Affine Schemes

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 collection of continuous functions on that space, and the $\mathcal{O}_{\Spec A}$ we are about to define is similar. However, if $\mathcal{O}_{\Spec A}$ were merely the sheaf of continuous functions on $\Spec A$, there would be no need to give it a new name. For the simplest example, since the only prime ideal of any field $\mathbb{K}$ is $(0)$, the topological space $\Spec \mathbb{K}$ is always a singleton, and there is only one topology on it. In other words, if we wish to distinguish the spectra of two non-isomorphic fields, that information must be encoded in the structure sheaf of $\Spec \mathbb{K}$. To ensure that the spectrum carries enough algebraic information, we define $\mathcal{O}_{\Spec A}$ as the sheaf of algebraic functions on $A$.

Locally ringed space

We have already treated sheaves on a topological space in [Topology] §Sheaves, but the definitions given there are insufficient to describe the structure sheaf we will define 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 $x$ of $X$, the stalk $\mathcal{O}_{X,x}$ at $x$ is always 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 $\Spec A$ to make $(\Spec A, \mathcal{O}_{\Spec A})$ a locally ringed space, and that $\Spec$ defined in this way enjoys the same functoriality as in §Spectra, ⁋Proposition 2 or §Spectra, ⁋Proposition 8. To write this down mathematically, we must 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 that additionally induces a local homomorphism $\varphi_x^\sharp:\mathcal{O}_{Y,\varphi(x)} \rightarrow \mathcal{O}_{X,x}$ for each $x\in X$.

Algebraic Functions on $\Spec A$

Now we must define $\mathcal{O}_{\Spec A}$. As mentioned at the beginning of this post, it is the sheaf of algebraic functions on $\Spec A$, and we saw in §Spectra, §§Classical algebraic geometry that when $A=\mathbb{K}[\x_1,\ldots, \x_n]$, these are functions that can be represented as rational functions on a suitable neighborhood. What played an important role in this process was that elements of $A$, i.e., polynomials, could be treated as functions on $\mathbb{A}_\mathbb{K}^n=\mSpec A$; however, in the general case, elements of $A$ are not polynomials, and moreover we cannot evaluate points of $\Spec A$ at elements of $A$.

Therefore, to generalize this discussion, we argue as follows. First, we regard an element of $A$ as a function $f$, just as in the preceding example. Then the function value of $f$ at a point $\mathfrak{p}\in\Spec A$ is the image of $f$ under the canonical projection $\pr_\mathfrak{p}: A \rightarrow A/\mathfrak{p}$. In particular, the condition that $f$ vanishes at the point $\mathfrak{p}$ is

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

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

From this perspective, we can describe what the algebraic functions on $\Spec A$ are. As in §Spectra, §§Classical algebraic geometry, these are defined as functions that can be represented as rational functions whose denominators are functions not vanishing on the given open set.

Now suppose a principal open set $D(f)$ is given. Then by definition, when an algebraic function on $D(f)$ is represented as a rational function $g/h$, the functions $h$ that can appear in the denominator 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

First, 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 equality

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

we know that $D(f)\subseteq D(h_1)\cap D(h_2)=D(h_1h_2)$. This equality is nothing but a geometric explanation of [Algebraic Structures] §Field of Fractions, ⁋Proposition 8.

It is intuitive that the collection of algebraic functions on the subset $D(f)$ of $\Spec A$ should be defined as $S(f)^{-1}A$, and indeed we will define it this way. Before doing so, we prove the following lemma.

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

Proof

That $D(f)\subseteq D(h)$ is equivalent to $Z(h)\subseteq Z(f)$, which by the third result of §Spectra, ⁋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 $f\in \sqrt{(h)}$, and thus there exists an integer $n\geq 1$ such that $f^n\in (h)$. Conversely, if there exists an integer $n\geq 1$ such that $f^n\in (h)$, then from $f\in \sqrt{(h)}$ we get $(f)\subseteq \sqrt{(h)}$, and therefore

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

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

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

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

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

$localizations$

commutes.

Proof

First, let us denote the canonical morphisms by $\epsilon(f): A \rightarrow S(f)^{-1}A$ and $\epsilon_f:A \rightarrow S_f^{-1}A$. Then since $D(f)=D(f^n)$ for any $n\geq 1$, we have $f^n\in S(f)$, and therefore the image of $S_f$ under $\epsilon(f)$ consists entirely of units in $S(f)^{-1}A$. Hence from [Commutative Algebra] §Localization, ⁋Proposition 6 we obtain the following commutative diagram

$universal_property-1$

Now observe the following equivalence

\[D(f)\subseteq D(g)\iff f^n\in (g)\text{ for some $n\geq 1$}\iff f^n=ag\text{ for some $n\geq 1$ and $a\in A$}\tag{$\ast$}\]

Then for any $g$ satisfying $D(f)\subseteq D(g)$, we can find suitable $n\geq 1$ and $a\in A$ such that $f^n=ag$, so from

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

we know that $g$ is a unit in $S_f^{-1}A$. Therefore, again from [Commutative Algebra] §Localization, ⁋Proposition 6 we obtain the following commutative diagram

$universal_property-2$

That $\overline{\epsilon(f)}$ and $\overline{\epsilon_f}$ are inverse to each other is obvious from uniqueness.

Now suppose $S(g)\subseteq S(f)$. Then $\widehat{\epsilon(f)}:S(g)^{-1}A \rightarrow S(f)^{-1}A$ is again obviously defined via [Commutative Algebra] §Localization, ⁋Proposition 6 by the following diagram

$universal_property-3$

and since $S(g)\subseteq S(f)\iff D(f)\subseteq D(g)$, from the equivalence ($\ast$) above we know that $g$ is a unit in $S_f^{-1}A$, and therefore so are all $g^k$. From this we obtain the following commutative diagram involving $\widecheck{\epsilon_f}: S_g^{-1}A \rightarrow S_f^{-1}A$:

$universal_property-4$

Then that the diagram in question commutes is obvious from considering the following diagram:

$universal_property-5$

That is, from

\[\epsilon_f=\widecheck{\epsilon_f}\circ\epsilon_g=\widecheck{\epsilon_f}\circ\overline{\epsilon_g}\circ\epsilon(g)\]

and

\[\epsilon_f=\overline{\epsilon_f}\circ\epsilon(f)=\overline{\epsilon_f}\circ\widehat{\epsilon(f)}\circ\epsilon(g)\]

we know that $\epsilon_f$ sends elements of $S(g)$ to units in $S_f^{-1}A$, and moreover from the uniqueness of $\widetilde{\epsilon_f}$ defined to satisfy $\epsilon_f=\widetilde{\epsilon_f}\circ\epsilon(g)$ via [Commutative Algebra] §Localization, ⁋Proposition 6, we obtain $\widecheck{\epsilon_f}\circ\overline{\epsilon_g}=\overline{\epsilon_f}\circ\widehat{\epsilon(f)}$.

Therefore, it suffices to regard algebraic functions on $D(f)$ as elements of $S_f^{-1}A$. In the previous post we agreed to write $S_f^{-1}A$ as $A_f$ for convenience.

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 $f_i,f_j\in A$ satisfying $D(f_i)\subseteq D(f_j)$, define the restriction map

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

as the map obtained by applying [Commutative Algebra] §Localization, ⁋Proposition 6 to the canonical morphism $A\rightarrow S(f_i)^{-1}(A)$. Then these data satisfy the two conditions of [Topology] §Sheaves, ⁋Proposition 8, and therefore the ($\cRing$-valued) sheaf on $\Spec A$ extending $\mathcal{F}$ is uniquely determined.

Proof

That the $\rho_{ji}$ satisfy the conditions for restriction maps ([Topology] §Presheaves, ⁋Definition 2) is obvious from the universal property of [Commutative Algebra] §Localization, ⁋Proposition 6. Here, by Lemma 5, the map $\rho_{ji}: S(f_j)^{-1}(A) \rightarrow S(f_i)^{-1}(A)$ simply takes an element of $S(f_j)^{-1}(A)$, written in the form

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

and views it as an element of $S(f_i)^{-1}(A)$; indeed, from the relation

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

we can regard ($\ast$) as an element of $S(f_i)^{-1}(A)$.

We now prove the two conditions of [Topology] §Sheaves, ⁋Proposition 8. For notational convenience, since $D(f)=\Spec A_f$, it suffices to consider the case $f=1$ after replacing $A$ by $A_f$. Fix elements $f_i\in A$ satisfying $\Spec A=\bigcup_{i\in I}D(f_i)$.

First, to show 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 $s$ is also $0$ as an element of $A$. Then by §Spectra, ⁋Lemma 12 we can choose $f_1,\ldots, f_n$ among the $f_i$ such that $\Spec A=\bigcup_{i=1}^n D(f_i)$, and by assumption there exist integers $m_i$ satisfying

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

for all $i=1,\ldots, n$. On the other hand, from the calculation after §Spectra, ⁋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 from this there exist elements $a_i\in A$ such that $1=\sum_{i=1}^n a_i f_i^{m_i}$. (See the proof of §Spectra, ⁋Lemma 12, or the proof of [Commutative Algebra] §Integral Extensions, ⁋Proposition 15)

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

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

But since $D(f_i)=D(f_i^{m_i})$ and $D(f_j)=D(f_j^{m_j})$, we have

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

and therefore there exists an integer $N_{ij}$ such that

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

Let $N=\max_{i,j}\{N_{ij}\}$, so that

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

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

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

there exist elements $b_i\in A$ such that

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

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 therefore on $D(f_j)$

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

From this we obtain the desired $s$.

If $I$ is infinite, choose a finite subset $J=\{1,\ldots, n\}$ of $I$ satisfying $\Spec A=\bigcup_{j\in J} D(f_j)$ and repeat the above to obtain $s\in \mathcal{F}(\Spec A)$; then it suffices to show that this also satisfies $s_\alpha=s\vert_{D(f_\alpha)}$ on $D(f_\alpha)$ for $\alpha\in I\setminus J$. To show 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 definition $s$ and $s'$ satisfy $s\vert_{D(f_i)}=s'\vert_{D(f_i)}$ for all $i=1,\ldots, n$, and since $\Spec A=\bigcup D(f_i)$, by the first condition of [Topology] §Sheaves, ⁋Proposition 8 shown above we know $s=s'$, and from this

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

Since this holds for every $\alpha$, the restriction of $s$ to any $D(f_\alpha)$ equals $s_\alpha$.

Definition 7 We write the sheaf on $\Spec A$ defined by Lemma 5 as $\mathcal{O}_{\Spec A}$, and call it the structure sheaf.

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 exists 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 $f\in A$ satisfying $\mathfrak{p}\in D(f)$, the following diagram

$stalk_and_localization-1$

commutes.

Proof

Since the $D(f)$ form a base for $\Spec A$ by [Topology] §Bases of Topological Spaces, ⁋Proposition 2, we have

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

by [Topology] §Bases of Topological Spaces, ⁋Proposition 5. On the other hand, since $\mathfrak{p}\in D(f)\iff f\not\in \mathfrak{p}$, we obtain the following diagram

$stalk_and_localization-2$

and therefore proving the given isomorphism is the same as proving the following algebraic isomorphism

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

which follows by using the universal property of localization ([Commutative Algebra] §Localization, ⁋Proposition 6) and the universal property of direct limits. For the diagram in question, one simply replaces $\varinjlim A_f$ by $A_\mathfrak{p}$ in the above diagram via the isomorphism ($\ast\ast$).

Now we are finally ready to write down 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 $\Spec\phi: \Spec B \rightarrow \Spec A$. (§Spectra, ⁋Proposition 8) Therefore 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 map on principal open sets

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

On the other hand, from the proof of §Spectra, ⁋Proposition 8 we know that

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

and therefore

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

Thus, 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 to the composition

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

Of course, we must show that this $(\Spec\phi)^\sharp$ defines the same map on the intersection $D(f)\cap D(g)$, but this follows from using the uniqueness result in [Commutative Algebra] §Localization, ⁋Proposition 6 on $D(f)\cap D(g)$.

From the above we know that $(\Spec\phi, (\Spec\phi)^\sharp): (\Spec B, \mathcal{O}_{\Spec B}) \rightarrow (\Spec A, \mathcal{O}_{\Spec A})$ is a morphism of ringed spaces. Now to show that this is a morphism of locally ringed spaces, it suffices to show 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 $(\Spec \phi)(\mathfrak{q})=\phi^{-1}(\mathfrak{q})$, and therefore by Lemma 8 the map $(\Spec\phi)^\sharp_\mathfrak{q}$ is a ring homomorphism from $A_{\phi^{-1}(\mathfrak{q})}$ to $B_{\mathfrak{q}}$ that sends 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 $B_{\mathfrak{q}}$.

Affine Schemes

Definition 10 We define the essential image of the functor $\Spec:\cRing^\op \rightarrow \LRS$ from Proposition 9 as the affine scheme.

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

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

and therefore this functor is necessarily faithful. Moreover, the following holds.

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

Proof

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

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

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

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

Now to complete the claim we must show that $\Spec\phi=(\varphi,\varphi^\sharp)$. This is shown by proving that for any $\mathfrak{q}\in \Spec B$,

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

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

$faithful$

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

$commuting_square$

are commuting squares. Therefore, in the above diagram the map $A \rightarrow \mathcal{O}_{\Spec B, \mathfrak{q}}$ is determined equally no matter which path we take, and applying [Commutative Algebra] §Localization, ⁋Proposition 6 to this map uniquely determines $A_{\varphi(\mathfrak{q})} \rightarrow \mathcal{O}_{\Spec B, \mathfrak{q}}$. From this we know 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 therefore $\phi^{-1}(\mathfrak{q})=\varphi(\mathfrak{q})$. Now that $\phi$ equals $\varphi^\sharp$ on the structure sheaf follows by considering only restriction maps, so the desired claim is proved from the above.

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

By [Category Theory] §Natural Transformations, ⁋Theorem 5, it suffices to show that $\Spec$ is full.

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

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

If a locally ringed space $(X, \mathcal{O}_X)$ is an affine scheme, then by examining $\mathcal{O}_X(X)$ in the same way we can determine whether $(X, \mathcal{O}_X)$ is isomorphic to the spectrum of some ring. That is, for an affine scheme $(X, \mathcal{O}_X)$, if we set $A=\mathcal{O}_X(X)$ then $(X, \mathcal{O}_X)\cong (\Spec A, \mathcal{O}_{\Spec A})$ holds. More generally, we define the following.

Definition 12 For any locally ringed space $(X, \mathcal{O}_X)$, we define the global section functor $\Gamma:\LRS \rightarrow \cRing^\op$ by $X\mapsto \Gamma(X, \mathcal{O}_X)=\mathcal{O}_X(X)$.¹

A notable fact in the proof of Proposition 11 is that the hypothesis that $(X, \mathcal{O}_X)$ is an affine scheme is 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 diagram in Proposition 11,

$adjoint$

we can carry out a similar argument, and in this case the $B$ in the conclusion is replaced by $\Gamma(X, \mathcal{O}_X)$. Since $\mathcal{O}_X$ is data determined by $X$ anyway, writing this briefly 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 $\Spec:\cRing^\op \rightarrow \LRS$.

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 write $\mathcal{F}(X)$ as $\Gamma(X, \mathcal{F})$. ↩

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