Integral Morphisms

Defn. A morphism $\pi: X \to Y$ of schemes is integral if $\pi$ is affine and for every affine open $\text{Spec} B \subset Y$ with $\pi^{-1}\text{Spec} B = \text{Spec} A$, the induced map $B \to A$ is an integral ring homomorphism.

Claim. Finite morphisms are integral.

Note: the converse is not true. e.g. $\text{Spec} \overline{\mathbb{Q}} \to \text{Spec}\mathbb{Q}.$

Claim. Integrality is an affine-local condition

Claim: Integrality is closed under composition.

Fibers. Unlike finite morphisms, fibers of integral morphisms might not be finite. Fibers of integral morphisms have the property that no point is in the closure of any other point.

Proposition. Integral morphisms are closed. (Thus finite morphisms are closed)

Proof.
First we prove the proposition for affine case. Suppose $\pi: \text{Spec} A \to \text{Spec} B$ is an integral morphism induced by $\phi: B \to A$. Let $\mathbb{V}(I) \subset \text{Spec} A$ be a closed subset. It's image is the set $S$ of $\phi^{-1}(\mathfrak{p})$ where $\mathfrak{p} \supseteq I$. We claim that is this just $\mathbb{V}(\phi^{-1}(I)).$

Indeed, every prime ideal in $S$ lies on $\mathbb{V}(\phi^{-1}(I)).$ It thus suffices to show that every prime ideal containing $J = \phi^{-1}(I)$ is the contraction of a prime ideal containing $I$.

Key point: Lying Over (Going-up).
We have $\phi$ induces an embedding $B/J \to A/I$, and $A/I$ is integral over $B/J$.  QED

Now let $\pi: X \to Y$ be an integral morphism between schemes. Let $Z$ be a closed subset of $X$. Let $V_{\alpha}$ be an affine cover of $Y$ and $U_{\alpha} = \pi^{-1}(V_{\alpha})$ be their affine open preimages. Then $\pi(Z) \cap V_{\alpha}= \pi(Z \cap U_{\alpha})$ is closed in $V_{\alpha}$ for each $\alpha$, so $\pi(Z)$ is closed in $Y$.

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Theorem. Lying Over. Suppose $\phi: B \to A$ is an integral extension. Then for any prime ideal $\mathfrak{q}$ of $B$ there is a prime ideal $\mathfrak{p}$ of $A$ such that $\varphi^{-1}(\mathfrak{p}) = \mathfrak{q}.$

In other words, $\text{Spec} A \to \text{Spec} B$ is surjective. 

(Its generalization is Going-up).

Lemma. Let $U_{\alpha}$ be an open cover of $X$. If $Z \cap U_{\alpha}$ is closed in every $U_{\alpha}$ then $Z$ is closed in $X$.
Proof.
Let $V = X - Z$. Then $V \cap U_{\alpha}$ is open in every $U_{\alpha}$ and thus must be open in $X$. So their union, $V$, must be open in $X$.