§8. Chevalley Schemes

8.1. Allied local rings

For every local ring $A$, we shall denote by $\mathfrak{m}(A)$ the maximal ideal of $A$.

Lemma (8.1.1). Let $A$, $B$ be two local rings such that $A\subset B$; the following conditions are equivalent: (i) $\mathfrak{m}(B)\cap A=\mathfrak{m}(A)$; (ii) $\mathfrak{m}(A)\subset \mathfrak{m}(B)$; (iii) $1$ does not belong to the ideal of $B$ generated by $\mathfrak{m}(A)$.

Proof. It is evident that (i) implies (ii) and that (ii) implies (iii); finally, if (iii) is verified, $\mathfrak{m}(B)\cap A$ contains $\mathfrak{m}(A)$ and does not contain $1$, hence is equal to $\mathfrak{m}(A)$.

When the equivalent conditions of (8.1.1) are satisfied, one says that $B$ dominates $A$; it comes to the same thing to say that the injection $A\to B$ is a local homomorphism. It is clear that, in the set of local subrings of a ring $R$, the relation of domination is an order relation.

(8.1.2) Let us now consider a field $R$. For every subring $A$ of $R$, we shall denote by $L(A)$ the set of local rings $A_{\mathfrak{p}}$, where $\mathfrak{p}$ runs through the prime spectrum of $A$; they are identified with subrings of $R$ containing $A$. Since $\mathfrak{p}=(\mathfrak{p}{A}_{\mathfrak{p}})\cap A$, the map $\mathfrak{p}\to A_{\mathfrak{p}}$ from $\mathrm{Spec}(A)$ to $L(A)$ is bijective.

Lemma (8.1.3). Let $R$ be a field, $A$ a subring of $R$. In order that a local subring $M$ of $R$ dominate a ring $A_{\mathfrak{p}}\in L(A)$, it is necessary and sufficient that $A\subset M$; the local ring $A_{\mathfrak{p}}$ dominated by $M$ is then unique and corresponds to $\mathfrak{p}=\mathfrak{m}(M)\cap A$.

Proof. Indeed, if $M$ dominates $A_{\mathfrak{p}}$, one has $\mathfrak{m}(M)\cap A_{\mathfrak{p}}=\mathfrak{p}{A}_{\mathfrak{p}}$ by (8.1.1), whence the uniqueness of $\mathfrak{p}$; on the other hand, if $A\subset M$, $\mathfrak{m}(M)\cap A=\mathfrak{p}$ is prime in $A$, and since $A-\mathfrak{p}\subset M$, one has $A_{\mathfrak{p}}\subset M$ and $\mathfrak{p}{A}_{\mathfrak{p}}\subset \mathfrak{m}(M)$, hence $M$ dominates $A_{\mathfrak{p}}$.

Lemma (8.1.4). Let $R$ be a field, $M$, $N$ two local subrings of $R$, $P$ the subring of $R$ generated by $M\cup N$. The following conditions are equivalent:

(i) There exists a prime ideal $\mathfrak{p}$ of $P$ such that $\mathfrak{m}(M)=\mathfrak{p}\cap M$, $\mathfrak{m}(N)=\mathfrak{p}\cap N$.

(ii) The ideal $\mathfrak{a}$ generated in $P$ by $\mathfrak{m}(M)\cup \mathfrak{m}(N)$ is distinct from $P$.

(iii) There exists a local subring $Q$ of $R$ dominating both $M$ and $N$.

Proof. It is clear that (i) implies (ii); conversely, if $\mathfrak{a}\ne P$, $\mathfrak{a}$ is contained in a maximal ideal $\mathfrak{n}$ of $P$, and since $1\notin \mathfrak{n}$, $\mathfrak{n}\cap M$ contains $\mathfrak{m}(M)$ and is distinct from $M$, hence $\mathfrak{n}\cap M=\mathfrak{m}(M)$ and likewise $\mathfrak{n}\cap N=\mathfrak{m}(N)$. It is clear that if $Q$ dominates $M$ and $N$, one has $P\subset Q$, and $\mathfrak{m}(M)=\mathfrak{m}(Q)\cap M=(\mathfrak{m}(Q)\cap P)\cap M$, $\mathfrak{m}(N)=(\mathfrak{m}(Q)\cap P)\cap N$, hence (iii) implies (i); the converse is evident on taking $Q=P_{\mathfrak{p}}$.

When the conditions of (8.1.4) are satisfied, one says, with C. Chevalley, that the local rings $M$ and $N$ are allied.

Proposition (8.1.5). Let $A$, $B$ be two subrings of a field $R$, $C$ the subring of $R$ generated by $A\cup B$. The following conditions are equivalent:

(i) For every local ring $Q$ containing $A$ and $B$, one has $A_{\mathfrak{p}}=B_{\mathfrak{q}}$, on setting $\mathfrak{p}=\mathfrak{m}(Q)\cap A$, $\mathfrak{q}=\mathfrak{m}(Q)\cap B$.

(ii) For every prime ideal $\mathfrak{r}$ of $C$, one has $A_{\mathfrak{p}}=B_{\mathfrak{q}}$, on setting $\mathfrak{p}=\mathfrak{r}\cap A$, $\mathfrak{q}=\mathfrak{r}\cap B$.

(iii) If $M\in L(A)$ and $N\in L(B)$ are allied, they are identical.

(iv) One has $L(A)\cap L(B)=L(C)$.

Proof. Lemmas (8.1.3) and (8.1.4) prove that (i) and (iii) are equivalent; it is clear that (i) implies (ii) on applying it to $Q=C_{\mathfrak{r}}$; conversely, (ii) implies (i), for if $Q$ contains $A\cup B$, it contains $C$, and if $\mathfrak{r}=\mathfrak{m}(Q)\cap C$, one has $\mathfrak{p}=\mathfrak{r}\cap A$ and $\mathfrak{q}=\mathfrak{r}\cap B$ by (8.1.3). It is immediate that (iv) implies (i), for if $Q$ contains $A\cup B$, it dominates a local ring $C_{\mathfrak{r}}\in L(C)$ by (8.1.3); by hypothesis one has $C_{\mathfrak{r}}\in L(A)\cap L(B)$, and (8.1.1) and (8.1.3) prove that $C_{\mathfrak{r}}=A_{\mathfrak{p}}=B_{\mathfrak{q}}$. Let us prove finally that (iii) implies (iv). Let $Q\in L(C)$; $Q$ dominates an $M\in L(A)$ and an $N\in L(B)$ (8.1.3), hence $M$ and $N$, being allied, are identical by hypothesis. Since one then has $C\subset M$, $M$ dominates a $Q^{\prime }\in L(C)$ (8.1.3), hence $Q$ dominates $Q^{\prime }$, which (8.1.3) necessarily entails $Q=Q^{\prime }=M$, hence $Q\in L(A)\cap L(B)$. Conversely, if $Q\in L(A)\cap L(B)$, one has $C\subset Q$, hence (8.1.3) $Q$ dominates a $Q^{\prime \prime }\in L(C)\subset L(A)\cap L(B)$; $Q$ and $Q^{\prime \prime }$ being allied are identical, hence $Q=Q^{\prime \prime }\in L(C)$, which completes the proof.

8.2. Local rings of an integral scheme

(8.2.1) Let $X$ be an integral prescheme, $R$ its field of rational functions, identical with the local ring of the generic point $a$ of $X$; for every $x\in X$, one knows that ${\mathcal{O}}_x$ is canonically identified with a subring of $R$ (7.1.5), and for every rational function $f\in R$, the domain of definition $\delta (f)$ of $f$ is the open set of $x\in X$ such that $f\in {\mathcal{O}}_x$. It follows (7.2.6) that for every open set $U\subset X$, one has

$$\Gamma (U,{\mathcal{O}}_X)=\bigcap _{x\in U}{\mathcal{O}}_x.\text{\hspace{1em}(8.2.1.1)}$$

Proposition (8.2.2). Let $X$ be an integral prescheme, $R$ its field of rational functions. In order that $X$ be a scheme, it is necessary and sufficient that the relation “${\mathcal{O}}_x$ and ${\mathcal{O}}_y$ are allied” (8.1.4) between points $x$, $y$ of $X$ imply $x=y$.

Proof. Suppose this condition verified, and let us show that $X$ is separated. Let $U$ and $V$ be two distinct affine opens of $X$, $A$ and $B$ their rings, identified with subrings of $R$; $U$ (resp. $V$) is therefore identified (8.1.2) with $L(A)$ (resp. $L(B)$), and the hypothesis entails (8.1.5) that if $C$ is the subring of $R$ generated by $A\cup B$, $W=U\cap V$ is identified with $L(A)\cap L(B)=L(C)$. Moreover, one knows ([1], p. 5-03, prop. 4 bis) that every subring $E$ of $R$ is equal to the intersection of the local rings belonging to $L(E)$; $C$ is therefore identified with the intersection of the rings ${\mathcal{O}}_z$ for $z\in W$, in other words (8.2.1.1) with $\Gamma (W,{\mathcal{O}}_X)$. Consider then the subprescheme induced by $X$ on $W$; to the identity homomorphism $\phi :C\to \Gamma (W,{\mathcal{O}}_X)$ corresponds (2.2.4) a morphism $\Phi =(\psi ,\theta ):W\to \mathrm{Spec}(C)$; we are going to see that $\Phi$ is an isomorphism of preschemes, whence it will follow that $W$ is an affine open. The identification of $W$ with $L(C)=\mathrm{Spec}(C)$ shows that $\psi$ is bijective. On the other hand, for every $x\in W$, ${\theta }_x^{♯}$ is the injection $C_{\mathfrak{r}}\to {\mathcal{O}}_x$ if $\mathfrak{r}={\mathfrak{m}}_x\cap C$, and by definition $C_{\mathfrak{r}}$ is identified with ${\mathcal{O}}_x$, hence ${\theta }_x^{♯}$ is bijective. It remains therefore to see that $\psi$ is a homeomorphism, in other words that for every closed part $F\subset W$, $\psi (F)$ is closed in $\mathrm{Spec}(C)$. Now $F$ is the trace on $W$ of a closed part of $U$, of the form $V(\mathfrak{a})$, where $\mathfrak{a}$ is an ideal of $A$; let us show that $\psi (F)=V(\mathfrak{a}C)$, which will prove our assertion. Indeed, the prime ideals of $C$ containing $\mathfrak{a}C$ are the prime ideals of $C$ containing $\mathfrak{a}$, hence the ideals of the form $\psi (x)={\mathfrak{m}}_x\cap C$ where $x\in W$; since $\mathfrak{a}\subset {\mathfrak{m}}_x$ is equivalent to $x\in V(\mathfrak{a})=W\cap F$ for $x\in U$, one has indeed $\psi (F)=V(\mathfrak{a}C)$.

It follows that $X$ is separated, for $U\cap V$ is affine and its ring $C$ is generated by the union $A\cup B$ of the rings of $U$ and $V$ (5.5.6).

Conversely, suppose $X$ separated, and let $x$, $y$ be two points of $X$ such that ${\mathcal{O}}_x$ and ${\mathcal{O}}_y$ are allied. Let $U$ (resp. $V$) be an affine open containing $x$ (resp. $y$), of ring $A$ (resp. $B$); one knows then that $U\cap V$ is affine and that its ring $C$ is generated by $A\cup B$ (5.5.6). If $\mathfrak{p}={\mathfrak{m}}_x\cap A$, $\mathfrak{q}={\mathfrak{m}}_y\cap B$, one has $A_{\mathfrak{p}}={\mathcal{O}}_x$, $B_{\mathfrak{q}}={\mathcal{O}}_y$, and since $A_{\mathfrak{p}}$ and $B_{\mathfrak{q}}$ are allied, there exists a prime ideal $\mathfrak{r}$ of $C$ such that $\mathfrak{p}=\mathfrak{r}\cap A$, $\mathfrak{q}=\mathfrak{r}\cap B$ (8.1.4). But then there exists a point $z\in U\cap V$ such that $\mathfrak{r}={\mathfrak{m}}_z\cap C$ since $U\cap V$ is affine, and one has evidently $x=z$ and $y=z$, whence $x=y$.

Corollary (8.2.3). Let $X$ be an integral scheme, $x$, $y$ two points of $X$. In order that $x\in \overline{\{y\}}$, it is necessary and sufficient that ${\mathcal{O}}_x\subset {\mathcal{O}}_y$, in other words that every rational function defined at $x$ be defined at $y$.

Proof. The condition is evidently necessary since the domain of definition $\delta (f)$ of a rational function $f\in R$ is open; let us show that it is sufficient. If ${\mathcal{O}}_x\subset {\mathcal{O}}_y$, there exists a prime ideal $\mathfrak{p}$ of ${\mathcal{O}}_x$ such that ${\mathcal{O}}_y$ dominates $({\mathcal{O}}_x{)}_{\mathfrak{p}}$ (8.1.3); now (2.4.2) there exists $z\in X$ such that $x\in \overline{\{z\}}$ and ${\mathcal{O}}_z=({\mathcal{O}}_x{)}_{\mathfrak{p}}$; since ${\mathcal{O}}_z$ and ${\mathcal{O}}_y$ are allied, one has $z=y$ by (8.2.2), whence the corollary.

Corollary (8.2.4). If $X$ is an integral scheme, the map $x\to {\mathcal{O}}_x$ is injective; in other words, if $x$, $y$ are two distinct points of $X$, there exists a rational function defined at one of these points and not at the other.

Proof. This results from (8.2.3) and from the axiom $(T_0)$ (2.1.4).

Corollary (8.2.5). Let $X$ be an integral scheme whose underlying space is Noetherian; when $f$ runs through the field $R$ of rational functions on $X$, the sets $\delta (f)$ generate the topology of $X$.

Proof. Indeed, every closed part of $X$ is then a finite union of irreducible closed sets, that is, of sets of the form $\overline{\{y\}}$ (2.1.5). Now, if $x\notin \overline{\{y\}}$, there exists a rational function $f$ defined at $x$ and not at $y$ (8.2.3), in other words, one has $x\in \delta (f)$ and $\delta (f)$ does not meet $\overline{\{y\}}$. The complement of $\overline{\{y\}}$ is consequently a union of sets of the form $\delta (f)$, and by virtue of the first remark, every open set of $X$ is a union of finite intersections of opens of the form $\delta (f)$.

(8.2.6) Corollary (8.2.5) shows that the topology of $X$ is entirely characterized by the datum of the family of local rings $({\mathcal{O}}_x{)}_{x\in X}$ having $R$ as field of fractions. It comes to the same thing, moreover, to say that the closed parts of $X$ are defined in the following way: given a finite part $\{x_1,\dots ,x_n\}$ of $X$, one considers the set of $y\in X$ such that ${\mathcal{O}}_y\subset {\mathcal{O}}_{x_i}$ for at least one index $i$, and these sets (for all the choices of $\{x_1,\dots ,x_n\}$) are the closed sets of $X$. Moreover, once the topology of $X$ is known, the structure sheaf ${\mathcal{O}}_X$ is just as well determined by the family of the ${\mathcal{O}}_x$, since $\Gamma (U,{\mathcal{O}}_X)={\bigcap }_{x\in U}{\mathcal{O}}_x$ (8.2.1.1). The family $({\mathcal{O}}_x{)}_{x\in X}$ therefore completely determines the prescheme $X$ when $X$ is an integral scheme whose underlying space is Noetherian.

Proposition (8.2.7). Let $X$, $Y$ be two integral schemes, $f:X\to Y$ a dominant morphism (2.2.6), $K$ (resp. $L$) the field of rational functions on $X$ (resp. $Y$). Then $L$ is identified with a subfield of $K$, and for every $x\in X$, ${\mathcal{O}}_{f(x)}$ is the unique local ring of $Y$ dominated by ${\mathcal{O}}_x$.

Proof. Indeed, if $f=(\psi ,\theta )$ and if $a$ is the generic point of $X$, $\psi (a)$ is the generic point of $Y$ (0, 2.1.5); ${\theta }_a^{♯}$ is consequently a monomorphism of the field $L={\mathcal{O}}_{\psi (a)}$ into the field $K={\mathcal{O}}_a$. Since every nonempty affine open $U$ of $Y$ contains $\psi (a)$, it follows from (2.2.4) that the homomorphism $\Gamma (U,{\mathcal{O}}_Y)\to \Gamma ({\psi }^{-1}(U),{\mathcal{O}}_X)$ corresponding to $f$ is the restriction of ${\theta }_a^{♯}$ to $\Gamma (U,{\mathcal{O}}_Y)$. Hence, for every $x\in X$, ${\theta }_x^{♯}$ is the restriction to ${\mathcal{O}}_{\psi (x)}$ of ${\theta }_a^{♯}$, and is consequently a monomorphism. One knows moreover that ${\theta }_x^{♯}$ is a local homomorphism, hence, if one identifies $L$ with a subfield of $K$ by ${\theta }_a^{♯}$, ${\mathcal{O}}_{\psi (x)}$ is dominated by ${\mathcal{O}}_x$ (8.1.1); it is moreover the only local ring of $Y$ dominated by ${\mathcal{O}}_x$ since two local rings of $Y$ which are allied are identical (8.2.2).

Proposition (8.2.8). Let $X$ be an irreducible prescheme, $f:X\to Y$ a local immersion (resp. a local isomorphism); suppose moreover that the morphism $f$ is separated. Then $f$ is an immersion (resp. an open immersion).

Proof. Let $f=(\psi ,\theta )$; it suffices, in both cases, to prove that $\psi$ is a homeomorphism of $X$ onto $\psi (X)$ (4.5.3). Replacing $f$ by $f_{\mathrm{red}}$ (5.1.6 and 5.5.1, (vi)), one may suppose $X$ and $Y$ reduced. If $Y^{\prime }$ is the reduced closed subprescheme of $Y$ having $\overline{\psi (X)}$ as underlying space, $f$ factors as $X\stackrel{f^{\prime }}{\to }{Y}^{\prime }\stackrel{j}{\to }Y$, where $j$ is the canonical injection (5.2.2). It follows from (5.5.1, (v)) that $f^{\prime }$ is again a separated morphism; moreover, $f^{\prime }$ is again a local immersion (resp. a local isomorphism), for the question being local on $X$ and $Y$, one may restrict oneself, in order to prove it, to the case where $f$ is a closed immersion (resp. an open immersion), and our assertion then follows at once from (4.2.2).

One may therefore suppose that $f$ is a dominant morphism, which entails that $Y$ is, like it, irreducible (0, 2.1.5), hence that $X$ and $Y$ are both integral. Moreover, the question being local on $Y$, one may suppose that $Y$ is an affine scheme; since $f$ is separated, $X$ is a scheme (5.5.1, (ii)), and one is finally in the hypotheses of (8.2.7). Then, for every $x\in X$, ${\theta }_x^{♯}$ is injective; but the hypothesis that $f$ is a local immersion implies that ${\theta }_x^{♯}$ is surjective (4.2.2), hence ${\theta }_x^{♯}$ is bijective, in other words (with the identification of (8.2.7)) one has ${\mathcal{O}}_{\psi (x)}={\mathcal{O}}_x$. This implies by (8.2.4) that $\psi$ is an injective map, which already proves the proposition when $f$ is a local isomorphism (4.5.3). When one supposes only that $f$ is a local immersion, for every $x\in X$ there exists an open neighborhood $U$ of $x$ in $X$ and an open neighborhood $V$ of $\psi (x)$ in $Y$ such that the restriction of $\psi$ to $U$ is a homeomorphism of $U$ onto a closed part of $V$. Now $U$ is dense in $X$, hence $\psi (U)$ is dense in $Y$ and a fortiori in $V$, which proves that $\psi (U)=V$; since $\psi$ is injective, ${\psi }^{-1}(V)=U$, and this completes the proof that $\psi$ is a homeomorphism of $X$ onto $\psi (X)$.

8.3. Chevalley schemes

(8.3.1) Let $X$ be an integral Noetherian scheme, $R$ its field of rational functions; let us denote by $X^{\prime }$ the set of local subrings ${\mathcal{O}}_x\subset R$, where $x$ runs through $X$. The set $X^{\prime }$ verifies the three following conditions:

(Sch. 1) For every $M\in X^{\prime }$, $R$ is the field of fractions of $M$.

(Sch. 2) There exists a finite set of Noetherian subrings $A_i$ of $R$ such that $X^{\prime }={\bigcup }_{i}L(A_i)$ and such that, for every pair of indices $i$, $j$, the subring $A_{ij}$ of $R$ generated by $A_i\cup A_j$ is an algebra of finite type over $A_i$.

(Sch. 3) Two elements $M$, $N$ of $X^{\prime }$ which are allied are identical.

One has indeed seen in (8.2.1) that (Sch. 1) is satisfied, and (Sch. 3) results from (8.2.2). To prove (Sch. 2), it suffices to cover $X$ by a finite number of affine opens $U_i$, of Noetherian rings, and to take $A_i=\Gamma (U_i,{\mathcal{O}}_X)$; the hypothesis that $X$ is a scheme entails that $U_i\cap U_j$ is affine and that $\Gamma (U_i\cap U_j,{\mathcal{O}}_X)=A_{ij}$ (5.5.6); moreover, since the space $U_i$ is Noetherian, the immersion $U_i\cap U_j\to U_i$ is of finite type (6.3.5), hence $A_{ij}$ is an $A_i$-algebra of finite type (6.3.3).

(8.3.2) The structures whose axioms are (Sch. 1), (Sch. 2), and (Sch. 3) generalize the “schemes” in the sense of C. Chevalley, who supposes moreover that $R$ is an extension of finite type of a field $K$ and that the $A_i$ are $K$-algebras of finite type (which renders part of (Sch. 2) useless) [1]. Conversely, if one has such a structure on a set $X^{\prime }$, one may associate with it an integral scheme $X$ by using the remarks of (8.2.6): the underlying space of $X$ is equal to $X^{\prime }$ equipped with the topology defined in (8.2.6), and with the sheaf ${\mathcal{O}}_X$ such that $\Gamma (U,{\mathcal{O}}_X)={\bigcap }_{x\in U}{\mathcal{O}}_x$ for every open set $U\subset X$, with an evident definition of the restriction homomorphisms. We leave to the reader the care of verifying that one obtains in this way an integral scheme, whose local rings are the elements of $X^{\prime }$; we shall not use this result in what follows.