Höchste Zeit, unsern Kurs zu ändern :

proper morphisms

Thu, 19 Jan 2012 13:11:00 -0500

This is the continuation of the last post. First I want to prove a couple of algebraic results that were glossed over last time.

Proposition 1 Noetherian value ring is a DVR.

Proof: By the above correspondence and Noetherianity there is an minimal element $l$ of the value group which is $> 0$. Take an arbitrary positive element $k$ of the value group and consider $k-1$, $k-2$, \ldots. Some $k-n$ must be $l$, otherwise $l$ is not minimal, that gives an isomorphism with ${\mathbb Z}$. ☐

Lemma 2 (Atiyah-MacDonald, Theorem~5.21) Let ${\mathcal O} \subset K$ be an integral local ring. Then there exists a valuation ring that dominates it. In fact, it is the maximal local ring that contains ${\mathcal O}$ with respect to the domination order.

Here is the easier dimension 1 version of this statement.

Proposition 3 Let $R$ be a dimension 1 local ring with maximal ideal $\mathfrak{m}$. Take the integral closure of $R$ in $Q(R)$ and let $\mathfrak{p}$ be a maximal ideal in $R^{int}$ such that $\mathfrak{p} \cap R = m$. Then the localisation $R^{int}_p$ is a valuation ring.

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).

valuative criterion of separatedness

Tue, 03 Jan 2012 12:20:00 -0500

Here are some notes on the valuative criterion of separatedness, annoyingly detailed at times.

Let $f: X \rightarrow Y$ be a morphism. Then by the universal property of the fibre product there is a diagonal morphism $\Delta: X \rightarrow X \times_Y X$ such that $p_1 \circ \Delta = p_2 \circ \Delta = id_X$.

The morphism $f$ is is called separated if $\Delta$ is a closed embedding.

Proposition 1 If $X, Y$ are affine, then $f$ is separated.

Proof: Let $X = \mathrm{Spec}{}S, Y = \mathrm{Spec}{}R$. Passing to the corresponding maps of rings we get $\Delta \circ p_1 = \Delta \circ p_2 = id$, $\Delta: S \otimes_R S \rightarrow S$. Then $\Delta$ is defined by $\Delta(s_1 \otimes s_2) = s_1 s_2$. This map is surjective, hence it corresponds to a closed embedding on $\mathrm{Spec}$-s. ☐

Example. Let $X$ be the double affine line (over a field), i.e. $X = {\mathbb A}^1 \sqcup {\mathbb A}^1/\sim$ where $x \sim y \textrm{ iff } x \neq 0 \textrm{ or } y \neq 0$. There is a surjective map $p: X \rightarrow {\mathbb A}^1$. Let $p^{-1}(0)=\{a_1, a_2\}$. Let $f_1, f_2$ be two different sections of $p$, $f_1(0)=a_1, f_2(0)=a_2$. Let $\varphi$ be the morphism such that $p_1 \circ \varphi = f_1, p_2 \circ \varphi = f_2$ that exists by the universal property of the fibre product.

Then $\varphi$ coincides with $\Delta$ on ${\mathbb A}^1 \setminus \{0\}$. In particular, $\varphi.(A^1 \setminus \{0\}) \subset \Delta(X)$. Since $\varphi$ is continuous,

$\displaystyle \overline{\varphi(A^1 \setminus \{0\})} \subset \overline{\Delta(X)} $

but $\varphi(0) = (a_1, a_2) \in \overline{\varphi.(A^1 \setminus \{0\})}$ (again by continuity of $\varphi$) and $(a_1, a_2)$ is clearly not in $\Delta(X)$.

Lemma 2 Let $f: X \rightarrow Y$ be a morphism. Then $f$ is separated if $\Delta(X)$ is closed in $X \times_Y X$.

Proof: Since we know that $\Delta(X)$ is closed, what is left to show is that the morphism of sheaves $\Delta: \mathcal{O}_{X \times_Y X} \rightarrow \Delta^* \mathcal{O}_X$ is surjective. Suffices to check it locally, in a neighbourhood of any point of $\Delta(X)$. But this follows from the affine case. In particular, we can cover $X$ and $Y$ by open affines, $X_i$ and $Y_j$ respectively, such that for any $X_i$ there is a $Y_j$ such that $f(X_i) \subset Y_j$ and apply the previous proposition to $X_i \times_{Y_j} X_i$ (indeed, these sets cover $X \times_Y X$). ☐

A subring $R$ of a field $K$ is called a valuation ring if for any $x \in K$ either $x \in R$ or $x^{-1} \in R$. Then $\mathfrak{m} = R \setminus R^\times \subset R$ is the maximal ideal of $R$. The factor $\Gamma = K^\times / R^\times$ is called the valuation group and the natural projection on the factor, $v: K^\times \rightarrow \Gamma$, is called the valuation map. By declaring the elements of $\mathfrak{m}$ positive one turns $\Gamma$ into a (totally) ordered group.

There is a correspondence

prime ideals of $R$

$\Longleftrightarrow$

subgroups $\Gamma'$ of $\Gamma$ such that $0 < y \leq x, x \in \Gamma'$ implies $y \in \Gamma'$

right to left is given by

$\displaystyle I(\Gamma') = \{ x \in R^\times \mid v(x) \notin \Gamma' \} $

In particular, if $\Gamma = \mathbb R$, $R$ has only one non-trivial prime ideal, $\mathfrak m$. We will further consider only valuation ring of this kind, i.e. $\mathrm{Spec}~R = \{(0), \mathfrak{m}\}$.

Let $X$ be a topological space. The specialisation pre-order on $X$ is defined as

$\displaystyle x \leq y \textrm{ iff } \overline{\{x\}} \subset \overline{\{y\}} $

If $X$ satisfies the $T_0$ separation axiom (for any $x,y \in X$ there is an open $O_x \ni x$, $y \notin O_x$ or there is an open $O_y \ni y, x \notin O_y$) then this is an order. If $x \leq y$ then one says that $x$ is a specilisation of $y$.

Theorem 3 (Valuative criterion of separatedness) Let $f: X \rightarrow Y$ be a morphism of schemes, where $X$ is Noetherian. The morphism $f$ is separated 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 is at most one morphism $\mathrm{Spec}~R \rightarrow X$ completing the diagram

This breaks down into several lemmas.

Lemma 4 Let $R \subset K$ be a valuation ring, $\mathrm{Spec}~R=\{\mathrm{Spec}~k, \mathrm{Spec}~K\}$. Let $\iota: \mathrm{Spec}\,k \rightarrow \mathrm{Spec}~R, \iota': \mathrm{Spec}~K \rightarrow \mathrm{Spec}~R $ be the natural morphisms. Let $f_1, f_2: \mathrm{Spec}~R \rightarrow X$ be two morphisms such that
$\displaystyle \begin{array}{lll} f_1 \circ \iota & = & f_1 \circ \iota\\ f_1 \circ \iota' & = & f_2 \circ \iota'\\ \end{array} $
Then $f_1 = f_2$.

Proof: Let $U=\mathrm{Spec}{}A$ be an open affine neighbourhood of $\iota(\mathrm{Spec}~k) \in X$. Since $f_1, f_2$ are continuous they factor through $U$ therefore we have maps $f_1^*: A \rightarrow R \hookrightarrow K, f_2^*: A \rightarrow R \hookrightarrow K$ that coincide as morphisms to $K$ (since $f_1 \circ \iota' = f_2 \circ \iota'$), but then they just coincide. ☐

Let $R, S$ be two local subrings in $K$ with maximal ideals $\mathfrak{m}, \mathfrak{n}$. Then $R$ dominates $S$ if $R \supset S$ and $\mathfrak{m} \cap S = \mathfrak{n}$.

The stronger version à la Hartshorne:

Lemma 5 Let $R \subset K$ be a valuation ring. To give a morphism $f: \mathrm{Spec}{}R \rightarrow X$ is the same as to give the following data:

- $x_0, x_1 \in X$, $x_0 \in \overline{\{x_1\}}$;

- inclusion $\kappa(x_1) \subset K$ such that $R$ dominates $\mathcal{O}_{\overline{\{x_1\}}, x_0}$.

Proof: Let $f: \mathrm{Spec}{}R \rightarrow X$ be a morphism. Let $x_0 = f(\mathrm{Spec}{}k)$, $x_1 = f(\mathrm{Spec}{}K)$. Take an affine $U \ni x$, $U = \mathrm{Spec}{}A$. Then the restriction of $f$ to $\mathrm{Spec}{}K$ gives a morphism $A \rightarrow K$. Let $\mathfrak{p}_0$ be the ideal of $A$ corresponding to $x_0$, and $\mathfrak{p}_1$ be the ideal corresponding to $x_1$, $\mathfrak{p}_1 \subset \mathfrak{p}_0$.

Remarks:

$(A/\mathfrak{p}_1) \otimes A_{\mathfrak{p}_1} = A_{\mathfrak{p}_1} / \mathfrak{p}_1 A_{\mathfrak{p}_1} = Q(A/\mathfrak{p}_1)$;
$A/\mathfrak{p}_1 \rightarrow R$ is an inclusion, and since $A/\mathfrak{p}_1$ is an integral domain, $A/\mathfrak{p}_1 \rightarrow (A/\mathfrak{p}_1)_{\mathfrak{p}_0}$ is an inclusion too;
we know that $\mathfrak{m} \cap A/\mathfrak{p}_1 = \mathfrak{p}_0$, it follows by tensoring with $(A/\mathfrak{p}_1)_{\mathfrak{p}_0}$ that $\mathfrak{m} \cap (A/\mathfrak{p}_1)_{\mathfrak{p}_0} = \mathfrak{p}_0 \otimes (A/\mathfrak{p}_1)_{\mathfrak{p}_0} = \mathfrak{m}_{x_0}$
$(A/\mathfrak{p}_1)_{\mathfrak{p}_0} \subset R$. Indeed, an element from the first ring is of the form $a/b$ where $a \in A_{\mathfrak{p}_1}, b \notin \mathfrak{p}_0$. Since, $a \in R$, the question boils down to whether $b$ is invertible in $R$. But it is, since it does not belong to the maximal ideal;

Therefore, $R$ dominates over $\mathcal{O}_{\overline{x_1},x_0}$.

Given the data as in the statement of the lemma, choose an affine $U = \mathrm{Spec}{}A$, $x_0 \in U$. We are given the inclusion $\kappa(x_1) = Q(A/\mathfrak{p}_1) \rightarrow K$, therefore we have a morphism $A \rightarrow K$. We need to prove that it factors through $R$ and that the preimage of $\mathfrak{m} \subset R$ is $\mathfrak{p}_0 \subset A$, the ideal corresponding to $x_0$.

Since $R$ dominates $(A/\mathfrak{p}_1)_{\mathfrak{p}_0}$, $R \supset (A/\mathfrak{p}_1)_{\mathfrak{p}_0} \supset A/\mathfrak{p}_1$, and therefore our morphism goes actually to $R$: $f: A \rightarrow R$. The other part of the definition of domination tells us that $\mathfrak{m} \cap (A/\mathfrak{p}_1)_{\mathfrak{p}_0} = \mathfrak{p}_0 \otimes (A/\mathfrak{p}_1)_{\mathfrak{p}_0}$. Taking intersection with $A/\mathfrak{p}_1$ we get $\mathfrak{m} \cap A/\mathfrak{p}_1 = \mathfrak{p}_0 \otimes A/\mathfrak{p}_1$, which was to be shown.

☐

Lemma 6 (EGA II.7.2.1, essentially uses EGA I.6.6.5) Let $f: X \rightarrow Y$ be a quasi-compact morphism. Then $f(X)$ is closed in $Y$ if and only if $f(X)$ is stable under specialisations (i.e. $x \leq y$, $y \in f(X)$ implies $x \in f(X)$).

Proof: Given $f: X \rightarrow Y$ there exists $f': X_{red} \rightarrow Y_{red}$ that commutes with the natural projections, so we might as well suppose $X$ and $Y$ reduced. We also can replace $Y$ by $\overline{f(X)}$. So suppose $f(X)$ is stable under specialisations and $y \in \overline{f(X)}$. We need to show that $y \in f(X)$. Consider an affine open $U \ni y$. It suffices to show $y \in f(X) \cap U$.

Since $f$ is quasi-compact, $X$ is covered by a finite number of open affines $X = \cup X_i$. Therefore $y \in \overline{f(\cup X_i)}$ and since there are finitely many $X_i$-s there is an $i$ such that $y \in \overline{f(X_i)}$ (that's where we have used quasi-compactness). $\overline{f(X_i)} \cap U$ is closed hence affine, say $\mathrm{Spec}{}B$, let $X_i = \mathrm{Spec}{}A$. We have an inclusion $B \hookrightarrow A$ (since $f$ is dominant). Let $\mathfrak{p}'$ be the minimal prime ideal contained in $\mathfrak{p}_y \in \mathrm{Spec}{}B$.

Consider the localisation of $A$ at $\mathfrak{p}'$ as a $B$-module. Then $B_{\mathfrak{p}'} \hookrightarrow B_{\mathfrak{p}'} \otimes A$. But $B_{\mathfrak{p}'}$ is a field. So any proper ideal $\mathfrak{q}' \subset A \otimes B_{\mathfrak{p}'}$ has the property $\mathfrak{q}' \cap B_{\mathfrak{p}'} = (0)$.

Let $\mathfrak{q}=h^{-1}(\mathfrak{q}')$ where $h$ is the localisation map. Then $\mathfrak{q} \cap B = \mathfrak{p}'$.

☐

Lemma 7 (Atiyah-MacDonald, Corollary 5.22) Let $\mathcal{O} \subset K$ be an integral local ring. Then there exists a valuation ring in $K$ that dominates $\mathcal{O}$. It the maximal ring among local subrings of $K$ ordered by domination relation.

Proof of the valuative criterion.
Necessity. Given that $\Delta(X)$ is stable under specialisations suppose for contradiction that $h_1, h_2: \mathrm{Spec} R \rightarrow X$ are two morphisms that make the diagram commute. By the universal property of fibre product there exists a morphism $\varphi: \mathrm{Spec} R \rightarrow X \times_Y X$ such that $p_1 \circ \varphi = h_1, p_2 \circ \varphi = h_2$. Since the upper triangle of the diagram commutes, $\mathrm{Spec} K \in \mathrm{Spec} R$ is mapped by $\varphi$ to $(\mathrm{Spec} K, \mathrm{Spec} K)$. But $\varphi(\mathrm{Spec} k)$ is a specialisation of this point, hence also belongs to $\Delta(X)$. Therefore restrictions of $h_1$ and $h_2$ to $\mathrm{Spec} k$ coincide and hence $h_1 = h_2$ (by our lemma).

Sufficiency. We need to prove that $\Delta(X)$ is stable under specialisations. Let $x_1 \in \Delta(X)$ and $x_0$ be its specialisation. Then let $K = \kappa(x_1)$ and let $\mathcal{O} \subset K$ be the local ring of $x_0$ on $\overline{\{x_1\}}$ with induced reduced scheme structure. There is a valuation ring $R$, a subring of $K$, dominationg $\mathcal{O}$. We therefore have a morphism $\varphi: \mathrm{Spec}{}R \rightarrow \Delta(X)$, and $\varphi(\mathrm{Spec}{}k)=x_0$. Therefore, $x_0 \in \Delta(X)$. ☐

Bijective holomorphic maps of complex spaces

Sat, 24 Dec 2011 11:13:00 -0500

Here is a another take on the problem discussed in the previous post, this time in holomrophic context. We want to show that

A bijective holomorphic map $f: U \to V$ is invertible in the neigbourhoods of smooth points of $V$.

First, the definitions: a complex space $X$ is a thing locally biholomorphically equivalent to zero sets of ideals of complex analytic functions defined on an open neighbourhood of $0 \in \mathbb{C}^m$. If a neighbourhood of a point $x \in X$ is isomorphic to an open of $\mathbb{C}^n$ then $x$ is called a smooth point. So it suffices to prove the invertibility for holomorphic maps between open sets of $\mathbb{C}^n$.

Theorem. (Inverse function theorem) If the Jacobian of a funciton $f: U \to V$ between open neighbourhoods of $0$ in $\mathbb{C}^n$ is non-degenerate then $f$ has a holomorphic inverse.

Propostion. (Griffiths-Harris "Principles of algebraic geometry", p.19). If $f: U \to V$ is a bijective holomorphic map between open neigbourhoods of $0$ in $\mathbb{C}^n$ then $f$ has a holomorphic inverse.

Proof. For n=1, suppose that $J(f)=0$ then after coordinate change $f(z)=z^2 g(z)$, or even $f(z)=z^k g(z)$, where $g(z)$ is non-zero on $U$. Then by possibly shrinking $U$ to a smaller neighbourhood of $0$ we can find a function $h(z)$ such that $h^k(z) = g(z)$ on $U$. Then $f(z)=(z\,h(z))^k$ and therefore $f$ is not bijective.

Suppose the statement is proved for $n-1$. Let $f$ be bijective and let the Jacobian $J(f)(0)$ be degenerate, i.e. $\mathrm{rk} J(f)(0) = k < n$. Suppose that the coordinate functions $f_1, \ldots, f_n$ are ordered so that $\left(\dfrac{\partial f_i}{\partial z_j}\right)_{i,j=1}^k$ is non-degenerate. Consider the restriction

$f': U \cap f^{-1}(\mathbb{C}^{n-k}) \to V \cap \mathbb{C}^{n-k}$

where $\mathbb{C}^{n-k}$ is the embedding $(z_1, \ldots, z_{n-k}) \mapsto (0, \ldots,0, z_1, \ldots, z_{n-k})$.

Let $g: (z_1, \ldots, z_n) \mapsto (f_1(z), \ldots, f_k(z), z_{k+1}, \dots, z_n)$. Since $J(g)(0)$ is non-degenerate, $g$ is invertible in some neighbourhood of $0$. Then $f'\circ g^{-1}$ maps $\{0, \ldots, 0\} \times U'$ to $\{0, \ldots, 0\} \times V'$, where $U',V' \subset \mathbb{C}^{n-k}$. Call $h$ the restriction of $f' \circ g^{-1}$ to last $n-k$ coordinates.

If $k > 0$ then we can apply the induction hypothesis to $h$. It is bijective, and therefore $J(h)(0)$ is non-degenerate. But $h$ coincides with last $n-k$ coordinates of $f$, and hence $J(h)(0)$ is a $(n-k)\times(n-k)$ minor in $J(f)(0)$. It follows that $J(f)(0)$ is non-degenerate, which contradicts the initial assumption.

Therefore $k=0$, i.e. $f$ is constant. This also cannot be as $f$ is bijective.

ZMT and bijective regular morphisms

Fri, 23 Dec 2011 07:33:00 -0500

Тут понадобилось понять, почему биективные морфизмы алгмногообразий бирегулярны. Точнее, бирегулярны они когда бьют в нормальное многообразие. Да, ещё надо потребовать, чтобы морфизм сепарабельный был (в том смысле, что расширение полей рацфункций должно таковым быть), потому что Фробениус очевидный контрпример.

Многообразия над полем характеристики 0 можно нарезать на гладкие куски (by generic smoothness), которые будут нормальны. Поэтому биективный морфизм будет кусочно бирегулярен. Кстати, это утверждение тривиально следуствие элиминации кванторов в $\textrm{ACF}_p$. А какое будет "честное" алггеометрическое доказательство? Это следствие из Zariski's Main Theorem в одной из её форм. С ZMT в учебниках какая-то путаница. В Хартсхорне этим называется одно, в EGA и у Милна другое, в красной книге Мамфорда -- штук пять разных утверждений, причём эквивалентности доказаны не между всеми. Далее — разбор (engl.)

A rational map $X \to Y$ is an equivalence class of maps defined on open subsets of $X$, modulo coincidence on open subsets. A birational morphism/map is a morphism which has the property that its restriction on some open subset is an isomorphism between open subsets of $X$ and $Y$. So, a rational map is not really a map, but a birational map is a real map. Of course, a birational map defines a rational map (which is invertible as a rational map; by the way, any rational map is a composition of an inverse of a birational map and a regular one).

A rational map can also be defined via its graph: it is a closed subscheme $\Gamma$ of $X \times Y$ such that for some open $U \subset X$ the projection on $U$ induces an isomorphism. There is a maximal such open $U$, the complement of it is called the set of fundamental points (or fundamental domain or indetemrinancy locus). The map is a real map on $U$, however, on the fundamental points it is a multi-valued map. An image of a fundamental point is then some set, it is obtained by composing $p_1^{-1}$ and $p_2$. It is called a total transform.

Zariski wanted to prove something about birational maps. Namely

Theorem. (original ZMT, I) Let $f: X \to Y$ be a birational map of projective varieties, and assume $X$ is normal. If x \in X is a fundamental point then its total transform is connected (and of dimension $> 0$).

Sometimes it is stated a bit more generally:

Theorem. (original ZMT, II) Let $f: X \to Y$ be a birational projective map of Noetherian integral schemes and assume that $Y$ is normal. Then for every $y \in Y$ $f^{-1}(y)$ is connected. (another verison is that you do not suppose the whole $Y$ to be normal, but just the local ring of your point $y$)

To get ZMT-I, apply ZMT-II it to the $p_1$ of the graph of $f$.

In EGA III there are two sections called "théorème de connexion de Zariski" (EGA III, 4.3) and "le Main Theorem de Zariski (EGA III, 4.4). The key statements are

Theorem. (4.3.1, Stein factorisation) Let $f: X \to Y$ be a proper morphism, $Y$ locally Noetherian. Then $f$ factors into a compositon of a proper morphism with connected fibres and a finite morphism.

Theorem. (4.4.1, "Main Theorem") Let $Y$ be a locally Noetherian prescheme and $f:X \to Y$ be a proper morphism. Let $X′$ be the set of points in $X$ which are isolated in their fibre $f^{−1}(f(x))$. Then $X′$ is an open subset of $X$ and if $f = g \circ f′: X \to Y′ \to Y$ is the Stein factorisation of f, the restriction of $f′$ to $X′$ is an isomorphism of $X′$ onto a induced subscheme of on an open $U \subset Y′$.

These theorems together imply the original ZMT.

In EGA IV, Grothendieck proves a similar statement ("something factors into an open immersion and and a finite morphism")

Theorem. (ZMT in Grothendieck's form) Let $f: X \to Y$ be a separated quasi-finite map and $Y$ be quasi-compact. Then $f$ factors into a composition of an open immersion and a finite morphism.

It is a bit confusing why this is a "Zariski's Main Theorem" as it doesn't say anything about connectedness. However, if $f$ is projective then the set of points isolated in their fibre is quasi-finite, so by applying this theorem we get EGA III, 4.4.1. On the other hand if one supposes $f$ to be quasi-projective and quasi-finite, the theorem follows by EGA III, 4.4.1.

Anyway, it's a powerful structure theorem and it's precisely what we need for our purposes.

Proposition. Let $f: X \to Y$ be a bijective morphism of varieties over an algebraically closed field. Suppose that $Y$ is normal. If char=0 then it is an isomorphism, if char > p then it is an isomorphism possibly composed with a power of the Frobenius.

Proof. By Grothendieck's ZMT, $f$ is finite. Suppose $X$ and $Y$ are affine. Consider the field extension $k(X)/k(Y)$.

If the extension is non-trivial and separable, take an element $h \in k(Y), a \notin k(X)$. Then the minimal polynomial $P(z)$ of $h$ will have distinct roots. The discriminant of $P(z)$ is therefore not identically zero. Then there will be an open subset of the set $U \subset Y$ where $h$ is regular and where $h$ takes more than one value. Hence $f$ is not bijective, contradiction.

Othewise, if the extension is trivial then we are done, as $k[Y]$ is integrally closed in $k(Y)=k(X)$ and $k[X]$ is an integral extension.

Since any field extension is a sequence of a separable and purely inseparable extensions, the only case left is a purely inseparable extension. In this case $f$ is just a power of the Frobenius morphism.

неконечнопорождённые кольца глобальных функций

Tue, 20 Dec 2011 07:22:00 -0500

фан: бывают квазипроективные многообразия с неконечнопорождённым кольцом глобальных функций.

надо взять эллиптическую кривую (или большего рода), обратимый пучок $N$ степени 0 на ней (но такой, чтобы он ни в какой степени не давал тривиальный) и какой-нибудь обратимый пучок положительной степени $P$. Взять тотальное пространство их прямой суммы. Шутка в том, что тотальное пространство это относительный $\textrm{Spec}$ симметрического произведения, глобальные сечения у него будут суммы глобальных сечений всевозможных тензорных произведений $N^m \otimes P^k$. У $N \otimes \mathcal{O}$ нет сечений, кроме констант, однако у разных произведений $N^m \otimes P$ сечения не те же, что и у $P$. Получается, что каждая степень $N$ даёт новые генераторы.

больше примеров здесь.
по следам вопроса на МО: раз, а так же два.

кря-кря

Tue, 07 Dec 2010 19:22:00 -0500

Еще вкруг солнцев не вращались / В превыспренных странах миры, / Еще в хаосе сокрывались / Сии висящие шары

$\sum_{n=0}^\infty$

Придумал тут сценарий для новогоднего огонька

Sun, 05 Dec 2010 10:40:00 -0500

(Сохраняю сие сочинение Димы Самвоара)

Это фильм про московских студентов, не дураков каких-нибудь, все герои фильма пепельные блондины, словно братья-близнецы, как в старом фильме ужасов про детей-убийц из космоса. Фильм начинается с того, что показывают цветы, замерзающие под снегом довольно тугого хвойно-лиственного леса. Голос за кадром говорит, что это подснежники, цветы, которые цветут под снегом. Он говорит холодно-поэтично, дикторским глубоким голосом. Мы назовем его диктор №1.

«Подснежники живут в толще снега и распускают свои нежные бледные бутоны тогда, когда все осальные растения давно мертвы или пребывают в зимнем сне, сладком как крыжовник. Подснежник найти — большая удача. Дети, найдите подснежник, найдите подснежник, пожалуста». Камера неловко отодвигается и упирается в какой-нибудь нейтральный по содержанию, но гармоничный по форме предмет — ворох хвороста, кусок крупной грубой древесной коры и смешно исчерченный птичьими лапками снег. В это время зритель должен подумать о том, что происходитв России и мире, для этого выводится специальная табличка на экран. Далее мы переносимся в пейзаж, выполненный на очень простой компьютерной программе трехмерного моделирования, такой, что его в принципе мог бы выполнить студент какого-нибудь технического провинциального вуза (лучше именно его и нанять для выполнения фильма). В пейзаже примитивная детализация, изображаются потоки лавы, извергающийся шипящий вулкан, бурлящие гейзеры. За кадром женский голос с немного северорусским акцентом, читает плохо, но не от того,что текст плохо выучен, а от волнения, текст наоборот идет как бы самсобой, «своими словами».В тексте говорится о недавнем открытии в глубине серного озера нового вида бактерий, которые в невероятно сложных условиях жизни в своем метаболизме заменили фосфор (ранее считавшийся необходимым компонентом всех живых организмов) на мышьяк. При этом при наличии фосфора они также способны использовать и его.

Потом за кадром появляется голос Диктора №1, который говорит: «Что нам далекие планеты, ледяные поля, кошмарная эта высь, бездна эта. Три саженя в землю, да две ручки к сердцу.»

Женский голос рассказывает о механизмах преобразования фосфора в организмах «Именно поэтому ...» Дальше идет сугубо научный и осмысленный текст о бактериях анаэробах.

Мужской голос снова делается слышен:«.. дыхание занавесок этих как паруса последнего корабля уходящего за кошмарное море, полное бесмысленной воды. Поднять паруса, отправляемся. Когда ваша планета превращается в облако раскаленного газа, нужно думать о потоках, и все пройдет. На нашем шаре всегда найдется две точки, где нет ветров, в силу свойств самого»

Потом оба диктора снова в своих обычных положениях: вулканном пейзаже и декабрьском ледяном лесу. Сначала первый диктор говорит «А теперь посмотрим, как плохие ребята дерутся на ножах, и камера сдвигаетсяк месту, где лес как-бы кончается, как декорация, и появляются прибрежные скалы, покоторым прыгают и остервенело деруться два очень похожих человека, одеты они в тонкие элегантные костюмы. Тут важно показать, что дерутсяони жестоко и отчаянно, а не красиво. Лучше всего поставить после прочтения книжки о сицилийском ножевом бое. При свете луны и ее отражении в гладком море. Дерущиеся издают лишь зверообразные звуки и напоминают огромных котов. Второй диктор отвлекается от анаэробных бактерий и предлагает так же посмотреть на двух людей, их играют те же два актера, только теперь они два брата, в голландском идиллическом пасторальном пейзаже, ослепительно, до идиотизма красивом и выполненном на очень качественной пленке, переход от дешевого к дорогому качеству происходит как-бы постепенно. Братья продали мельницу и теперь пьют водку в кабаке. В кабаке к ним подходит человек и говорит с белорусским дзекающим акцентом: «С чего бы эта склонность к противопоставлению, эти апологии конспирологии, бесконечное множество мысленных схем сводится лишь к противопоставлению двух фундаментальных сил, все от этого отказалисьуже давно, потому что тупо и устарело, да и не было никогда умно-то». Потом берет гармошку в руки, начинает приплясывать и петь: «Евразизм-Атлантизм // Цивилизация вина - цивилизация пива // Цивилизация вины - цивилизация стыда // Городское - сельское // Сложное - простецкое». Потом мы видим пышные облака, они медленно плывут и кроме них ничего не видно на экране. Изображение нужно как-то испорить, чтобы появились разные цвета - красный, голубой, картинка чтобы была немного расплывчатой, но не мутной, а как бы наложенной из 15-20 чуть различных картинок. Становится ясно, что последний корабль уплыл, что паруса вдали, они белые. Из Эгейского моря выныривает старик, лучше взять какую-нибудь архивную пленку для этой цели. Резкий конец. Титров нет никаких, звучит ветер, гудящий под крышей, такой трубный звук.

Становится ясно, что диктор №1 и диктор №2 это Дед Мороз и Снегурочка и также становится ясно, что ничего хорошего не будет.

Это должен быть очень злой фильм, и очень некрасивый, роль разбойника должен играть Гоша Куценко, вообще лучше всего набрать самых популярных актеров -- Сергея Безрукова, Сергея Дроботенко, Леонида Утесова, Актера Который Играл в Фильме «Ирония Судьбы». И дожен быть суровый фильм, как сурово смотрит человек, ополаскивающийся после бани на человека, не ополаскивающегося после бани, но при этом не понимающего что не так. Это должен быть фильм о скверне и об интуициях скверны. Фильм о средневековье и странных обычаях.

Morphisms of schemes

Tue, 19 Oct 2010 17:58:00 -0400

Recall that a scheme is a locally ringed space ${(X, {\mathscr O}_x)}$, i.e. a topological space and a sheaf of rings, such that for any ${x \in X}$ the stalk ${{\mathscr O}_{x,X}}$ is a local ring. Recall that the stalk is ${\left.\varprojlim_{U \ni x}\right. {\mathscr O}_X(U)}$.

An affine scheme is the scheme one gets from a commutative ring ${A}$ as follows:

$\displaystyle \begin{array}{l} X = \mathrm{Spec} A = \{\mathfrak{p} \subseteq A \textrm{, prime }\}\\ \quad \quad \quad \textrm{ with the base of topology } X_f = \{\mathfrak{p} \mid f \notin \mathfrak{p}\} \textrm{ where } f \in A\\ {\mathscr O}_X(X_f)=A_{(f)} \end{array} $

Let ${f: Y \rightarrow X}$ be a map of ringed spaces and ${{\mathscr F}}$ a sheaf on ${X}$. A pullback sheaf ${f^{-1}{\mathscr F}}$ is defined as follows:

$\displaystyle f^{-1}{\mathscr F}(U) = \varprojlim_{V \supseteq f(U)} {\mathscr F}(V) $

One has the natural map ${f^*: {\mathscr F} \rightarrow f^{-1}(F)}$. It is called inverse image map. One similarly defines ${f_*: Shv(X) \rightarrow Shv(Y)}$ and ${f^{-1}: \Gamma({\mathscr F},U) \rightarrow \Gamma(f_*({\mathscr F}),U)}$ for all ${U \subseteq X}$.

A morphism of locally ringed spaces is a morphism of underlying spaces ${f: Y \rightarrow X}$ and a morphism of sheaves ${f^*: f^{-1}{\mathscr O}_X \rightarrow {\mathscr O}_Y}$ (or ${f^*: {\mathscr O}_Y \rightarrow f_*{\mathscr O}_X}$), it is called local if the induced stalk morphisms map the maximal ideal to the maximal ideal.

Morphisms of affine schemes are induced by morphisms of rings (this is a theorem).

Non-local morphisms of ring spaces are not induced by ring morphisms. Example: a local ring and its field of fractions.

A morphism ${f: Y \rightarrow X}$ of schemes is called a closed immersion if it induces a homeomorphism between the underlying topological spaces of ${Y}$ and a closed subset of ${X}$ and is surjective on sections.

A scheme ${X}$ is called separated in the diagonal morphism ${\Delta: X \rightarrow X \times X}$ is a closed immersion (relative definition over some scheme ${S}$ is also possible).

Remark A category of schemes over a scheme ${S}$ in general does not have products, only fibred products, but it does have products if ${S}$ is separated.
A category of schemes over a scheme ${S}$ in general does not have a final object, but it does have one if ${S}$ is affine.

Imaginaries 1

Tue, 19 Oct 2010 17:40:00 -0400

Imaginaries after Poizat («Une théorie de Galois imaginaire».

A theory is said to weakly eliminate imaginaries if every formula ${\varphi}$ has a minmal algebraically closed set of parameters (i.e. a set ${A=\mathrm{acl}(A)}$ such that ${\varphi(x,b) = \psi(x,a)}$ with ${a\in A}$).
A theory is said to eliminate imaginaries if every formula ${\varphi}$ has the (definably closed) canonical set of parameters or canonical base, i.e. any automorphism preserves ${\varphi}$ set-wise iff it preserves ${B=\mathrm{dcl}(B)}$ point-wise.

Clearer approach to imaginaries.

A theory ${T}$ is said to weakly eliminate imaginaries if for every ${a \in M^\mathrm{eq}}$ there exists ${b \in M^n}$ such that ${b \in \mathrm{acl}(a)}$, ${a \in \mathrm{dcl}(M)}$
A theory ${T}$ is said to eliminate imaginaries if for every ${a \in M^\mathrm{eq}}$ there exists ${b \in M^n}$ such that ${b \in \mathrm{dcl}(a)}$, ${a \in \mathrm{dcl}(M)}$

The first is equivalent to the second (proved in Poizat).

Strong elimination of imaginaries is the weak elimination of imaginaries plus elimination of finite imaginaries, i.e. those such that the equivalence classes are finite.

Notable example of weak elimination:

Theorem 1 Let ${T}$ be a strongly minimal theory with infinite ${\mathrm{acl}{}(\emptyset)}$. Then ${T}$ weakly eliminates imaginaries.

Proof: Let ${e \in M^{eq}}$ be the image of a 0-definable function, ${f(\bar{b})=e}$, ${\bar{b} \in M^n}$. We need to prove that there are only finitely many such ${\bar{c}}$ that ${f(\bar{c})=e}$. It boils down to finding one such ${\bar{c}}$ in ${\mathrm{acl}{}(e)}$. Suppose we have already found such ${c_0}$ that ${\exists x_1\ldots x_n f(c_0,x_1,\ldots,x_n)=e}$. Then let ${g(x_1)=\exists x_2\ldots x_n f(c_0,x_1,x_2,\ldots,x_n)}$. Since ${T}$ is strongly minimal, ${g}$ defines either a finite or cofinite set in ${M}$. If it is finite, then it belongs to ${\mathrm{acl}{}(e)}$ and we take on of its elements as ${c_1}$. Otherwise, it is cofinite and it intersects with ${\mathrm{acl}{}(M)}$, because the latter is infinite, and we can pick a ${c_1}$ from the intersection. By iterating this process, we get ${\bar{c}}$. ☐

Thus, ACF weakly eliminates imaginaries.

Theorem 2 ACF eliminates finite imaginaries as well.

Proof: Symmetric polynomials ${p_1(\bar{x}), \ldots, p_k(\bar{x})}$ are by definition the generators of ${k[x_1,\ldots,x_n]^{\Sigma_n}}$. It follows that ${p_i(\bar{a})=p_i(\bar{b})}$ for all ${i}$ iff ${\bar{a}}$ is a permutation of ${\bar{b}}$.

A finite imaginary is 0-interdefinable with an imaginary for the equivalence relation ${(\bar{x}_1, \dots, \bar{x}_n) \sim_E (\bar{y}_1, \dots, \bar{y}_n)}$ where ${\bar{y}_i=\bar{x}_{\sigma(i)}}$ for some permutation ${\sigma}$.

The function ${f(\bar{z})=\langle p_i(\ \prod_j (z-x_{1j}),\ldots, \prod_j (z-x_{nj})\ ) \rangle}$ eliminates such an imaginary. ☐

Affine schemes and Nullstellensatz

Tue, 19 Oct 2010 15:07:00 -0400

(These are the notes I made for a talk at Junior Logic Seminar at Oxford.)

The category of affine schemes is the category opposite to the category of commutative rings.

Let $A$ be a ring. One associates to it a topological space called $\mathrm{Spec} A$:

$$\begin{array}{l}\mathrm{Spec} A = \{ \textrm{prime ideals of } A \} \textrm{ with closed sets } \\ \quad \quad \quad X_\mathfrak{a} = \{\mathfrak{p} \subseteq \mathfrak{a} \mid \mathfrak{p} \in \mathrm{Spec} A\} \textrm{ for prime } \mathfrak{a}\in A\end{array}$$

Clearly, the points that correspond to maximal ideals are closed. If $\mathfrak{m} \subset A$ is a maximal ideal, $A/\mathfrak{m}$ is called the residue field of $\mathfrak{m}$. In classical algebraic geometry one deals with schemes over a field. Affine scheme over a field $ k$ is $ \mathrm{Spec} A$, where $ A=k[x_1,\ldots,x_n]/\mathfrak{a}$.

Proposition. Let $ A$ be the coordinate algebra of an affine scheme over $ k$. Let $ \mathrm{Max}_k\,A$ be the subset of maximal ideals with the residue field $ k$. Then $ \mathrm{Max}_k\,A\cong V(\mathfrak{a})=\{\bar{a}\in k^n \mid\forall f(\bar{x}) \in\mathfrak{a} f(\bar{a})=0\}$. Note that in general one cannot recover $ A$ from $ \mathrm{Spec} A$: consider $ k[x]/(x^2)$.

Theorem. Let $ K \supset k$ be a field extension, and $ K$ is finitely generated over $ k$ with generators $x_1, \ldots, x_n \in K$. Then $K$ is algebraic over $k$.

Proof. Regroup $x_1, \ldots, x_n$ so that $ K_0=k(x_1, \ldots, x_r)$ is a transcendental extension of $ k$ and $ K_{i-r}=K_0\langle x_{r+1}, \ldots, x_{i-1} \rangle$, $ x_i$ algebraic over $ K_{i-r}$ for $ i > r$. Let $ P_i(x)$-s be minimal polynomials of $ x_i$-s. They are polynomials over $ K_0$, i.e. have rational functions of $ x_1,\ldots,x_r $ as their coefficients, so there is a least common denominator of these, a polynomial $ g \in k[x_1,\ldots, x_r]$, such that $ P_i$ has coefficients in the ring $B=k \langle x_1,\ldots, x_r, g^{-1} \rangle$. Any element of the field $K$ has the form

$$ \sum b_\alpha\displaystyle\mathop{\prod}_{r+1 \leq e_i \leq d_i} x^{e_i}$$

where $\alpha$ is a multi-index, and $d_i$ are degrees of polynomials $P_i$, and $b_\alpha \in B$. From this one can see that if $I$ is a proper ideal of $ B$ (as a ring), then $ IK$ is a proper ideal of the field $ K$. Since $ I$ can only be trivial, $ B$ is a field. In other words it contains every rational function of $ x_1, \ldots, x_r, g^{-1}$, which cannot be true. ☐

Corollary. If $ A$ is finitely generated over an algebraically closed field then $ \mathrm{Max}_k A=\mathrm{Max} A$. ☐

Corollary. [Nullstellensatz] Let $ k$ be an a.c.f. and $ \mathfrak{a} \subset k[x_1,\ldots,x_n]$ be an ideal. Then $V(\mathfrak{a})$ is empty iff $ \mathfrak{a}=k[x_1,\ldots,x_n]$.

Model-theoretical proof of Nullstellensatz.

Theorem. [Nullstellensatz, weaker form. Together with Hilbert's basis theorem implies the version in the previous section] Let $ k$ be an algebraically closed field, $ I \subset k[x_1, \ldots,x_n]$ a proper ideal, generated by polynomials $ f_1, \ldots,f_m$. Then these polynomials have at least one common zero in $k^n$. Proof. First prove that

Claim. $ACF_p$ has quantifier elimination.

Suffices to construct a family of partial isomorphisms $ I$ between finitely generated substructures of two sufficiently saturated models $ ACF_p \models k_1, k_2$. $ I$ is non-empty: there is at least the partial isomorphism that identifies prime fields of $ k_1$ and $ k_2$. Construct the rest by induction. Let $ f$ be a partial isomorphism between substructures $ R_1 \subset k_1$, $ R_2 \subset k_2$ (remember, the language is just the language of rings) related by the family of partial isomorphisms. Let $ L_1$, $ L_2$ be the fraction fields of $ R_1$, $ R_2$. Pick an element $ a \in k_1-L_1$. There are two cases to consider:

$ a$ is algebraic over $ L_1$ ; then pick the minimal polynomial of $ a$ and extend $ f$ by putting $ f(a)=b$ where $ b$ is a solution of the minimal polynomial of $ a$ in $ k_2$;

$ a$ is transcendental over $ L_1$; one can choose any transcendental (over $ L_2$) element $ b$ of $ k_2$ and extend $ f$: $ f(a)=b$. The question is if there is any such element. Since any polynomial only has finitely many roots, the set of formulae $ \{ P(\bar{x}) \neq 0 \mid P \in L_2[\bar{x}]\}$ is finitely satisfiable (and is a type over a finite set, since $ L_2$ is finitely generated). By saturatedness (in fact $ \aleph_0$-saturatedness would saffice), it has a solution in $ L_2$.

Now take the smallest prime ideal $ J$ that contains $ I$. It is not the whole of $ k[\bar{x}]$ unless $ I$ is. Let $ L$ be the field of fractions of the quotient $ k[x_1, \ldots, x_n]/J$, then $ k$ canonically identifies with a substructure of $ K$. Clearly, $ f_i(\bar{x}) = 0$ in $ K$, and all the more in the algebraic closure of $ K$, $ \bar{K}$. By quantifier elimination, $ k$ is an elementary substructure of $ \bar{K}$ and so has a solution to $ f_1,\ldots,f_n$. ☐