I am trying to understand the proof of Theorem 4.3 in Hartshorne's Algebraic Geometry (the valuative criterion for separatedness). I'm having trouble justifying the highlighted line below:
Conversely, let us suppose the condition of the theorem satisfied. To show that $f$ is separated, it is sufficient by $(4.2)$ to show that $\Delta(X)$ is a closed subset of $X\times_Y X$. And since we have assumed that $X$ is noetherian, the morphism $\Delta$ is quasi-compact, so by $(4.5)$ it will be sufficient to show that $\Delta(X)$ is stable under speicalization. So let $\xi_1\in\Delta(X)$ be a point, and let $\xi_1\rightsquigarrow\xi_0$ be a specialization. Let $K=k(\xi_1)$ and let $\mathcal{O}$ be the local ring of $\xi_0$ on the subscheme $\{\xi_1\}^-$ with its reduced induced structure. Then $\mathcal{O}$ is a local ring contained in $K$, so by $(\text{I}, 6.1\text{A})$ there is a valuation ring $R$ of $K$ which dominated $\mathcal{O}$. Now by $(4.4)$ we obtain a morphism $T=\operatorname{Spec}R$ to $X\times_YX$ sending $t_0$ and $t_1$ to $\xi_0$ and $\xi_1$. Composing with the projections $p_1,p_2$ gives two morphisms of $T$ to $X$, which give the same morphism to $Y$, and $\color{red}{\text{whose restrictions to $U=\operatorname{Spec}K$ are the same, since $\xi_1\in\Delta(X)$.}}$ So by the condition, these two morphisms of $T$ to $X$ must be the same. Therefore the morphism $T\to X\times_YX$ factors through the diagonal morphism $\Delta:X\to X\times_YX$, and so $\xi_0\in\Delta(X)$. This completes the proof. Note in the last step it would not be sufficient to know only that $p_1(\xi_0)=p_2(\xi_0)$. For in general if $\xi\in X\times_YX$ then $p_1(\xi)=p_2(\xi)$ does not imply $\xi\in\Delta(X)$.
The author claims that the two morphisms $\operatorname{Spec} K \to X$ are the same since they have the same image $\xi_1$. However, it is possible for two morphisms from the spectrum of a field to have the same image but be distinct. (For example, if $F$ is a field with a nontrivial automorphism $F \to F$, then this automorphism induces a morphism of schemes which is not the identity.) What am I missing here?
Best Answer
For an arbitrary scheme $Z$, a morphism $U\to Z$ is specified by a point $z\in Z$ and an inclusion of fields $k(z)\subset K$. In this case, we're looking at the map $U\to X\times_Y X$ specified by $\xi_1\in X\times_Y X$ and the identity $k(\xi_z)\subset K$.
What does it mean for $\xi_1\in\Delta(X)$? It means there's some $x\in X$ so that $\Delta(x)=\xi_1$ (set-theoretically). Since $p_1\circ \Delta = p_2\circ\Delta = id_X$ as morphisms of schemes by the definition of the fiber product and the diagonal morphism, we get $p_1(\xi_1)=p_2(\xi_1)=x$, so we have induced morphisms of residue fields $k(x)\stackrel{p_i^*}{\to} K \stackrel{\Delta^*}{\to} k(x)$ so that the composition $k(x)\to k(x)$ is the identity. Since morphisms of fields are injective, this implies that $K\cong k(x)$, and further $k(x)\stackrel{p_1^*}{\to} K$ is equal to $k(x)\stackrel{p_2^*}{\to} K$. This is enough to say that the morphisms $U\to X$ are the same.