Let \(X\) be a scheme. For any \(x \in X\), let \(\OO_x\) be the local ring at \(x\) and \(\mfm_x\) its maximal ideal. Define the residue field of \(x\) on \(X\) to be the field \(k(x) = \OO_x / \mfm_x\).
Now let \(K\) be any field. Show that to give a morphism \(\Spec K \to X\) is equivalent to giving a point \(x \in X\) together with an inclusion map \(k(x) \to K\).
Solution
Let \[ S=\operatorname{Spec}K. \] Since \(K\) is a field, \(S\) has a single point, which we denote by \(\eta\), and \[ \mathcal O_{S,\eta}=K \] with maximal ideal \((0)\).
1. A morphism\[ f:S\longrightarrow X \]determines a point\[ x=f(\eta)\in X \]and a local ring homomorphism\[ f_\eta^\sharp: \mathcal O_{X,x} \longrightarrow K. \]
The underlying continuous map sends the unique point \(\eta\) to a point \(x\in X\). Since a morphism of schemes is a morphism of locally ringed spaces, its induced map on stalks at \(\eta\) is \[ \mathcal O_{X,f(\eta)} \longrightarrow \mathcal O_{S,\eta}=K, \] and this homomorphism is local by definition.
2. The kernel of the local homomorphism in 1 is exactly\[ \mathfrak m_x. \]
For a local ring homomorphism \[ \phi:(A,\mathfrak m_A) \longrightarrow (B,\mathfrak m_B), \] one has \[ \phi^{-1}(\mathfrak m_B)=\mathfrak m_A. \] Here the target is the field \(K\), whose maximal ideal is \[ \mathfrak m_K=(0). \] Thus \[ \ker f_\eta^\sharp =(f_\eta^\sharp)^{-1}(0) =\mathfrak m_x. \]
3. Hence \(f\) determines an injective field homomorphism\[ \boxed{ \kappa(x) =\mathcal O_{X,x}/\mathfrak m_x \hookrightarrow K. } \]
By 2, the stalk map factors through the quotient by its kernel: \[ \mathcal O_{X,x} \longrightarrow \mathcal O_{X,x}/\mathfrak m_x \longrightarrow K. \] The second map has zero kernel, again by 2, and is therefore injective. Both source and target are fields.
4. Conversely, suppose we are given a point \(x\in X\) and an injective field homomorphism\[ \iota:\kappa(x)\hookrightarrow K. \]Define a continuous map\[ f:S\longrightarrow X \]by\[ f(\eta)=x. \]
The only nonempty subset of the one-point space \(S\) is \(S\) itself. For an open set \(U\subseteq X\), \[ f^{-1}(U) = \begin{cases} S,&x\in U,\\ \varnothing,&x\notin U. \end{cases} \] Both subsets are open, so \(f\) is continuous.
5. Define a sheaf morphism\[ f^\sharp:\mathcal O_X\longrightarrow f_*\mathcal O_S \]as follows. For an open \(U\subseteq X\) containing \(x\), put\[ f_U^\sharp: \mathcal O_X(U) \longrightarrow \mathcal O_{X,x} \longrightarrow \kappa(x) \xrightarrow{\iota} K =(f_*\mathcal O_S)(U), \]where the first map takes a section to its germ at \(x\). If \(x\notin U\), use the unique map\[ \mathcal O_X(U)\longrightarrow0 =(f_*\mathcal O_S)(U). \]
If \[ V\subseteq U \] and both contain \(x\), taking the germ at \(x\) after restriction gives the same element of \(\mathcal O_{X,x}\) as taking the germ directly. Thus the displayed maps commute with restrictions.
If \(x\notin V\), the target over \(V\) is the zero ring, so compatibility is automatic. Hence the maps \(f_U^\sharp\) define a morphism of sheaves of rings.
6. The pair \((f,f^\sharp)\) from <1>4–<1>5 is a morphism of schemes.
The only stalk map to check is at the point \(\eta\in S\). It is \[ \mathcal O_{X,x} \longrightarrow \kappa(x) \xrightarrow{\iota} K. \] Its kernel is \[ \mathfrak m_x, \] because the quotient map has that kernel and \(\iota\) is injective. Since the maximal ideal of \(K\) is zero, \[ (f_\eta^\sharp)^{-1}(0) =\mathfrak m_x. \] Thus the stalk map is local, so \((f,f^\sharp)\) is a morphism of locally ringed spaces and hence a morphism of schemes.
7. Starting with a morphism \(f:S\to X\), extracting the pair\[ (x,\kappa(x)\hookrightarrow K) \]and applying the construction of <1>4–<1>6 recovers the original morphism.
5. On opens not containing \(x\), both maps land in the zero ring. Hence the entire morphism is recovered.
8. Starting with a pair\[ (x,\iota:\kappa(x)\hookrightarrow K), \]constructing a morphism and then extracting its residue-field map recovers the same pair.
The constructed continuous map sends the unique point of \(S\) to \(x\). Its stalk map is \[ \mathcal O_{X,x} \to\kappa(x) \xrightarrow{\iota}K. \] Passing to the quotient by the maximal ideal therefore gives exactly the original embedding \(\iota\).
9. Therefore there is a natural bijection\[ \boxed{ \operatorname{Hom}_{\mathrm{Sch}}(\operatorname{Spec}K,X) \cong \left\{ (x,\iota): x\in X, \ \iota:\kappa(x)\hookrightarrow K \right\}. } \]
Steps 7 and 8 show that the two constructions are inverse.
10. Q.E.D.
Step 9 is the equivalence requested in the exercise.