# uniqueness of morphism $Spec(K) \to X$ of schemes

let $K$ be a field and $X$ a scheme. I'd like to understand the bijection

$Hom_{Sch}(Spec(K), X) \cong \{x \in X | \exists \kappa(x) \to K \}$

That map is given by sending a morphism $f: Spec(K) \to X$ to the value $f((0))$ of the only prime ideal in $K$. Why is this map injective? I don't understand why $f^\#$ is determined by $f((0))$.

• As you note below, the RHS needs to be modified a little. – Hoot Dec 29 '14 at 17:15

Let $f :$ Spec$K \rightarrow X$ be given and let $f((0)) = x \in X$. Then it will induce a map at the local ring level: $\mathcal{O}_{X,x} \rightarrow \mathcal{O}_{K, (0)}=K.$ So we have an induced map $k(x) \rightarrow K.$
On the other hand, suppose a map $k(x) \rightarrow K$ is given, for some $x \in X.$ Define $f:$ Spec$K \rightarrow X$ by $(0) \mapsto x.$ Let $U \subseteq X$ be open. Now define $f^\#: \mathcal{O}_X(U) \rightarrow \mathcal{O}_{Spec K}(f^{-1}(U))$ as follows: if $x \in U,$ then $\mathcal{O}_X(U) \rightarrow \mathcal{O}_{X,x} \rightarrow k(x) \rightarrow K = \mathcal{O}_{Spec K}(f^{-1}(U)),$ and if $x \notin U,$ then $\mathcal{O}_X(U) \rightarrow 0 \cong \mathcal{O}_{Spec K}(f^{-1}(U)).$ This is a morphism between two sheaves: $\mathcal{O}_X$ and $f_* \mathcal{O}_{SpecK}$ (why?). The only thing that remains to check is that it's a morphism between between local rings. For that use the homomorphism $k(x) \rightarrow K.$
• $(f^\#_{(0)})^{-1}((0)) = \{[s] | s \in \mathcal{O}_X(U), f^\#_U(s) = 0\} = \{s | (\mathcal{O}_{X,x} \to \kappa(x))([s]) = 0\} = \mathcal{m}_x$. That follows from the fact that $\kappa(x) \to K$ is injective. – Leon Lang Dec 29 '14 at 16:50
• The main problem was that I did not know that on the right hand side the morphism $\kappa(x) \to K$ is also important. So on the right side we have $\{ (x, f) | x \in X, f: \kappa(x) \to K \text{ a homomorphism } \}$. Thank you! – Leon Lang Dec 29 '14 at 17:01