})=0\]

that is,

\[a_if_i^{Nm_i}f_j^{Nm_j+m_j}=a_jf_j^{Nm_j}f_i^{Nm_i+m_i}\]

From the given assumption

\[\Spec A=\bigcup_{i=1}^n D(f_i)=\bigcup_{i=1}^n D(f_i^{Nm_i+m_i})\]

we can choose suitable $b_i\in A$ such that

\[1=\sum_{i=1}^n b_if_i^{Nm_i+m_i}\]

Now set $s=\sum_{i=1}^n b_ia_i f_i^{Nm_i}$; then

\[sf_j^{Nm_j+m_j}=\sum_{i=1}^n b_ia_i f_i^{Nm_i} f_j^{Nm_j+m_j}=\sum_{i=1}^nb_ia_jf_j^{Nm_j}f_i^{Nm_i+m_i}=a_jf_j^{Nm_j}\]

so $f_j^{Nm_j}(sf_j^{m_j}-a_j)=0$ holds for all $j$, and hence on $D(f_j)$

\[\frac{s}{1}=\frac{a_j}{f_j^{m_j}}\]

This gives the desired $s$.

If $I$ is infinite, choose a finite subset $J=\{1,\ldots, n\}$ of $I$ with $\Spec A=\bigcup_{j\in J} D(f_j)$ and repeat the above to obtain $s\in \mathcal{F}(\Spec A)$; then we need only show that this also satisfies $s_\alpha=s\vert_{D(f_\alpha)}$ for $D(f_\alpha)$ with $\alpha\in I\setminus J$. To see this, repeat the same process for the finite set

\[J\cup\{\alpha\}=\{1,2,\ldots, n,\alpha\}\subseteq I\]

to obtain $s'\in \mathcal{F}(\Spec A)$. Then by construction $s$ and $s'$ satisfy $s\vert_{D(f_i)}=s'\vert_{D(f_i)}$ for $i=1,\ldots, n$, and since $\Spec A=\bigcup D(f_i)$, the first condition of [[Topology] §Sheaves, ⁋Proposition 8]](/en/math/topology/sheaves#prop8) shown above gives $s=s'$, whence

\[s\vert_{D(f_\alpha)}=s'\vert_{D(f_\alpha)}=s_\alpha\]

This holds for every $\alpha$, so $s$ restricts to $s_\alpha$ on any $D(f_\alpha)$.

Definition 7 The sheaf on $\Spec A$ defined by Lemma 6 is denoted $\mathcal{O}_{\Spec A}$ and called the structure sheaf.

Although this definition was made only on principal open sets, since the $D(f)$ form a base for $\Spec A$, sections over arbitrary open sets $U$ are determined by the extension in [[Topology] §Sheaves, ⁋Proposition 8]](/en/math/topology/sheaves#prop8). That is, an element of $\mathcal{O}_{\Spec A}(U)$ is data given on $D(f)$ covering $U$ in the form $g/h$ that agree on intersections, which corresponds to the local definition mentioned earlier.

Then $(\Spec A,\mathcal{O}_{\Spec A})$ is a locally ringed space.

Lemma 8 For $(\Spec A,\mathcal{O}_{\Spec A})$ and any point $\mathfrak{p}\in \Spec A$, there is an isomorphism

\[A_\mathfrak{p}\cong \mathcal{O}_{\Spec A, \mathfrak{p}}=\varinjlim_\text{\scriptsize $U\ni\mathfrak{p}$ open} \mathcal{O}_{\Spec A}(U)\]

Moreover, for any $f\in A$ with $\mathfrak{p}\in D(f)$, the following diagram

commutes.

Proof

By [[Topology] §Topological Bases, ⁋Proposition 2]](/en/math/topology/topological_bases#prop2), the $D(f)$ form a base for $\Spec A$, so by [[Topology] §Topological Bases, ⁋Proposition 5]](/en/math/topology/topological_bases#prop5),

\[\mathcal{O}_{\Spec A, \mathfrak{p}}=\varinjlim_{D(f)\ni\mathfrak{p}} \mathcal{O}_{\Spec A}(D(f))\]

On the other hand, $\mathfrak{p}\in D(f)\iff f\not\in \mathfrak{p}$, so we obtain the following diagram

and thus proving the given isomorphism reduces to showing the algebraic isomorphism

\[A_\mathfrak{p}\cong \varinjlim_{\mathfrak{p}\not\ni f} A_f\tag{$\ast\ast$}\]

which follows from using the universal property of [[Commutative Algebra] §Localization, ⁋Proposition 6]](/en/math/commutative_algebra/localization#prop6) and the universal property of direct limits. The diagram in the claim is obtained by replacing $\varinjlim A_f$ with $A_\mathfrak{p}$ in the above diagram via the isomorphism ($\ast\ast$).

We are now finally ready to write the functoriality of $\Spec$ in the form we want.

Proposition 9 The correspondence $A\mapsto (\Spec A, \mathcal{O}_{\Spec A})$ defines a contravariant functor $\Spec: \cRing^\op \rightarrow \LRS$.

Proof

We already know that a ring homomorphism $\phi: A \rightarrow B$ induces a continuous map $\Spec\phi: \Spec B \rightarrow \Spec A$. ([§Spectrums, ⁋Proposition 8]) Thus it suffices to describe

\[(\Spec\phi)^\sharp: \mathcal{O}_{\Spec A} \rightarrow (\Spec\phi)_\ast \mathcal{O}_{\Spec B}\]

For this we look at the functions on principal open sets

\[(\Spec\phi)^\sharp(D(f)): \mathcal{O}_{\Spec A}(D(f)) \rightarrow \mathcal{O}_{\Spec B}((\Spec \phi)^{-1}(D(f)))\]

Now from the proof of [§Spectrums, ⁋Proposition 8],

\[(\Spec\phi)^{-1}(Z(f))=Z(\phi(f))\]

\[(\Spec\phi)^{-1}(D(f))=D(\phi(f))\]

Therefore, by the definition of the structure sheaf, defining $(\Spec\phi)^\sharp(D(f))$ is the same as defining

\[A_f \rightarrow B_{\phi(f)}\]

and this is obtained by applying [[Commutative Algebra] §Localization, ⁋Proposition 6]](/en/math/commutative_algebra/localization#prop6) to the composite

\[A \overset{\phi}{\longrightarrow}B \overset{\epsilon}{\longrightarrow} B_{\phi(f)}\]

Of course we must show that these maps commute with restriction maps. If $D(g)\subseteq D(f)$ then $D(\phi(g))\subseteq D(\phi(f))$, and the two composites

\[A_f \longrightarrow B_{\phi(f)} \longrightarrow B_{\phi(g)},\qquad A_f \longrightarrow A_g \longrightarrow B_{\phi(g)}\]

both compose with $\epsilon_f$ to give $A \rightarrow B \rightarrow B_{\phi(g)}$, so by the uniqueness above they are equal.

Now for an arbitrary open set $U\subseteq \Spec A$ and $s\in \mathcal{O}_{\Spec A}(U)$, consider the sections $(\Spec\phi)^\sharp(D(f))(s\vert_{D(f)})$ for each principal open set $D(f)$ contained in $U$. Since the intersection $D(f)\cap D(g)=D(fg)$ of two principal open sets is again principal, the commutativity above shows these sections agree on intersections, and hence by the gluability and identity axioms of the sheaf $(\Spec\phi)_\ast\mathcal{O}_{\Spec B}$ they glue to a unique $(\Spec\phi)^\sharp(U)(s)$. That this $(\Spec\phi)^\sharp(U)$ is a ring homomorphism commuting with restriction maps also follows from the identity axiom, and thus we obtain the sheaf morphism $(\Spec\phi)^\sharp$.

From the above, $(\Spec\phi, (\Spec\phi)^\sharp): (\Spec B, \mathcal{O}_{\Spec B}) \rightarrow (\Spec A, \mathcal{O}_{\Spec A})$ is a morphism of ringed spaces. To show this is a morphism of locally ringed spaces, we need that for any $\mathfrak{q}\in \Spec B$,

\[(\Spec\phi)^\sharp_\mathfrak{q}:\mathcal{O}_{\Spec A, (\Spec \phi)(\mathfrak{q})} \rightarrow\mathcal{O}_{\Spec B, \mathfrak{q}}\]

is a local homomorphism. But $(\Spec \phi)(\mathfrak{q})=\phi^{-1}(\mathfrak{q})$, so by Lemma 8, $(\Spec\phi)^\sharp_\mathfrak{q}$ is a ring homomorphism from $A_{\phi^{-1}(\mathfrak{q})}$ to $B_{\mathfrak{q}}$ sending the unique maximal ideal $\phi^{-1}(\mathfrak{q})A_{\phi^{-1}(\mathfrak{q})}$ of $A_{\phi^{-1}(\mathfrak{q})}$ to the unique maximal ideal $\mathfrak{q}B_\mathfrak{q}$ of $B_\mathfrak{q}$.

Finally, let us verify functoriality. For the map on points this was already checked in [§Spectrums, ⁋Proposition 2], so we only need to check the structure sheaf side. When $\phi=\id_A$, the above construction gives, for each $D(f)$, the unique map $A_f \rightarrow A_f$ extending $\epsilon_f$, which is the identity; hence $\Spec(\id_A)=\id$. Also, for two ring homomorphisms $\phi: A \rightarrow B$ and $\psi: B \rightarrow C$, the map $\Spec(\psi\circ\phi)^\sharp(D(f))$ is the unique map extending the composite $A \rightarrow C \rightarrow C_{\psi(\phi(f))}$, and the composite $A_f \rightarrow B_{\phi(f)} \rightarrow C_{\psi(\phi(f))}$ also extends the same map, so by uniqueness they are equal. Since two sheaf morphisms agreeing on a base are equal, $\Spec(\psi\circ\phi)=(\Spec\phi)\circ(\Spec\psi)$.

Affine scheme

Definition 10 The essential image of the functor $\Spec:\cRing^\op \rightarrow \LRS$ from Proposition 9 is called an affine scheme.

We write $\AffSch$ for the category of affine schemes. Then the contravariant functor $\Spec:\cRing^\op \rightarrow \AffSch$ is essentially surjective by definition. ([[Category Theory] §Natural Transformations, ⁋Theorem 5]](/en/math/category_theory/natural_transformations#thm5)) Moreover, if $(\varphi, \varphi^\sharp): (\Spec B, \mathcal{O}_{\Spec B}) \rightarrow (\Spec A, \mathcal{O}_{\Spec A})$ is induced from some ring homomorphism $\phi$, then taking $1=f\in A$ in the proof of Proposition 9 gives

\[\varphi^\sharp(D(1))= \bigl(A \overset{\phi}{\longrightarrow} B \overset{\id_B}{\longrightarrow} B_{\phi(1)}=B\bigr)=\phi\]

so this functor is necessarily faithful. Furthermore, the following holds.

Proposition 11 The functor $\Spec: \cRing^\op \rightarrow \LRS$ is fully faithful.

Proof

Suppose given any two affine schemes $(X, \mathcal{O}_{X})$, $(Y, \mathcal{O}_{Y})$ and a morphism

\[(X, \mathcal{O}_{X}) \rightarrow (Y, \mathcal{O}_{Y})\]

between them. Via isomorphisms $(\Spec B, \mathcal{O}_{\Spec B})\cong (X, \mathcal{O}_X)$ and $(\Spec A, \mathcal{O}_{\Spec A})\cong (Y, \mathcal{O}_Y)$, we can view this as a morphism

\[(\varphi, \varphi^\sharp): (\Spec B, \mathcal{O}_{\Spec B}) \rightarrow (\Spec A, \mathcal{O}_{\Spec A})\]

between spectra (as locally ringed spaces). Thus it suffices to prove that this morphism of locally ringed spaces comes from some ring homomorphism $\phi$. Taking a hint from the proof of faithfulness above, define a ring homomorphism $\phi:A \rightarrow B$ by

\[\phi=\varphi^\sharp(D(1)):A \rightarrow B\]

To complete the claim we must show $\Spec\phi=(\varphi,\varphi^\sharp)$. First, for any $\mathfrak{q}\in \Spec B$ we show

\[(\Spec \phi)(\mathfrak{q})=\phi^{-1}(\mathfrak{q})=\varphi(\mathfrak{q})\]

Setting $f=1$ in Lemma 8, we obtain the following diagram

In this diagram, the vertical maps are all isomorphisms, and we know that all faces except the following one

are commuting squares. Therefore, in the above diagram $A \rightarrow \mathcal{O}_{\Spec B, \mathfrak{q}}$ is determined identically no matter which path we take, and applying [[Commutative Algebra] §Localization, ⁋Proposition 6]](/en/math/commutative_algebra/localization#prop6) to this map determines $A_{\varphi(\mathfrak{q})} \rightarrow \mathcal{O}_{\Spec B, \mathfrak{q}}$ uniquely. From this we see that all faces of the above diagram are commuting squares. That is, $\phi_\mathfrak{q}:A_{\varphi(\mathfrak{q})}\rightarrow B_\mathfrak{q}$ is also a local homomorphism, and hence $\phi^{-1}(\mathfrak{q})=\varphi(\mathfrak{q})$.

Now we show the two sheaf morphisms are equal. Since $\varphi^\sharp$ commutes with restriction maps, for any $f\in A$

\[\varphi^\sharp(D(f))\circ\epsilon_f=\epsilon_{\phi(f)}\circ\phi\]

But in the construction of Proposition 9, $(\Spec\phi)^\sharp(D(f))$ was exactly the unique map satisfying this equation, so by the uniqueness in [[Commutative Algebra] §Localization, ⁋Proposition 6]](/en/math/commutative_algebra/localization#prop6), $\varphi^\sharp(D(f))=(\Spec\phi)^\sharp(D(f))$. Since two sheaf morphisms agreeing on the base $\{D(f)\}_{f\in A}$ are equal, by the identity axiom for $\varphi_\ast\mathcal{O}_{\Spec B}$ we have $\varphi^\sharp=(\Spec\phi)^\sharp$.

Thus, viewing $\Spec$ as a contravariant functor from $\cRing$ to $\AffSch$, it is a categorical equivalence between $\cRing^\op$ and $\AffSch$. Moreover, by Proposition 11, $\AffSch$ is a full subcategory of $\LRS$.

On the other hand, for any spectrum $(\Spec A, \mathcal{O}_{\Spec A})$, by definition we know

\[\mathcal{O}_{\Spec A}(\Spec A)=\mathcal{O}_{\Spec A}(D(1))\cong A\]

If a locally ringed space $(X, \mathcal{O}_X)$ were an affine scheme, we could similarly examine $\mathcal{O}_X(X)$ to see which ring’s spectrum $(X, \mathcal{O}_X)$ is isomorphic to. That is, for an affine scheme $(X, \mathcal{O}_X)$, setting $A=\mathcal{O}_X(X)$ gives $(X, \mathcal{O}_X)\cong (\Spec A, \mathcal{O}_{\Spec A})$. More generally, we define:

Definition 12 The global section functor $\Gamma:\LRS \rightarrow \cRing^\op$ is defined, for any locally ringed space $(X, \mathcal{O}_X)$, by the correspondence

\[X\mapsto \Gamma(X, \mathcal{O}_X)=\mathcal{O}_X(X)\]

and for any morphism $(\varphi,\varphi^\sharp):(X,\mathcal{O}_X) \rightarrow (Y, \mathcal{O}_Y)$, by

\[\Gamma(\varphi,\varphi^\sharp)=\varphi^\sharp(Y):\Gamma(Y, \mathcal{O}_Y) \rightarrow (\varphi_\ast\mathcal{O}_X)(Y)=\Gamma(X, \mathcal{O}_X)\]

Here, $\varphi^\sharp(Y)$ is a map in $\cRing$ from $\Gamma(Y)$ to $\Gamma(X)$, which we can read as a map $\Gamma(X) \rightarrow \Gamma(Y)$ in $\cRing^\op$. That this correspondence preserves identities and composition follows from the fact that composition of morphisms of ringed spaces is given by composition of the $\varphi^\sharp$, so this is indeed a functor.

Meanwhile, a notable fact from the proof of Proposition 11 is that the assumption that $(X, \mathcal{O}_X)$ is an affine scheme was unnecessary. That is, even if we drop the assumption $(X, \mathcal{O}_X)\cong(\Spec B, \mathcal{O}_{\Spec B})$ and use the following diagram instead of the one in Proposition 11

we can carry out a similar argument, where the conclusion’s $B$ is replaced by $\Gamma(X, \mathcal{O}_X)$. Since $\mathcal{O}_X$ is data determined by $X$ anyway, abbreviating this as $\Gamma(X)$, we obtain the following theorem.

Theorem 13 For any locally ringed space $(X, \mathcal{O}_X)$ and any ring $A$, there exists a natural isomorphism

\[\Hom_\LRS(X, \Spec A)\cong \Hom_{\cRing^\op}(\Gamma(X), A)=\Hom_{\cRing}(A, \Gamma(X))\]

That is, the global section functor $\Gamma: \LRS \rightarrow \cRing^\op$ is the left adjoint of the $\Spec$ functor $\Spec:\cRing^\op \rightarrow \LRS$.

Proof

Through the isomorphism $\mathcal{O}_{\Spec A}(D(f))\cong A_f$ from Lemma 6, we identify $\mathcal{O}_{\Spec A}(\Spec A)=\mathcal{O}_{\Spec A}(D(1))$ with $A$. Under this identification, the restriction map $\mathcal{O}_{\Spec A}(\Spec A) \rightarrow \mathcal{O}_{\Spec A}(D(f))$ of $\mathcal{O}_{\Spec A}$ is the canonical morphism $\epsilon_f: A \rightarrow A_f$.

First define two correspondences $\Phi$ and $\Psi$, then show they are inverses of each other. Given a morphism $(\varphi,\varphi^\sharp): X \rightarrow \Spec A$ of locally ringed spaces, evaluating $\varphi^\sharp$ on the open set $\Spec A$ yields the ring homomorphism

\[\Phi(\varphi,\varphi^\sharp)=\varphi^\sharp(\Spec A): A=\mathcal{O}_{\Spec A}(\Spec A) \rightarrow (\varphi_\ast\mathcal{O}_X)(\Spec A)=\mathcal{O}_X(X)=\Gamma(X)\]

Conversely, suppose a ring homomorphism $\phi:A \rightarrow \Gamma(X)$ is given. For each $x\in X$, write $\phi_x:A \rightarrow \mathcal{O}_{X,x}$ for the ring homomorphism obtained by composing $\phi$ with the germ map $\Gamma(X) \rightarrow \mathcal{O}_{X,x}$. That is, $\phi_x(a)=\phi(a)_x$. Since $(X,\mathcal{O}_X)$ is a locally ringed space, $\mathcal{O}_{X,x}$ is a local ring with unique maximal ideal $\mathfrak{m}_x$, so by [[Algebraic Structures] §Field of Fractions, ⁋Proposition 10]](/en/math/algebraic_structures/field_of_fractions#prop10),

\[\varphi(x)=\phi_x^{-1}(\mathfrak{m}_x)\]

is a prime ideal of $A$, i.e., a point of $\Spec A$.

We show this function $\varphi: X \rightarrow \Spec A$ is continuous. To do this, we first show that for any $s\in \Gamma(X)$,

\[X_s=\{x\in X\mid \text{$s_x\not\in \mathfrak{m}_x$}\}\]

is an open set of $X$. By [[Commutative Algebra] §Localization, ⁋Proposition 2]](/en/math/commutative_algebra/localization#prop2), $\mathfrak{m}_x$ consists of all non-units of the local ring $\mathcal{O}_{X,x}$, so $x\in X_s$ is equivalent to $s_x$ being a unit in $\mathcal{O}_{X,x}$. Now if $x\in X_s$, there exists $t\in \mathcal{O}_{X,x}$ with $s_xt=1$, and choosing a suitable open neighborhood $W$ of $x$ and a section $u\in \mathcal{O}_X(W)$ representing $t$, we have $(s\vert_Wu)_x=1_x$, so by shrinking $W$ if necessary we can ensure $s\vert_Wu=1$ in $\mathcal{O}_X(W)$. Then for any $y\in W$, $s_yu_y=1$ so $s_y$ is a unit in $\mathcal{O}_{X,y}$, and hence $W\subseteq X_s$. That is, $X_s$ is open.

On the other hand, for any $f\in A$,

\[\varphi^{-1}(D(f))=\{x\in X\mid f\not\in \varphi(x)\}=\{x\in X\mid \phi(f)_x\not\in \mathfrak{m}_x\}=X_{\phi(f)}\]

and since principal open sets form a base for $\Spec A$ ([§Spectrums, ⁋Lemma 11]), $\varphi$ is continuous.

Now we define the sheaf morphism $\varphi^\sharp: \mathcal{O}_{\Spec A} \rightarrow \varphi_\ast \mathcal{O}_X$. For each $f\in A$, set $V_f=\varphi^{-1}(D(f))=X_{\phi(f)}$, and write

\[\theta_f: A\overset{\phi}{\longrightarrow} \Gamma(X) \longrightarrow \mathcal{O}_X(V_f)\]

for the ring homomorphism obtained by composing $\phi$ with the restriction map. Our claim is that $\theta_f(f)=\phi(f)\vert_{V_f}$ is a unit in $\mathcal{O}_X(V_f)$. Indeed, by the definition of $V_f$, for every $y\in V_f$ the element $\phi(f)_y$ is a unit in $\mathcal{O}_{X,y}$, so repeating the argument above we can find an open neighborhood $W_y\subseteq V_f$ of $y$ and $u_y\in \mathcal{O}_X(W_y)$ with $\phi(f)\vert_{W_y}u_y=1$. Then on the intersection $W_y\cap W_{y'}$, the restrictions of $u_y$ and $u_{y'}$ are both multiplicative inverses of $\phi(f)\vert_{W_y\cap W_{y'}}$ and hence equal, so by the gluability axiom of [[Topology] §Sheaves, ⁋Definition 1]](/en/math/topology/sheaves#def1) they glue to a single $u\in \mathcal{O}_X(V_f)$. Now $\phi(f)\vert_{V_f}u$ and $1$ agree on each $W_y$, so by the identity axiom $\phi(f)\vert_{V_f}u=1$.

In particular, $\theta_f$ sends all elements of the multiplicative subset $S_f=\{1,f,f^2,\ldots\}$ to units in $\mathcal{O}_X(V_f)$, so by [[Commutative Algebra] §Localization, ⁋Proposition 6]](/en/math/commutative_algebra/localization#prop6), there exists a unique ring homomorphism

\[\varphi^\sharp(D(f)): A_f=\mathcal{O}_{\Spec A}(D(f)) \rightarrow \mathcal{O}_X(V_f)=(\varphi_\ast\mathcal{O}_X)(D(f))\]

satisfying

\[\varphi^\sharp(D(f))\circ \epsilon_f=\theta_f\]

If $D(g)\subseteq D(f)$ then $V_g\subseteq V_f$, and the two composites

\[A_f \overset{\varphi^\sharp(D(f))}{\longrightarrow} \mathcal{O}_X(V_f) \longrightarrow \mathcal{O}_X(V_g),\qquad A_f \longrightarrow A_g \overset{\varphi^\sharp(D(g))}{\longrightarrow} \mathcal{O}_X(V_g)\]

both compose with $\epsilon_f$ to give $\theta_g$, so again by the uniqueness in [[Commutative Algebra] §Localization, ⁋Proposition 6]](/en/math/commutative_algebra/localization#prop6) they are equal. That is, the $\varphi^\sharp(D(f))$ commute with restriction maps.

Then, just as in the proof of Proposition 9, the maps given on the base extend uniquely to a sheaf morphism $\varphi^\sharp$ by the gluability and identity axioms of $\varphi_\ast\mathcal{O}_X$.

Finally, we show $(\varphi,\varphi^\sharp)$ is a morphism of locally ringed spaces. For any $x\in X$ and $\mathfrak{p}=\varphi(x)$, by Lemma 8 we have $\mathcal{O}_{\Spec A,\mathfrak{p}}\cong A_\mathfrak{p}$, and under this identification the stalk morphism $\varphi_x^\sharp: A_\mathfrak{p} \rightarrow \mathcal{O}_{X,x}$ induced by $\varphi^\sharp$ satisfies, taking germs at $x$ of both sides of the equation $\varphi^\sharp(D(f))\circ\epsilon_f=\theta_f$,

\[\varphi^\sharp_x\circ\epsilon=\phi_x\]

where $\epsilon: A \rightarrow A_\mathfrak{p}$ is the canonical morphism. Now for any $a/s\in A_\mathfrak{p}$, since $s\not\in \mathfrak{p}=\phi_x^{-1}(\mathfrak{m}_x)$, the element $\phi_x(s)$ is a unit, and hence

\[\varphi_x^\sharp(a/s)=\phi_x(a)\phi_x(s)^{-1}\]

In particular, if $a\in \mathfrak{p}$ then $\phi_x(a)\in \mathfrak{m}_x$, so $\varphi_x^\sharp(a/s)\in \mathfrak{m}_x$, and thus the ideal $(\varphi_x^\sharp)^{-1}(\mathfrak{m}_x)$ contains $\mathfrak{p}A_\mathfrak{p}$. On the other hand, $\varphi_x^\sharp(1)=1\not\in \mathfrak{m}_x$ so this ideal is not all of $A_\mathfrak{p}$, and since $\mathfrak{p}A_\mathfrak{p}$ is the unique maximal ideal of $A_\mathfrak{p}$ ([[Commutative Algebra] §Localization, ⁋Proposition 8]](/en/math/commutative_algebra/localization#prop8)),

\[(\varphi_x^\sharp)^{-1}(\mathfrak{m}_x)=\mathfrak{p}A_\mathfrak{p}\]

That is, $\varphi_x^\sharp$ is a local homomorphism, and by Definition 2, $\Psi(\phi)=(\varphi,\varphi^\sharp)$ is a morphism of locally ringed spaces.

Now we show $\Phi$ and $\Psi$ are inverses of each other. First, for $\Psi(\phi)=(\varphi,\varphi^\sharp)$, taking $f=1$ gives $D(1)=\Spec A$, $V_1=X$, and $\epsilon_1=\id_A$, so the above construction yields

\[\Phi(\Psi(\phi))=\varphi^\sharp(\Spec A)=\theta_1=\phi\]

Conversely, given a morphism $(\varphi,\varphi^\sharp): X \rightarrow \Spec A$ of locally ringed spaces, let $\phi=\Phi(\varphi,\varphi^\sharp)=\varphi^\sharp(\Spec A)$ and $\Psi(\phi)=(\varphi',(\varphi')^\sharp)$. For any $x\in X$, since $\varphi^\sharp$ commutes with restriction maps, the stalk morphism $\varphi_x^\sharp: \mathcal{O}_{\Spec A, \varphi(x)}\cong A_{\varphi(x)} \rightarrow \mathcal{O}_{X,x}$ induced by $\varphi^\sharp$ satisfies $\varphi_x^\sharp\circ\epsilon=\phi_x$, where $\epsilon: A \rightarrow A_{\varphi(x)}$ being the canonical morphism follows from Lemma 8. On the other hand, since $(\varphi,\varphi^\sharp)$ is a morphism of locally ringed spaces, $\varphi_x^\sharp$ is a local homomorphism, so the ideal $(\varphi_x^\sharp)^{-1}(\mathfrak{m}_x)$ is a proper ideal containing $\varphi(x)A_{\varphi(x)}$, i.e., $\varphi(x)A_{\varphi(x)}$ itself. Therefore, by [[Commutative Algebra] §Localization, ⁋Proposition 8]](/en/math/commutative_algebra/localization#prop8),

\[\varphi'(x)=\phi_x^{-1}(\mathfrak{m}_x)=\epsilon^{-1}\left((\varphi_x^\sharp)^{-1}(\mathfrak{m}_x)\right)=\epsilon^{-1}\left(\varphi(x)A_{\varphi(x)}\right)=\varphi(x)\]

and the two continuous maps $\varphi$ and $\varphi'$ are equal. Now since $\varphi^\sharp$ commutes with restriction maps, for any $f\in A$

\[\varphi^\sharp(D(f))\circ\epsilon_f=\theta_f\]

holds, and this is exactly the equation defining $(\varphi')^\sharp(D(f))$, so by the uniqueness in [[Commutative Algebra] §Localization, ⁋Proposition 6]](/en/math/commutative_algebra/localization#prop6), $\varphi^\sharp(D(f))=(\varphi')^\sharp(D(f))$. Since two sheaf morphisms agreeing on the base $\{D(f)\}_{f\in A}$ are equal, by the identity axiom for $\varphi_\ast\mathcal{O}_X$ we have $\varphi^\sharp=(\varphi')^\sharp$, and hence $\Psi(\Phi(\varphi,\varphi^\sharp))=(\varphi,\varphi^\sharp)$.

Finally, we verify that this bijection is natural. Given a morphism $\psi: X' \rightarrow X$ of locally ringed spaces, computing the composite $(\varphi\circ\psi)^\sharp$ on $\Spec A$ gives $\psi^\sharp(X)\circ\varphi^\sharp(\Spec A)$, so

\[\Phi(\varphi\circ\psi)=\Gamma(\psi)\circ\Phi(\varphi)\]

Also, given a ring homomorphism $\theta: A \rightarrow A'$, from the construction in Proposition 9 with $f=1$ we have $(\Spec\theta)^\sharp(\Spec A)=\theta$, so for any $\varphi: X \rightarrow \Spec A'$,

\[\Phi((\Spec\theta)\circ\varphi)=\Phi(\varphi)\circ\theta\]

That is, the given bijection is natural in both $X$ and $A$, yielding the claimed natural isomorphism.

References

[Har] R. Hartshorne, Algebraic geometry. Graduate texts in mathematics. Springer, 1977.
[Vak] R. Vakil, The rising sea: Foundation of algebraic geometry. Available online.

In general, for any sheaf $\mathcal{F}$ on $X$ we denote $\mathcal{F}(X)$ by $\Gamma(X, \mathcal{F})$. ↩

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