# Closed points of a locally of finite type $k$-scheme and transcendence degree

Let $k$ be a field, and let $R$ be an integral domain and finitely generated $k$-algebra. It is a basic result from commutative algebra that the dimension of $R$ is equal to the transcendence degree of the quotient field of $R$ over $k$.

In particular, if $X = \textrm{Spec } A$ is an affine scheme of finite type over $k$, and $\mathfrak p \in X$, then the dimension of $A/\mathfrak p$ is equal to the transcendence degree of the residue field $\kappa(x) = A_{\mathfrak p}/\mathfrak p A_{\mathfrak p}$ over $k$. In particular, the point $\{\mathfrak p\}$ is closed if and only if $\mathfrak p$ is a maximal ideal of $A$, if and only if $\textrm{tr.deg}_k\kappa(x)=\textrm{Dim } A/\mathfrak p = 0$, if and only if $\kappa(x)$ is algebraic over $k$.

Now, assume that $X$ is a scheme which is locally of finite type over $k$. This means that $X$ has an open cover $U_i : i \in I$ by affine open sets, such that each $\mathcal O_X(U_i)$ is finitely generated as a $k$-algebra. In fact, every affine open subset of $X$ is the spectrum of a ring which is finitely generated as a $k$-algebra.

Proposition: The following are equivalent for $x \in X$:

(i): $x$ is a closed point.

(ii): $x$ is a closed point in some affine open set containing it.

(iii): $x$ is closed in every affine open set containing it.

(iv): $\kappa(x)$ is algebraic over $k$.

Proof: (i) $\Rightarrow$ (iii) $\Rightarrow$ (ii) is clear, and (ii) $\Rightarrow$ (iv) $\Rightarrow$ (iii) follow from the commutative algebra result above. Assuming (iii), we get (i) from the fact that $E \subseteq X$ is closed if and only if there is an open cover $U_i$ of $X$ such that $E \cap U_i$ is closed in $U_i$ for all $i$.