So I was reading the book "Algebraic Geometry" by Görtz and Wedhorn.
I have trouble with Corollary 3.36 on Page 80. The statement is:
Let $X$ be a scheme locally of finite type over an algebraically closed field $k$. Let $x \in X$ and $\kappa(x)$ be the residue field at $x$. Then the following equalities hold:
$$\{x \in X\; |\; x\; \text{is closed}\} = \{x \in X\; |\; k = \kappa(x)\} = \text{Hom}_{k}(\text{Spec}(k),X)$$
Now what I can prove is the following:
If $x$ is closed, then by Proposition 3.33, $\kappa(x)$ is finite extension of $k$ and hence $k = \kappa(x)$ as $k$ is algebraically closed.
If $k = \kappa(x)$, then by Proposition 3.8 we have a map from $\text{Spec}(k)$ to $X$.
Now the problem is:
If we have a morphism from $\text{Spec}(k)$ to $X$ with image $x$, then we have again by Proposition 3.8 a map from $\kappa(x)$ to $k$. However, I do not see why this should imply that $\kappa(x) = k$ or that $x$ is closed.
Ideas:
Since we have $k \to \kappa(x) \to k$ and both maps are injective, I thought that this might imply that they are isomorphic. However, that turns out to be not true. Is it true in the special case where $k$ is algebraically closed?
Best Answer
If $x\in X$ is a closed point hence $\{x\}\cong\textrm{Spec }k(x)$.