Höchste Zeit, unsern Kurs zu ändern

Proof: It is not hard to see that $R^{int}_\mathfrak{p}$ is integral over $R$ (and integrally closed), hence Cohen-Seidenerg (going down) theorem applies, and we conclude that $R^{int}_\mathfrak{p}$ is a local ring of dimension 1. In other words it has a unique prime ideal. Take an $x \in Q(R^{int}_\mathfrak{p}) = Q(R^{int})$. Let us denote $\mathfrak{q} = pR^{int}_\mathfrak{p}$. Consider $x \mathfrak{q}$. It is an (non-zero, as $x$ is) ideal, so either $x \cdot \mathfrak{q} = \mathfrak{q}$ or $x \cdot \mathfrak{q} = R$.

In the first case, take some generators of $\mathfrak{q}$, $a_1, \ldots, a_n$ ($\mathfrak{q}$ is finitely generated, since $R^{int}_\mathfrak{p}$ is a finitely generated $R$-module). Since $x \cdot a_i \in \mathfrak{q}$

$$ x \cdot a_i = \sum b_{ij} a_j $$

or $(x \cdot I - B) \bar{a} = 0$. Multiply $(x \cdot I - B)$ by the matrix consisting of its minors. It is "almost" the inverse, in fact we get:

$$ \det (x \cdot I - B) \cdot I \cdot \bar{a} = 0 $$

so $\det (x \cdot I - B) = 0$. The latter is a monic polinomial with coefficients in $\mathfrak{q}$, hence $x \in (R^{int}_\mathfrak{p})^{int}=R^{int}_\mathfrak{p}$.

In the second case, there are some $a_1, \ldots, a_n \in \mathfrak{q}$ such that $\sum_{i=1}^n a_i x^i = 1$. Muliply it by $x^{-n}$, then

$\displaystyle (1-a_0)x^{-n} + \sum a_i x^{i-n} = 0 $

But since $a_0$ belongs to the maximal ideal of the local ring $R^{int}_\mathfrak{p}$, $(1-a_0)$ is invertible, and so $x^{-n} + \sum a_i' x^{i-n} = 0$, i.e. $x^{-1}$ is integral over $R^{int}_\mathfrak{p}$, i.e. belongs to $R^{int}_\mathfrak{p}$. ☐

Note that in the proof of valuative criterion of separatedness we could have chosen $x_1$ and $x_0$ in such a way that ${\mathcal O}$ is one-dimensional, and therefore $R$ in the valuative criterion can be assumed to be a DVR (in this case, we only need to prove Lemma 2 for one-dimensional rings).

Proper morphisms

A morphism $f: X \rightarrow Y$ is called universally closed if for any morphism $g: Y' \rightarrow Y$ (base change), the morphism $f': X \times_{Y'} X \rightarrow Y'$ is a closed morphism, i.e. sends closed sets to closed sets.

A morphism $f: X \rightarrow Y$ is locally of finite type if $Y$ can be covered by affine $Y_i$ and $f^{-1}(Y_i)$ by affine $X_{ij}$-s with corresponding maps of rings $B_i \rightarrow A_{ij}$ such that $A_{ij}$ is finitely generated $B_i$-algebra. If there are finitely many $X_{ij}$ for each $Y_i$ then $f$ is called a morphism of finite type.

A morphism is called proper if it is of finite type, separated and universally closed.

Example. Let $f: {\mathbb A}^2 \rightarrow {\mathbb A}^1$ be the projection on one of the axes. It is not a closed morphism because the projection of $\mathrm{Spec}~ k[x,y]/(xy-1)$ is ${\mathbb A}^1 \setminus \{0\}$ which is open.

Recall that finite type = locally of finite type + quasi-compact. Violating the second condition contradicts properness.

Example. Let $f: X \rightarrow \mathrm{Spec}~{}k$ be a morphism which is not quasi-compact. Then $f$ is not universally closed.

Let $Y=\mathrm{Spec}~\,k[y_1, \ldots, y_n, \ldots]$. Let $Y_i$-s be the principal opens corresponding to $y_i$-s and let $\cup X_i$ be an infinite cover of $X$ which doesn't have a finite subcover. Let $Z$ be the complement in $X \times Y$ of $\cup X_i \times Y_i$ and $f': X \times Y \rightarrow Y$ be the projection onto the second coordinate corresponding to the base change $Y \rightarrow \mathrm{Spec}~\,k$. Then $f'(Z)$ consists of precisely such $y=(y_1, \ldots, y_n, \ldots)$ that $X \setminus \cup_{i\in I} X_i$ is non-empty where $I$ the set of $i$ such that $y_i$ is non-zero.

In particular, since $X$ is not quasi-compact, any $y$ with finitely many non-zero coordinates belongs to $f'(Z)$. Call the set of such $y$-s $V$, then $V \subset Z$. But $V$ is dense in $Y$ since it is not contained in any hypersurface. Indeed, for any hypersurface defined by a polynomial $g(y_1, \ldots, y_n)$ one can find a point $(\alpha_i)$ such that $g(\alpha_1, \ldots, \alpha_n) \neq 0$ and put all the coordinates starting from $n+1$ to 0. Since $Z$ is a proper subset of $Y$, it is not closed.

Theorem 4 (Valuative criterion of properness) Let $f: X \rightarrow Y$ be a morphism of schemes, where $X$ is Noetherian. The morphism $f$ is proper if and only if for any valuation ring $R \subset K$ and any commuting diagram

where $\iota^*$ is the morphism of schemes induced by the inclusion $R \subset K$, there exists a unique morphism $\mathrm{Spec}~ R \rightarrow X$ completing the diagram. The morphism $f$ is universally closed if and only if for any diagram as above there is at least one morphism that completes it.

The proof does not present much difficulty given the auxilary lemmas that we have proved last time. It is worth mentioning that the finite type requirement in the definition of properness is superfluous, and the universal closedness and separatedness criteria work for quasi-compact morphisms (for universal closedness it is crucial that quasi-compact is invariant under a base change).

Theorem 5 An integral morphism is universally closed.  

Proof: It follows from the conjunction of two statements: finite morphisms are closed, being integral is preserved under base change. Let us show the first.

Work locally in an affine open neighbourhood of some point on the destination: we have $R \subset S$ integral extension. So let $Z$ be a closed set and let $E=\overline{f(Z)}$. Consider the restriction of $f$ to $Z$. We need to show that $f$ is surjective onto $E$. Pick a point $x \in E$ and consider its local ring ${\mathcal O}_{E,x}$. We need to show that $\mathfrak{m}_x S \neq S$. We might as well replace $S$ by $S_x$. Note that $S_x$ is integral over $R_x$. Suppose $\mathfrak{m}_x S = S$, and take a finitely generated submodule $S' \subset S$ such that $1 \in \mathfrak{m}_x S'$. Now we can apply Nakayama's lemma to $S'$, $m_x S' = S'$ and we get $S'=0$, which is absurd, and we are done. ☐

This theorems yields an example of a morphism not (locally) of finite type, quasi-compact, universally closed: take a morphism corresponding to an infinite algebraic extension of fields.

Theorem 6 Let $R$ be a subring of a field $K$. Then the integral closure of $R$ is the intersection of all valuation rings that contain $R$.

Proof: It is not hard to see that a valuation ring is integral, from which one inclusion follows. We will prove the other one.

Let $x \notin R^{int}$. Then $x \notin R[1/x]$, for otherwise $x=\sum a_i x^{-i}$ and by multiplying by some $x^n$ we get a monic polynomial that $x$ satisfies. So $1/x$ is not a unit in $R[1/x]$. Then there is a maximal ideal $\mathfrak{p} \subset R[1/x], 1/x \in \mathfrak{p}$. By a result mentioned previously, $R[1/x]_{\mathfrak{p}}$ is dominated by a valuation ring $S$ with a maximal ideal $\mathfrak{q} \not\ni 1/x$. But since $1/x$ is not a unit, $x \notin S$. ☐

Theorem 7 Let $f: \mathrm{Spec}~{}A \rightarrow \mathrm{Spec}~{}B$ be a proper morphism of spectra of integral domains. Then it is finite.

Proof: Apply the criterion for $K=Q(B)$ and let $R$ range through all valuation rings that contain $A$. Then by the criterion $B$ embeds into every such $R$, hence $B \subset A^{int}$. ☐

However useful the valuative criterion may be, a generalisation of the last theorem can in fact be proved directly.

Lemma 8 Let $f: X \rightarrow S$ be universally closed and $g: Y \rightarrow S$ be separated. Then $h: X \rightarrow Y$ such that $f = g \circ h$ is universally closed.

Proof: By universal property of fibre product $h$ factors as $X \rightarrow X \times_S Y \rightarrow Y$ where the first map corresponds to $id$ and $h$ in the universal property diagram and the second one is projection no the second factor in the fibre product.

Consider the base change of the diagonal morphism $Y \rightarrow Y \times_S Y$ by the morphism $X \times_S Y \rightarrow Y \times_S Y$. By general nonsense, the base change is $X \times_Y Y$, since it satisfies the universal property.

Since $\Delta$ is universally closed (as a closed embedding), the base changed morphism $X \times_{h,Y,id} Y \rightarrow X\times_{f,S,g} Y$ is also universally closed. Now notice that $X \times_{h,Y,id} Y \cong X$ and get back to factorisation of $h$ into $X \rightarrow X \times_S Y \rightarrow Y$. We have just shown that the first morphism is universally closed, and the second one is the base change of a universally closed morphism, hence their composition is universally closed. ☐

Theorem 9 Let $f: \mathrm{Spec}\ A \rightarrow \mathrm{Spec}\ B$ be a universally closed morphism of affine schemes. Then it is integral.

Proof: Pick an element $a \in A$ and consider the following diagram

  

where $h$ is the morphism such that $h^*(P)=P(a), P \in B[x]$. We need to show that the ideal $I=(h^*)^{-1}(0)$, which geometrically corresponds to $\overline{f(\mathrm{Spec}~\,A)}$ contains a monic polynomial, it will witness the integrality of $a$.

Since $f$ is universally closed and $h$ is separated, $g$ is universally closed by the Lemma 8. But the same Lemma can be applied to deduce that the composition of $g$ with the inclusion ${\mathbb A}^1_B \hookrightarrow \mathbb{P}_B^1$ is (universally) closed. Therefore $\mathrm{Spec}~{}B[x]/I$ is closed in $\mathbb{P}_B^1$.

The projective line over $B$ is glued together from two affine schemes $U_0=\mathrm{Spec}~\,B[x]$ and $U_1=\mathrm{Spec}~\,B[y]$ glued along $U_{01}=\mathrm{Spec}~\,B[x,1/x]$, the embedding $U_{01} \hookrightarrow U_1$ given by the morphism of rings $B[y] \rightarrow B[x,1/x], y \mapsto 1/x$. For $Z=\mathrm{Spec}~{}B[x]/I$ to be closed in $\mathbb{P}_B^1$ the closure of $Z|_{U_{01}}$ in $U_1$ must be a proper subset of $U_{01}=D(y)$. So,

$$ Z|_{U_{01}}=\mathrm{Spec}~\,B[x]/I'\textrm{, where }I'=\{\dfrac{P(x)}{x^n} \mid P(x) \in I, n \geq 0\} $$

and the closure $Z'=\overline{Z|_{U_{01}}}$ in $U_1$ is $\mathrm{Spec}~\,B[y]/J$, where

$$ J=\{Q(1/y) \mid Q(x) \in I'\} \cap B[y]=\{Q(1/y)y^n \mid Q(x) \in I, n \geq \mathrm{deg}\,Q\} $$

For $Z'$ to be a proper subset of $D(y)$, $J$ and $y$ must generate $B[y]$. Then there exists a polynomial $Q(y)\in J$ such that $\alpha Q(y) + \beta y =1, \alpha, \beta \in B[y], n \geq \mathrm{deg}\,P$. It follows that there exists $P(x) \in I$ such that $\alpha P(1/x)x^n+\beta x=1$ or just $P(1/x)x^n+\beta x=1$ (since $\alpha P \in I$ too). The constant term of $P(1/x)x^n$ is then 1 and $n=\mathrm{deg}\,P$, but this means that the leading term coefficient in $P$ is 1, so we are done. ☐

(I have used the material from Atiyah-Macdonald, Stacks Project and some posts on MO).