Separated morphism

Remarks

Example: every morphism $\operatorname{Spec} A \to \operatorname{Spec} B$ is separated. The diagonal is $\operatorname{Spec}$ of the multiplication map $A \otimes_B A \to A$, $a \otimes a' \mapsto a a'$, which is surjective, so the diagonal is a closed immersion. Non-example: let $X$ be the affine line over a field $k$ with a doubled origin, the gluing of two copies of $\mathbb{A}^1_k$ along the identity of $\mathbb{A}^1_k \setminus \set{0}$. Then $X \to \operatorname{Spec} k$ is not separated. Let $j_1, j_2 \colon \mathbb{A}^1_k \to X$ be the two inclusions, and let $(j_1, j_2) \colon \mathbb{A}^1_k \to X \times_k X$. The preimage of the diagonal under $(j_1, j_2)$ is $\mathbb{A}^1_k \setminus \set{0}$, the set where $j_1$ and $j_2$ agree. This set is not closed in $\mathbb{A}^1_k$, so the image of the diagonal is not closed. A morphism is separated if and only if its diagonal has closed image, since the diagonal is always an immersion.