separated morphisms

The definition of a separated morphism resembles the definition of a Hausdorff space.

DEFINITION 9.1   A morphism $f:X\to S$ of schemes is separated if the diagonal

$$\Delta_X \subset X\times_S X$$

(That means, the image of the diagonal map $X \to X\times_S X$ ) is closed in $X\times_S X$ . In other words, $f$ is closed if and only if there exists an ideal sheaf $\mathcal I_{\Delta}$ of $X\times_S X$ such that $f$ induces an isomorphism $X \cong V(\mathcal I_{\Delta})$ of schemes.

LEMMA 9.2   A morphism $\operatorname{Spec}(B)\to \operatorname{Spec}(A)$ of affine schemes is always separated. More generally, an affine morphism is always separated.

PROOF.. Let $I$ be the kernel of a ring homomorphism

$$B\otimes_A B \ni b_1 \otimes b_2 \mapsto b_1 b_2 \in B.$$

Then it is easy to see that $I$ gives the defining equation of the diagonal $\Delta$ .

For the general affine morphism case, let $X=\operatorname{Spec}(\mathcal A)$ be a scheme which is affine over $Y$ . Then we have $X\times_Y X=\operatorname{Spec}(\mathcal A \otimes \mathcal{O}_Y \mathcal A)$ . We may then see the situation locally and reduce the problem to the first case. $\qedsymbol$

LEMMA 9.3   Separated morphism is stable under base extension. That is, assume $f:X\to S$ be a separated morphism. Let $g:T\to S$ be a morphism of schemes. Then $f_T:X_T \to T$ is separated.

PROOF..

$$X_T \times_T X_T \cong (X\times_S T) \times_T (T\times_S X)\cong (X\times_S X) \times_S T$$

The diagonal $\Delta_{X/T}$ is isomorphic to $\Delta_{X/S}\times_S T$ , and is therefore closed.

$\qedsymbol$

LEMMA 9.4   A composition of separated morphisms is again separated. Namely, if $f:X\to Y$ $g:Y\to S$ are separated morphism of schemes, then $g\circ f: X\to S$ is also separated.

PROOF.. We first claim the following sublemma:

SUBLEMMA 9.5   Under the assumption of the lemma above, we have

$$X\times_Y X \cong (X\times_S X) \times_{(Y\times_S Y)} \Delta_{Y/S}$$

The proof of the sublemma above is given by showing that the right hand side satisfies the same universal property as the left hand side.

Now, let us prove the lemma. Since $Y$ is separated over $S$ , we have a closed immersion

$$\Delta_{Y/S}\hookrightarrow Y\times_S Y.$$

By taking a base extension, we obtain a closed immersion

$$(X\times_S X) \times_{(Y\times_S Y)} \Delta_{Y/S} \hookrightarrow X\times_S X.$$

$$X\times_Y X \cong (X\times_S X) \times_{(Y\times_S Y)} \Delta_{Y/S} \hookrightarrow X\times_S X$$

Then $\Delta_{X/S}$ is identified with the composition of closed immersions

$$X\cong \Delta_{X/Y} \hookrightarrow X\times_Y X \hookrightarrow X\times_S X$$

$\qedsymbol$

PROPOSITION 9.6   Let $p:X\to S$ be a separated morphism of schemes. Let $q:Y\to S$ be a morphism of schemes. Then any $S$ -morphism $f:X\to Y$ is separated.

PROOF.. Since $X$ is separated over $S$ ,

$$X \to \Delta_{X/S}\hookrightarrow X\times_S X$$

is a closed immersion. Now $\Delta_{X/Y} \to X\times_ Y X$ may be identified with a pullback of the morphism above

$$\Delta_{X/S} \times_{(X\times_S X)} (X\times_Y X) \to (X\times_Y X)$$

$\qedsymbol$