The Stacks project

26.13 Points of schemes

Given a scheme $X$ we can define a functor

\[ h_ X : \mathit{Sch}^{opp} \longrightarrow \textit{Sets}, \quad T \longmapsto \mathop{\mathrm{Mor}}\nolimits (T, X). \]

See Categories, Example 4.3.4. This is called the functor of points of $X$. A fun part of scheme theory is to find descriptions of the internal geometry of $X$ in terms of this functor $h_ X$. In this section we find a simple way to describe points of $X$.

Let $X$ be a scheme. Let $R$ be a local ring with maximal ideal $\mathfrak m \subset R$. Suppose that $f : \mathop{\mathrm{Spec}}(R) \to X$ is a morphism of schemes. Let $x \in X$ be the image of the closed point $\mathfrak m \in \mathop{\mathrm{Spec}}(R)$. Then we obtain a local homomorphism of local rings

\[ f^\sharp : \mathcal{O}_{X, x} \longrightarrow \mathcal{O}_{\mathop{\mathrm{Spec}}(R), \mathfrak m} = R. \]

Lemma 26.13.1. Let $X$ be a scheme. Let $R$ be a local ring. The construction above gives a bijective correspondence between morphisms $\mathop{\mathrm{Spec}}(R) \to X$ and pairs $(x, \varphi )$ consisting of a point $x \in X$ and a local homomorphism of local rings $\varphi : \mathcal{O}_{X, x} \to R$.

Proof. Let $A$ be a ring. For any ring homomorphism $\psi : A \to R$ there exists a unique prime ideal $\mathfrak p \subset A$ and a factorization $A \to A_{\mathfrak p} \to R$ where the last map is a local homomorphism of local rings. Namely, $\mathfrak p = \psi ^{-1}(\mathfrak m)$. Via Lemma 26.6.4 this proves that the lemma holds if $X$ is an affine scheme.

Let $X$ be a general scheme. Any $x \in X$ is contained in an open affine $U \subset X$. By the affine case we conclude that every pair $(x, \varphi )$ occurs as the end product of the construction above the lemma.

To finish the proof it suffices to show that any morphism $f : \mathop{\mathrm{Spec}}(R) \to X$ has image contained in any affine open containing the image $x$ of the closed point of $\mathop{\mathrm{Spec}}(R)$. In fact, let $x \in V \subset X$ be any open neighbourhood containing $x$. Then $f^{-1}(V) \subset \mathop{\mathrm{Spec}}(R)$ is an open containing the unique closed point and hence equal to $\mathop{\mathrm{Spec}}(R)$. $\square$

As a special case of the lemma above we obtain for every point $x$ of a scheme $X$ a canonical morphism
\begin{equation} \label{schemes-equation-canonical-morphism} \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) \longrightarrow X \end{equation}

corresponding to the identity map on the local ring of $X$ at $x$. We may reformulate the lemma above as saying that for any morphism $f : \mathop{\mathrm{Spec}}(R) \to X$ there exists a unique point $x \in X$ such that $f$ factors as $\mathop{\mathrm{Spec}}(R) \to \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) \to X$ where the first map comes from a local homomorphism $\mathcal{O}_{X, x} \to R$.

In case we have a morphism of schemes $f : X \to S$, and a point $x$ mapping to a point $s \in S$ we obtain a commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) \ar[r] \ar[d] & X \ar[d] \\ \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}) \ar[r] & S } \]

where the left vertical map corresponds to the local ring map $f^\sharp _ x : \mathcal{O}_{S, s} \to \mathcal{O}_{X, x}$.

Lemma 26.13.2. Let $X$ be a scheme. Let $x, x' \in X$ be points of $X$. Then $x' \in X$ is a generalization of $x$ if and only if $x'$ is in the image of the canonical morphism $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) \to X$.

Proof. A continuous map preserves the relation of specialization/generalization. Since every point of $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$ is a generalization of the closed point we see every point in the image of $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) \to X$ is a generalization of $x$. Conversely, suppose that $x'$ is a generalization of $x$. Choose an affine open neighbourhood $U = \mathop{\mathrm{Spec}}(R)$ of $x$. Then $x' \in U$. Say $\mathfrak p \subset R$ and $\mathfrak p' \subset R$ are the primes corresponding to $x$ and $x'$. Since $x'$ is a generalization of $x$ we see that $\mathfrak p' \subset \mathfrak p$. This means that $\mathfrak p'$ is in the image of the morphism $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) = \mathop{\mathrm{Spec}}(R_{\mathfrak p}) \to \mathop{\mathrm{Spec}}(R) = U \subset X$ as desired. $\square$

Now, let us discuss morphisms from spectra of fields. Let $(R, \mathfrak m, \kappa )$ be a local ring with maximal ideal $\mathfrak m$ and residue field $\kappa $. Let $K$ be a field. A local homomorphism $R \to K$ by definition factors as $R \to \kappa \to K$, i.e., is the same thing as a morphism $\kappa \to K$. Thus we see that morphisms

\[ \mathop{\mathrm{Spec}}(K) \longrightarrow X \]

correspond to pairs $(x, \kappa (x) \to K)$. We may define a preorder on morphisms of spectra of fields to $X$ by saying that $\mathop{\mathrm{Spec}}(K) \to X$ dominates $\mathop{\mathrm{Spec}}(L) \to X$ if $\mathop{\mathrm{Spec}}(K) \to X$ factors through $\mathop{\mathrm{Spec}}(L) \to X$. This suggests the following notion: Let us temporarily say that two morphisms $p : \mathop{\mathrm{Spec}}(K) \to X$ and $q : \mathop{\mathrm{Spec}}(L) \to X$ are equivalent if there exists a third field $\Omega $ and a commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(\Omega ) \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(L) \ar[d]^ q \\ \mathop{\mathrm{Spec}}(K) \ar[r]^ p & X } \]

Of course this immediately implies that the unique points of all three of the schemes $\mathop{\mathrm{Spec}}(K)$, $\mathop{\mathrm{Spec}}(L)$, and $\mathop{\mathrm{Spec}}(\Omega )$ map to the same $x \in X$. Thus a diagram (by the remarks above) corresponds to a point $x \in X$ and a commutative diagram

\[ \xymatrix{ \Omega & L \ar[l] \\ K \ar[u] & \kappa (x) \ar[l] \ar[u] } \]

of fields. This defines an equivalence relation, because given any set of field extensions $K_ i/\kappa $ there exists some field extension $\Omega /\kappa $ such that all the field extensions $K_ i$ are contained in the extension $\Omega $.

Lemma 26.13.3. Let $X$ be a scheme. Points of $X$ correspond bijectively to equivalence classes of morphisms from spectra of fields into $X$. Moreover, each equivalence class contains a (unique up to unique isomorphism) smallest element $\mathop{\mathrm{Spec}}(\kappa (x)) \to X$.

Proof. Follows from the discussion above. $\square$

Of course the morphisms $\mathop{\mathrm{Spec}}(\kappa (x)) \to X$ factor through the canonical morphisms $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) \to X$. And the content of Lemma 26.13.2 is in this setting that the morphism $\mathop{\mathrm{Spec}}(\kappa (x')) \to X$ factors as $\mathop{\mathrm{Spec}}(\kappa (x')) \to \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) \to X$ whenever $x'$ is a generalization of $x$. In case we have a morphism of schemes $f : X \to S$, and a point $x$ mapping to a point $s \in S$ we obtain a commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(\kappa (x)) \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) \ar[r] \ar[d] & X \ar[d] \\ \mathop{\mathrm{Spec}}(\kappa (s)) \ar[r] & \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}) \ar[r] & S. } \]

Comments (2)

Comment #2645 by Manuel Hoff on

In 25.13.1, the last sentence mentions a map of local rings, but this map should go in the other direction right? Namely .

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