Why is a $k$-group scheme separated?

Question

Let $G$ be a group scheme over a field $k$.

Why is the unit section $e: Spec(k) \rightarrow G$ a closed immersion ?

I know someone asked that on this forum but it's still not clear for me.

Answer 1

If $e: \mathrm{Spec}(k) \to G$ is the identity section, then we have $\mathrm{Spec}(k) \to G \to \mathrm{Spec}(k)$ is the identity map. Pick an affine open $\mathrm{Spec}(A) \subset G$ containing the image of $e$. Then we get $\mathrm{Spec}(k) \to \mathrm{Spec}(A) \to \mathrm{Spec}(k)$ is the identity map, and dually $k \to A \to k$ is the identity map. In particular $A \to k$ is surjective, so $e(\mathrm{Spec}(k)) = \{\mathfrak{m}\}$ where $\mathfrak{m} = \text{ker}(A \to k)$ is a maximal ideal, i.e. a closed point of $\mathrm{Spec}(A)$. In particular this works for any affine open $\mathrm{Spec}(A) \subset G$ containing the image of $e$.

Fact: if $X$ is a topological space with an open cover $\{U_i\}_{i \in I}$ and $x \in X$ is a point such that $\{x\}$ is closed in every $U_i$ containing $x$, then $\{x\}$ is closed in $X$.

Apply the fact to $X = G$ and $\{U_i\}_{i \in I}$ any open cover of $G$ by affines.

So the image of $e$ is a closed point, thus $e$ is a closed immersion.