§6. Integral morphisms and finite morphisms

6.1. Preschemes integral over another

Definition.

Let $X$ be an $S$ -prescheme and $f : X \to S$ its structure morphism. We say that $X$ is integral over $S$ , or that $f$ is an integral morphism, if there exists a cover $(S_{α})$ of $S$ by affine opens such that, for every $α$ , the induced prescheme $f^{- 1} (S_{α})$ is an affine scheme whose ring $B_{α}$ is an integral algebra over the ring $A_{α}$ of $S_{α}$ . We say that $X$ is finite over $S$ , or that $f$ is a finite morphism, if $X$ is integral and of finite type over $S$ .

When $S$ is affine of ring $A$ , we shall also say "integral (resp. finite) over $A$ " instead of "integral (resp. finite) over $S$ ".

(6.1.2)

It is clear that if $X$ is integral over $S$ , then it is affine over $S$ . For an affine prescheme $X$ over $S$ to be integral (resp. finite) over $S$ , it is necessary and sufficient that the associated quasi-coherent $O_{S}$ -algebra $A (X)$ be such that there exists a cover $(S_{α})$ of $S$ by affine opens with the property that, for every $α$ , $Γ (S_{α}, A (X))$ is an integral (resp. integral and of finite type) algebra over $Γ (S_{α}, O_{S})$ . A quasi-coherent $O_{S}$ -algebra with this property is said to be integral (resp. finite) over $O_{S}$ . To give an integral (resp. finite) prescheme over $S$ is therefore the same (1.3.1) as to give a quasi-coherent $O_{S}$ -algebra that is integral (resp. finite) over $O_{S}$ . Note that a quasi-coherent $O_{S}$ -algebra $B$ is finite if and only if it is an $O_{S}$ -module of finite type (I, 1.3.9); this is equivalent to saying that $B$ is an integral $O_{S}$ -algebra of finite type, since an algebra that is integral and of finite type over a ring $A$ is an $A$ -module of finite type.

Proposition.

Let $S$ be a locally Noetherian prescheme. For an affine prescheme $X$ over $S$ to be finite over $S$ , it is necessary and sufficient that the $O_{S}$ -algebra $A (X)$ be coherent.

Proof. Taking the preceding remark into account, this comes down to noting that, if $S$ is locally Noetherian, the quasi-coherent $O_{S}$ -modules of finite type are exactly the coherent $O_{S}$ -modules (I, 1.5.1).

Proposition.

Let $X$ be an integral (resp. finite) prescheme over $S$ and $f : X \to S$ its structure morphism. Then, for every affine open $U \subset S$ of ring $A$ , $f^{- 1} (U)$ is an affine scheme whose ring $B$ is an integral (resp. finite) algebra over $A$ .

Proof. We first prove the following lemma.

Lemma.

Let $A$ be a ring, $M$ an $A$ -module, and $(g_{i})_{1 \leq i \leq m}$ a finite system of elements of $A$ such that the $D (g_{i})$ ( $1 \leq i \leq m$ ) cover $Spec (A)$ . If, for every $i$ , $M_{g_{i}}$ is an $A_{g_{i}}$ -module of finite type, then $M$ is an $A$ -module of finite type.

Proof. We may assume that each $M_{g_{i}}$ admits a finite system of generators $(m_{ij} / g_{i}^{n})$ with $m_{ij} \in M$ , with $n$ the same for all indices $i$ . We show that the $m_{ij}$ form a system of generators of $M$ . Let $M^{'}$ be the sub- $A$ -module of $M$ generated by these elements, and let $m$ be an element of $M$ . By hypothesis, for each $i$ , there exist $a_{ij} \in A$ and an integer $p$ (independent of $i$ ) such that, in $M_{g_{i}}$ , $m /1 = (\sum_{j} a_{ij} m_{ij}) / g_{i}^{p}$ ; this implies that there exists an integer $r \geq p$ such that, for every $i$ , $g_{i}^{r} m \in M^{'}$ . Now, since the $D (g_{i}^{r}) = D (g_{i})$ cover $Spec (A)$ , the ideal of $A$ generated by the $g_{i}^{r}$ is equal to $A$ ; in other words, there exist elements $a_{i} \in A$ such that $\sum_{i} a_{i} g_{i}^{r} = 1$ , and consequently $m = (\sum_{i} a_{i} g_{i}^{r}) m \in M^{'}$ , whence the lemma.

This being so, we already know (1.3.2) that $f^{- 1} (U)$ is affine. If $ϕ$ is the homomorphism $A \to B$ corresponding to $f$ , there exists a finite cover of $U$ by opens $D (g_{i})$ ( $g_{i} \in A$ ) such that, if $h_{i} = ϕ (g_{i})$ , then $B_{h_{i}}$ is an integral (resp. integral and finite) algebra over $A_{g_{i}}$ . Indeed, there is a cover of $U$ by affine opens $V_{α} \subset U$ such that, if $A_{α} = A (V_{α})$ , $B_{α} = A (f^{- 1} (V_{α}))$ is an integral (resp. finite) algebra over $A_{α}$ . Every $x \in U$ belongs to some $V_{α}$ , so there exists $g \in A$ such that $x \in D (g) \subset V_{α}$ ; if $g_{α}$ is the image of $g$ in $A_{α}$ , then $A (D (g)) = A_{g} = (A_{α})_{g_{α}}$ . Let $h = ϕ (g)$ and let $h_{α}$ be the image of $g_{α}$ in $B_{α}$ ; we have

$A (D (h)) = B_{h} = (B_{α})_{h_{α}}$

and, since $B_{α}$ is integral (resp. finite) over $A_{α}$ , $(B_{α})_{h_{α}}$ is integral (resp. finite) over $(A_{α})_{g_{α}}$ . It now suffices to use the fact that $U$ is quasi-compact to obtain the desired cover.

If we first suppose that the $B_{h_{i}}$ are integral and finite over the $A_{g_{i}}$ , then since $B_{h_{i}}$ , viewed as an $A_{g_{i}}$ -module, also coincides with $B_{g_{i}}$ , Lemma (6.1.4.1) shows that in this case $B$ is an $A$ -module of finite type.

Now suppose only that each $B_{h_{i}}$ is integral over $A_{g_{i}}$ ; let $b \in B$ , and let $C$ be the sub- $A$ -algebra of $B$ generated by $b$ . For every $i$ , $C_{h_{i}}$ is the algebra over $A_{g_{i}}$ generated by $b /1$ in $B_{h_{i}}$ ; it follows from the hypothesis that each $C_{h_{i}}$ is an $A_{g_{i}}$ -module of finite type, hence (6.1.4.1) $C$ is an $A$ -module of finite type, which proves that $B$ is integral over $A$ .

Proposition.

(i) A closed immersion is finite (and a fortiori integral).
(ii) The composition of two finite (resp. integral) morphisms is finite (resp. integral).
(iii) If $f : X \to Y$ is a finite (resp. integral) $S$ -morphism, then $f_{(S^{'})} : X_{(S^{'})} \to Y_{(S^{'})}$ is finite (resp. integral) for every base extension $S^{'} \to S$ .
(iv) If $f : X \to Y$ and $g : X^{'} \to Y^{'}$ are two finite (resp. integral) $S$ -morphisms, then so is $f \times_{S} g : X \times_{S} X^{'} \to Y \times_{S} Y^{'}$ .
(v) If $f : X \to Y$ and $g : Y \to Z$ are morphisms such that $g \circ f$ is finite (resp. integral) and $g$ is separated, then $f$ is finite (resp. integral).
(vi) If $f : X \to Y$ is a finite (resp. integral) morphism, then so is $f_{re d}$ .

Proof. By (I, 5.5.12), it suffices to prove (i), (ii), and (iii). To prove that a closed immersion $X \to S$ is finite, we may restrict to the case $S = Spec (A)$ , and everything then comes down to noting that a quotient ring $A / j$ is a monogeneous $A$ -module. To prove that the composition of two finite (resp. integral) morphisms $X \to Y$ , $Y \to Z$ is finite (resp. integral), we may again assume that $Z$ (and hence $X$ and $Y$ (1.3.4)) is affine, and the claim then amounts to: if $B$ is a finite (resp. integral) $A$ -algebra and $C$ a finite (resp. integral) $B$ -algebra, then $C$ is a finite (resp. integral) $A$ -algebra, which is immediate. Finally, to prove (iii), we may restrict to the case $S = Y$ , since $X_{(S^{'})}$ is identified with $X \times_{Y} Y_{(S^{'})}$ (I, 3.3.11); we may further assume that $S = Spec (A)$ and $S^{'} = Spec (A^{'})$ ; then $X$ is affine of ring $B$ (1.3.4), $X_{(S^{'})}$ is affine of ring $A^{'} \otimes_{A} B$ , and it suffices to remark that if $B$ is a finite (resp. integral) $A$ -algebra, then $A^{'} \otimes_{A} B$ is a finite (resp. integral) $A^{'}$ -algebra.

We also note that if $X$ and $Y$ are two $S$ -preschemes that are finite (resp. integral) over $S$ , their sum $X ⊔ Y$ is a finite (resp. integral) prescheme over $S$ , since this comes down to seeing that if $B$ and $C$ are finite (resp. integral) $A$ -algebras, then so is $B \times C$ .

Corollary.

If $X$ is an integral (resp. finite) prescheme over $S$ , then, for every open $U \subset S$ , $f^{- 1} (U)$ is integral (resp. finite) over $U$ .

Proof. This is a particular case of (6.1.5, (iii)).

Corollary.

Let $f : X \to Y$ be a finite morphism. Then, for every $y \in Y$ , the fibre $f^{- 1} (y)$ is a finite algebraic scheme over $κ (y)$ , and a fortiori its underlying space is discrete and finite.

Proof. Indeed, as a $κ (y)$ -prescheme, $f^{- 1} (y)$ is identified with $X \times_{Y} Spec (κ (y))$ (I, 3.6.1), which is finite over $Spec (κ (y))$ (6.1.5, (iii)); it is thus an affine scheme whose ring is an algebra of finite rank over $κ (y)$ (6.1.4). The corollary then follows from (I, 6.4.4).

Corollary.

Let $X$ , $S$ be two integral preschemes and $f : X \to S$ a dominant morphism. If $f$ is integral (resp. finite), then the field $R (X)$ of rational functions on $X$ is algebraic (resp. algebraic of finite degree) over the field $R (S)$ of rational functions on $S$ .

Proof. Let $s$ be the generic point of $S$ ; the $κ (s)$ -prescheme $f^{- 1} (s)$ is integral (resp. finite) over $Spec (κ (s))$ (6.1.5, (iii)) and contains, by hypothesis, the generic point $x$ of $X$ ; the local ring of $x$ in $f^{- 1} (s)$ , being equal to $κ (x)$ (I, 3.6.5), is a local ring of an integral (resp. finite) algebra over $κ (s)$ (6.1.4), whence the corollary.

Remark.

The hypothesis that $g$ is separated is essential for the validity of (6.1.5, (v)): indeed, if $Y$ is not separated over $Z$ , then the identity 1_Y is the composite morphism $Y \to^{Δ_{Y}} Y \times_{Z} Y \to^{p_{1}} Y$ , but $Δ_{Y}$ is not an integral morphism, as follows from (6.1.10):

Proposition.

Every integral morphism is universally closed.

Proof. Let $f : X \to Y$ be an integral morphism; by (6.1.5, (iii)), it suffices to show that $f$ is closed. Let $Z$ be a closed part of $X$ ; there exists a subprescheme of $X$ whose

underlying space is $Z$ (I, 5.2.1), and it therefore follows from (6.1.5, (i) and (ii)) that we may restrict to proving that $f (X)$ is closed in $Y$ . By (6.1.5, (vi)), we may assume that $X$ and $Y$ are reduced; furthermore, if $T$ is the closed reduced subprescheme of $Y$ whose underlying space is $f (X)$ (I, 5.2.1), we know that $f$ factors as $X \to T \to^{j} Y$ , where $j$ is the injection morphism (I, 5.2.2), and since $j$ is separated (I, 5.5.1, (i)), it follows from (6.1.5, (v)) that $g$ is an integral morphism. We may therefore assume that $f (X)$ is dense in $Y$ . Finally, since the question is local on $Y$ , we may restrict to the case $Y = Spec (A)$ . Then $X = Spec (B)$ , where $B$ is an $A$ -algebra that is integral over $A$ (6.1.4); moreover, $A$ is reduced (I, 5.1.4) and the hypothesis that $f (X)$ is dense in $Y$ implies that the homomorphism $ϕ : A \to B$ corresponding to $f$ is injective (I, 1.2.7). To say that under these conditions $f (X) = Y$ means that every prime ideal of $A$ is the trace on $A$ of a prime ideal of $B$ , which is none other than the first theorem of Cohen–Seidenberg ([13], t. I, p. 257, th. 3).

Corollary.

Every finite morphism $f : X \to Y$ is projective.

Proof. Since $f$ is affine, $O_{X}$ is a very ample $O_{X}$ -module relative to $f$ (5.1.2); furthermore, $f_{*} (O_{X})$ is a quasi-coherent $O_{Y}$ -module of finite type (6.1.2); finally, $f$ is separated, of finite type, and universally closed (6.1.10), so we are in the conditions of application of criterion (5.5.4, (i)).

Proposition.

Let $f : X^{'} \to X$ be a finite morphism, and let $B = f_{*} (O_{X^{'}})$ (which is a quasi-coherent $O_{X}$ -algebra and an $O_{X}$ -module of finite type). Let $F^{'}$ be a quasi-coherent $O_{X^{'}}$ -module; for $F^{'}$ to be locally free of rank $r$ , it is necessary and sufficient that $f_{*} (F^{'})$ be a locally free $B$ -module of rank $r$ .

Proof. It is clear that if $f_{*} (F^{'}) ∣ U$ is isomorphic to $B^{r} ∣ U$ ( $U$ open in $X$ ), then $F^{'} ∣ f^{- 1} (U)$ is isomorphic to $O_{X^{'}}^{r} ∣ f^{- 1} (U)$ (1.4.2). Conversely, suppose that $F^{'}$ is locally free of rank $r$ , and let us show that $f_{*} (F^{'})$ is locally isomorphic to $B^{r}$ as a $B$ -module. Let $x$ be a point of $X$ ; as $U$ runs over a fundamental system of affine neighbourhoods of $x$ , $f^{- 1} (U)$ runs over a fundamental system of affine neighbourhoods (1.2.5) of the finite set $f^{- 1} (x)$ , since $f$ is closed (6.1.10). The proposition then follows from the following lemma:

Lemma.

Let $Y$ be a prescheme, $E$ a locally free $O_{Y}$ -module of rank $r$ , and $Z$ a finite part of $Y$ contained in an affine open $V$ . Then there exists a neighbourhood $U \subset V$ of $Z$ such that $E ∣ U$ is isomorphic to $O_{Y}^{r} ∣ U$ .

Proof. We may evidently assume $Y$ affine; for every $z_{i} \in Z$ , there exists in the closure $z_{i}$ at least one closed point $z_{i}^{'}$ (0, 2.1.3); if $Z^{'}$ is the set of the $z_{i}^{'}$ , every neighbourhood of $Z^{'}$ is a neighbourhood of $Z$ , and we may therefore assume that $Z$ is closed in $Y$ and discrete. Consider the closed reduced subprescheme of $Y$ having $Z$ as its underlying space (I, 5.2.1) and let $j : Z \to Y$ be the canonical injection; $j * (E) = E \otimes_{Y} O_{Z}$ is locally free of rank $r$ on the discrete scheme $Z$ , and is therefore isomorphic to $O_{Z}^{r}$ ; in other words, there exist $r$ sections $s_{i}$ ( $1 \leq i \leq r$ ) of $E \otimes_{Y} O_{Z}$ over $Z$ such that the homomorphism $O_{Z}^{r} \to E \otimes_{Y} O_{Z}$ defined by these sections is bijective. But $Y = Spec (A)$ is affine, $Z$ is defined by an ideal $j$ of $A$ , and we have $E = \tilde{M}$ , where $M$ is an $A$ -module; the $s_{i}$ are elements

of $M \otimes_{A} (A / j)$ and are therefore images of $r$ elements $t_{i} \in M = Γ (Y, E)$ . For every $z_{j} \in Z$ there is then a neighbourhood $V_{j}$ of $z_{j}$ such that the restrictions of the $t_{i}$ to $V_{j}$ define an isomorphism $O_{Y}^{r} ∣ V_{j} \sim E ∣ V_{j}$ (0, 5.5.4); the neighbourhood $U$ given by the union of the $V_{j}$ therefore meets the question.

Proposition.

Let $g : X^{'} \to X$ be an integral morphism of preschemes, $Y$ a normal locally integral prescheme, and $f$ a rational map from $Y$ to $X^{'}$ such that $g \circ f$ is an everywhere defined rational map (I, 7.2.1). Then $f$ is everywhere defined.

Proof. If $f_{1}$ and $f_{2}$ are two morphisms (from dense opens of $Y$ to $X^{'}$ ) in the class $f$ , it is clear that $g \circ f_{1}$ and $g \circ f_{2}$ are equivalent morphisms, which justifies the notation $g \circ f$ for their equivalence class. We recall also that, if we further suppose $Y$ to be locally Noetherian, then the hypothesis that $Y$ is normal already implies that $Y$ is locally integral (I, 6.1.13).

To prove (6.1.13), let us first note that the question is local on $Y$ , and so we may assume that there exists in the class $g \circ f$ a morphism $h : Y \to X$ . Consider the inverse image $Y^{'} = X_{(h)}^{'} = X_{(Y)}^{'}$ , and note that the morphism $g^{'} = g_{(Y)} : Y^{'} \to Y$ is integral (6.1.5, (iii)). In view of the correspondence between rational maps from $Y$ to $X^{'}$ and rational $Y$ -sections of $Y^{'}$ (I, 7.1.2), we see that we are reduced to proving the particular case of (6.1.13) in which $X = Y$ ; in other words:

Corollary.

Let $X$ be a normal locally integral prescheme, $g : X^{'} \to X$ an integral morphism, and $f$ a rational $X$ -section of $X^{'}$ . Then $f$ is everywhere defined.

Proof. Since the question is local on $X$ , we may assume $X$ integral, and then $f$ is identified with a morphism from an open $U$ of $X$ to $X^{'}$ (I, 7.2.2) which is a $U$ -section of $g^{- 1} (U)$ . Since $g$ is separated, $f$ is a closed immersion of $U$ into $g^{- 1} (U)$ (I, 5.4.6); let $Z$ be the closed subprescheme of $g^{- 1} (U)$ associated to $f$ (I, 4.2.1), which is isomorphic to $U$ , hence integral; let $X_{1}$ be the reduced subprescheme of $X^{'}$ whose underlying space is the closure $\overset{ˉ}{Z}$ of $Z$ in $X^{'}$ (I, 5.2.1); then $Z$ is a subprescheme induced on an open of $X_{1}$ (I, 5.2.3), and, being irreducible, so is $X_{1}$ , which is therefore integral. The morphism $f$ can then be considered as a rational $X$ -section of $X_{1}$ ; since the restriction of $g$ to $X_{1}$ is an integral morphism (6.1.5, (i) and (ii)), we are finally reduced to proving (6.1.14) in the particular case $X^{'} = X_{1}$ ; in other words:

Corollary.

Let $X$ be a normal integral prescheme, $X^{'}$ an integral prescheme, and $g : X^{'} \to X$ an integral morphism. If there exists a rational $X$ -section $f$ of $X^{'}$ , then $g$ is an isomorphism.

Proof. Since the question is local on $X$ , we may assume $X$ affine of integral ring $A$ , and then $X^{'}$ is affine of ring $A^{'}$ integral over $A$ (6.1.4) and integral; furthermore, the argument of (6.1.14) shows that there exists a dense open in $X$ isomorphic to a dense open of $X^{'}$ , hence $A$ and $A^{'}$ have the same field of fractions. On the other hand, by (I, 8.2.1.1) and the hypothesis that the $O_{x}$ are integrally closed, the ring $A$ is integrally closed, hence $A^{'} = A$ , which completes the proof of (6.1.13).

6.2. Quasi-finite morphisms

Proposition.

Let $f : X \to Y$ be a morphism locally of finite type, $x$ a point of $X$ . The following conditions are equivalent:

a) The point $x$ is isolated in its fibre $f^{- 1} (f (x))$ .
b) The ring $O_{x}$ is a quasi-finite $O_{f (x)}$ -module (0, 7.4.1).

Proof. Since the question is evidently local on $X$ and on $Y$ , we may assume $X = Spec (A)$ and $Y = Spec (B)$ affine, with $A$ a $B$ -algebra of finite type (I, 6.3.3). Moreover, we may replace $X$ by $X \times_{Y} Spec (O_{f (x)})$ without changing the fibre $f^{- 1} (f (x))$ or the local ring $O_{x}$ (I, 3.6.5); we may therefore assume that $B$ is a local ring (equal to $O_{f (x)}$ ); if $n$ is the maximal ideal of $B$ , then $f^{- 1} (f (x))$ is an affine scheme of ring $A / n A$ , of finite type over $κ (f (x)) = B / n$ (I, 6.4.11). This being so, if a) is satisfied, we may further assume that $f^{- 1} (f (x))$ is reduced to the point $x$ ; then $A / n A$ is of finite rank over $B / n$ (I, 6.4.4), that is, $A$ is a quasi-finite $B$ -module. Conversely, if this is so, $f^{- 1} (f (x))$ is an Artinian affine scheme, hence discrete (I, 6.4.4); consequently $x$ is isolated in its fibre, which shows that b) implies a).

Corollary.

Let $f : X \to Y$ be a morphism of finite type. The following conditions are equivalent:

a) Every point $x \in X$ is isolated in its fibre $f^{- 1} (f (x))$ (in other words, the subspace $f^{- 1} (f (x))$ is discrete).
b) For every $x \in X$ , the prescheme $f^{- 1} (f (x))$ is a finite $κ (f (x))$ -prescheme.
c) For every $x \in X$ , the ring $O_{x}$ is a quasi-finite $O_{f (x)}$ -module.

Proof. The equivalence of a) and c) follows from (6.2.1). On the other hand, since $f^{- 1} (f (x))$ is a $κ (f (x))$ -algebraic prescheme (I, 6.4.11), the equivalence of a) and b) follows from (I, 6.4.4).

Definition.

When $f : X \to Y$ is a morphism of finite type satisfying the equivalent conditions of (6.2.2), we say that $f$ is quasi-finite, or that $X$ is quasi-finite over $Y$ .

It is clear that every finite morphism is quasi-finite (6.1.8).

Proposition.

(i) An immersion $X \to Y$ which is closed, or such that $X$ is Noetherian, is a quasi-finite morphism.
(ii) If $f : X \to Y$ and $g : Y \to Z$ are quasi-finite morphisms, then $g \circ f$ is quasi-finite.
(iii) If $X$ and $Y$ are $S$ -preschemes and $f : X \to Y$ a quasi-finite $S$ -morphism, then $f_{(S^{'})} : X_{(S^{'})} \to Y_{(S^{'})}$ is quasi-finite for every base extension $g : S^{'} \to S$ .
(iv) If $f : X \to Y$ and $g : X^{'} \to Y^{'}$ are two quasi-finite $S$ -morphisms, then
```
  f ×_S g : X ×_S X' → Y ×_S Y'
```
is quasi-finite.
(v) Let $f : X \to Y$ and $g : Y \to Z$ be morphisms such that $g \circ f$ is quasi-finite; if, further, $g$ is separated, or $X$ is Noetherian, or $X \times_{Z} Y$ is locally Noetherian, then $f$ is quasi-finite.
(vi) If $f$ is quasi-finite, so is $f_{re d}$ .

Proof. If $f : X \to Y$ is an immersion, then every fibre is reduced to a point, and assertion (i) therefore follows from (I, 6.3.4 (i) and 6.3.5). To prove (ii), we remark first that $h = g \circ f$ is of finite type (I, 6.3.4 (ii)); furthermore, if $z = h (x)$ and $y = f (x)$ , then $y$ is isolated in $g^{- 1} (z)$ , so there exists an open neighbourhood $V$ of $y$ in $Y$ meeting no point of $g^{- 1} (z)$ other than $y$ ; therefore $f^{- 1} (V)$ is an open neighbourhood of $x$ meeting

none of the $f^{- 1} (y^{'})$ , where $y^{'} \neq = y$ is in $g^{- 1} (z)$ ; since $x$ is isolated in $f^{- 1} (y)$ , it is isolated in $h^{- 1} (z) = f^{- 1} (g^{- 1} (z))$ . To prove (iii), we may restrict to the case $Y = S$ (I, 3.3.11); we remark again first that $f^{'} = f_{(S^{'})}$ is of finite type (I, 6.3.4, (iii)); on the other hand, if $x^{'} \in X^{'} = X_{(S^{'})}$ and if we set $y^{'} = f^{'} (x^{'})$ and $y = g (y^{'})$ , then $f^{' - 1} (y^{'})$ is identified with $f^{- 1} (y) \otimes_{κ (y)} κ (y^{'})$ (I, 3.6.5); since $f^{- 1} (y)$ is of finite rank over $κ (y)$ by hypothesis, $f^{' - 1} (y^{'})$ is of finite rank over $κ (y^{'})$ , hence discrete. The assertions (iv), (v), (vi) follow from (i), (ii), (iii) by the general method (I, 5.5.12), except when the hypotheses in (v) are other than " $g$ separated"; to treat this case, we remark first that if $x$ is isolated in $f^{- 1} (g^{- 1} (g (f (x))))$ , it is a fortiori isolated in $f^{- 1} (f (x))$ ; the fact that $f$ is of finite type then follows from (I, 6.3.6).

Proposition.

Let $A$ be a complete local Noetherian ring, $X$ an $A$ -scheme locally of finite type, $x$ a point of $X$ over the closed point $y$ of $Y = Spec (A)$ , and suppose that $x$ is isolated in its fibre $f^{- 1} (y)$ (where $f$ is the structure morphism $X \to Y$ ). Then $O_{x}$ is an $A$ -module of finite type, and $X$ is $Y$ -isomorphic to the sum (I, 3.1) of $X^{'} = Spec (O_{x})$ (which is a finite $Y$ -scheme) and an $A$ -scheme X''.

Proof. It follows from (6.2.1) that $O_{x}$ is a quasi-finite $A$ -module. Since $O_{x}$ is Noetherian (I, 6.3.7) and the homomorphism $A \to O_{x}$ is local, the hypothesis that $A$ is complete implies that $O_{x}$ is an $A$ -module of finite type (0, 7.4.3). Let $X^{'} = Spec (O_{x})$ be the local scheme of $X$ at the point $x$ (I, 2.4.1), and let $g : X^{'} \to X$ be the canonical morphism. Since the composite $X^{'} \to X \to Y$ is finite (6.1.1) and $f$ is separated, $g$ is finite (6.1.5, (v)), hence $g (X^{'})$ is closed in $X$ (6.1.10); on the other hand, since $g$ is of finite type, $g$ is a local isomorphism at the closed point $x^{'}$ of $X^{'}$ by the definition of $g$ and (I, 6.5.4); but since $X^{'}$ is the only open neighbourhood of $x^{'}$ , this means that $g$ is an open immersion, hence $g (X^{'})$ is also open in $X$ , which completes the proof.

Corollary.

Let $A$ be a complete local Noetherian ring, $Y = Spec (A)$ , $f : X \to Y$ a separated quasi-finite morphism. Then $X$ is $Y$ -isomorphic to a sum $X^{'} ⊔ X^{''}$ , where $X^{'}$ is a finite $Y$ -scheme and X'' a quasi-finite $Y$ -scheme such that, if $y$ is the closed point of $Y$ , $X^{''} \cap f^{- 1} (y) = \emptyset$ .

Proof. Indeed, the fibre $f^{- 1} (y)$ is finite and discrete by hypothesis, and the corollary therefore follows, by induction on the number of points of this fibre, from (6.2.5).

Remark.

In Chapter V, we will see that if $Y$ is locally Noetherian, then a separated quasi-finite morphism $X \to Y$ is necessarily quasi-affine.

6.3. Integral closure of a prescheme

Proposition.

Let $(X, A)$ be a ringed space, $B$ a (commutative) $A$ -algebra, $f$ a section of $B$ over $X$ . The following properties are equivalent:

a) The sub- $A$ -module of $B$ generated by the $f^{n}$ for $n \geq 0$ (0, 5.1.1) is of finite type.
b) There exists a sub- $A$ -algebra $C$ of $B$ that is an $A$ -module of finite type, such that $f \in Γ (X, C)$ .
c) For every $x \in X$ , $f_{x}$ is integral over the fibre $A_{x}$ .

Proof. Since the sub- $A$ -module of $B$ generated by the $f^{n}$ is an $A$ -algebra, it is clear that a) implies b). On the other hand, b) implies that for every $x \in X$ , the $A_{x}$ -module $C_{x}$ is of finite type, which implies that every element of the algebra $C_{x}$ , and in particular $f_{x}$ , is integral over $A_{x}$ . Finally, if for some $x \in X$ we have a relation of the form

  f_x^n + (a₁)_x f_x^{n−1} + … + (a_n)_x = 0

where the $a_{i}$ ( $1 \leq i \leq n$ ) are sections of $A$ over an open neighbourhood $U$ of $x$ , then the section $f^{n} ∣ U + a_{1} \cdot f^{n - 1} ∣ U + \dots + a_{n}$ is zero over a neighbourhood $V \subset U$ of $x$ , whence it follows immediately that all the $f^{k} ∣ V$ ( $k \geq 0$ ) are linear combinations with coefficients in $Γ (V, A)$ of the $f^{j} ∣ V$ for $0 \leq j \leq n - 1$ ; we conclude that c) implies a).

When the equivalent conditions of (6.3.1) are satisfied, we say that the section $f$ is integral over $A$ .

Corollary.

Under the hypotheses of (6.3.1), there exists a (unique) sub- $A$ -module $A^{'}$ of $B$ such that, for every $x \in X$ , $A_{x}^{'}$ is the set of germs $f_{x} \in B_{x}$ that are integral over $A_{x}$ . For every open $U \subset X$ , the sections of $A^{'}$ over $U$ are the sections $f \in Γ (U, B)$ that are integral over $A ∣ U$ .

Proof. The existence of $A^{'}$ is immediate, by taking $Γ (U, A^{'})$ to be the set of $f \in Γ (U, B)$ such that $f_{x}$ is integral over $A_{x}$ for every $x \in U$ . The second assertion follows immediately from (6.3.1).

It is clear that $A^{'}$ is a sub- $A$ -algebra of $B$ ; we say that it is the integral closure of $A$ in $B$ .

(6.3.3)

Let $(X, A)$ , $(Y, B)$ be two ringed spaces and

  g = (ψ, θ) : X → Y

a morphism. Let $C$ (resp. $D$ ) be an $A$ -algebra (resp. a $B$ -algebra) and let

$u : D \to C$

be a $g$ -morphism (0, 4.4.1). Then, if $C^{'}$ (resp. $B^{'}$ ) is the integral closure of $A$ (resp. $B$ ) in $C$ (resp. $D$ ), the restriction of $u$ to $B^{'}$ is a $g$ -morphism

$u^{'} : B^{'} \to C^{'} .$

Indeed, if $j$ is the canonical injection $B^{'} \to D$ , it suffices to show that

  v = u^♯ ∘ g*(j) : g*(ℬ') → 𝒞

sends $g * (B^{'})$ into $C^{'}$ . Now, an element of $g * (B^{'})_{x} = B_{ψ (x)}^{'} \otimes_{B_{ψ (x)}} A_{x}$ is integral over $A_{x}$ by the definition of $B^{'}$ , hence so is its image under $v_{x}$ , which establishes our assertion.

Proposition.

Let $X$ be a prescheme, $A$ a quasi-coherent $O_{X}$ -algebra. Then the integral closure $O_{X}^{'}$ of $O_{X}$ in $A$ is a quasi-coherent $O_{X}$ -algebra, and for every affine open $U$ of $X$ , $Γ (U, O_{X}^{'})$ is the integral closure of $Γ (U, O_{X})$ in $Γ (U, A)$ .

Proof. We may restrict to the case where $X = Spec (B)$ is affine and $A = \tilde{A}$ , where $A$ is a

$B$ -algebra; let $B^{'}$ be the integral closure of $B$ in $A$ . Everything comes down to seeing that for every $x \in X$ , an element of $A_{x}$ integral over $B_{x}$ necessarily belongs to $B_{x}^{'}$ , which follows from the fact that, for a commutative ring $C$ , the operations of integral closure in a $C$ -algebra and of passage to a ring of fractions (with respect to a multiplicative part of $C$ ) commute ([13], t. I, p. 261 and 257).

The $X$ -scheme $X^{'} = Spec (O_{X}^{'})$ is then called the integral closure of $X$ relative to $A$ (or relative to $Spec (A)$ ); it is clear that $X^{'}$ is integral over $X$ (6.1.2).

We immediately deduce from (6.3.4) that if $f : X^{'} \to X$ is the structure morphism, then, for every open $U$ of $X$ , $f^{- 1} (U)$ is the integral closure of the prescheme induced by $X$ on $U$ , relative to $A ∣ U$ .

(6.3.5)

Let $X$ , $Y$ be two preschemes, $f : X \to Y$ a morphism, $A$ (resp. $B$ ) a quasi-coherent $O_{X}$ -algebra (resp. quasi-coherent $O_{Y}$ -algebra), and let $u : B \to A$ be an $f$ -morphism. We have seen (6.3.3) that one deduces from it an $f$ -morphism $u^{'} : O_{Y}^{'} \to O_{X}^{'}$ , where $O_{X}^{'}$ (resp. $O_{Y}^{'}$ ) is the integral closure of $O_{X}$ (resp. $O_{Y}$ ) in $A$ (resp. $B$ ). Consequently, if $X^{'}$ (resp. $Y^{'}$ ) is the integral closure of $X$ (resp. $Y$ ) relative to $A$ (resp. $B$ ), one canonically deduces from $u$ a morphism $f^{'} = Spec (u^{'}) : X^{'} \to Y^{'}$ (1.5.6) making the diagram

  X' ─────f'────→ Y'
  │                │                                                       (6.3.5.1)
  ↓                ↓
  X ──────f─────→ Y

commute.

(6.3.6)

Suppose that $X$ has only a finite number of irreducible components $X_{i}$ ( $1 \leq i \leq r$ ), with generic points $ξ_{i}$ , and consider in particular the integral closure of $X$ relative to a quasi-coherent $R (X)$ -algebra $A$ (quasi-coherent as an $O_{X}$ -algebra or as an $R (X)$ -algebra, which amounts to the same thing). We know (I, 7.3.5) that $A$ is a direct sum of $r$ quasi-coherent $O_{X}$ -algebras $A_{i}$ , with the support of $A_{i}$ contained in $X_{i}$ , and the sheaf induced by $A_{i}$ on $X_{i}$ being a simple sheaf whose fibre $A_{i}$ is an algebra over $O_{ξ_{i}}$ . It is then clear (6.3.4) that the integral closure $O_{X}^{'}$ of $O_{X}$ in $A$ is the direct sum of the integral closures $O_{X}^{(i)}$ of $O_{X}$ in each of the $A_{i}$ , and consequently the integral closure $X^{'} = Spec (O_{X}^{'})$ of $X$ relative to $A$ is an $X$ -scheme that is the sum of the $Spec (O_{X}^{(i)}) = X_{i}^{'}$ ( $1 \leq i \leq r$ ).

Suppose further that the $O_{X}$ -algebra $A$ is reduced, or, equivalently, that each of the algebras $A_{i}$ is reduced, and hence may be considered as an algebra over the field $κ (ξ_{i})$ (equal to the field of rational functions of the reduced subprescheme of $X$ having $X_{i}$ as its underlying space); then (1.3.8) each of the $X_{i}^{'}$ is a reduced $X$ -scheme and $X^{'}$ is also the integral closure of $X_{re d}$ . Suppose moreover that each of the algebras $A_{i}$ is a direct sum of a finite number of fields $K_{ij}$ ( $1 \leq j \leq s_{i}$ ); if $K_{ij}$ is the sub-algebra of $A_{i}$ corresponding to $K_{ij}$ , it is clear that $O_{X}^{(i)}$ is the direct sum of the integral closures $O_{X}^{(ij)}$ of $O_{X}$ in each of the $K_{ij}$ . Consequently, $X_{i}^{'}$ is then the sum of the $X$ -schemes $X_{ij}^{'} = Spec (O_{X}^{(ij)})$ ( $1 \leq j \leq s_{i}$ ). Furthermore, under these hypotheses and with this notation:

Proposition.

Each of the $X_{ij}^{'}$ is an integral normal $X$ -prescheme, and its field of rational functions $R (X_{ij}^{'})$ is canonically identified with the algebraic closure $K_{ij}^{'}$ of $κ (ξ_{i})$ in $K_{ij}$ .

Proof. By the above, we may assume $X$ integral, so $r = 1$ and $s_{1} = 1$ , so that the unique algebra A_1 is a field $K$ ; let $ξ$ be the generic point of $X$ , and let $f : X^{'} \to X$ be the structure morphism. For every non-empty affine open $U$ of $X$ , $f^{- 1} (U)$ is identified with the spectrum of the integral closure $B_{U}^{'}$ in the field $K$ of the integral ring $B_{U} = Γ (U, O_{X})$ (6.3.4); since the ring $B_{U}^{'}$ is integral and integrally closed, so are the local rings of the points of its spectrum, hence $f^{- 1} (U)$ is by definition an integral and normal scheme ((0, 4.1.4) and (I, 5.1.4)). Moreover, since (0) is the only prime ideal of $B_{U}^{'}$ above the prime ideal (0) of B_U ([13], t. I, p. 259), $f^{- 1} (ξ)$ reduces to a single point $ξ^{'}$ , and $κ (ξ^{'})$ is the field of fractions $K^{'}$ of $B_{U}^{'}$ , which is none other than the algebraic closure of $κ (ξ)$ in $K$ . Finally, $X^{'}$ is irreducible, because as $U$ runs over the non-empty affine opens of $X$ , the $f^{- 1} (U)$ constitute an open cover of $X^{'}$ by irreducible opens; moreover, the intersection $f^{- 1} (U \cap V)$ of two such opens contains $ξ^{'}$ , hence is non-empty, and we conclude by (0, 2.1.4).

Corollary.

Let $X$ be a reduced prescheme having only a finite number of irreducible components $X_{i}$ ( $1 \leq i \leq r$ ), and let $ξ_{i}$ be the generic point of $X_{i}$ . The integral closure $X^{'}$ of $X$ relative to $R (X)$ is the sum of $r$ $X$ -schemes $X_{i}^{'}$ which are integral and normal. If $f : X^{'} \to X$ is the structure morphism, then $f^{- 1} (ξ_{i})$ reduces to the generic point $ξ_{i}^{'}$ of $X_{i}^{'}$ and we have $κ (ξ_{i}^{'}) = κ (ξ_{i})$ ; in other words, $f$ is birational.

In this case we say that $X^{'}$ is the normalisation of the reduced prescheme $X$ ; note that $f$ , being birational and integral, is surjective (6.1.10). In order to have $X^{'} = X$ , it is necessary and sufficient that $X$ be normal. When $X$ is an integral prescheme, it follows from (6.3.8) that its normalisation $X^{'}$ is integral.

(6.3.9)

Let $X$ , $Y$ be two integral preschemes, $f : X \to Y$ a dominant morphism, $L = R (X)$ , $K = R (Y)$ the fields of rational functions of $X$ and $Y$ ; there canonically corresponds to $f$ an injection $K \to L$ , and when we identify $K$ (resp. $L$ ) with the simple sheaf $R (Y)$ (resp. $R (X)$ ), this injection is an $f$ -morphism. Let $K_{1}$ (resp. $L_{1}$ ) be an extension of $K$ (resp. $L$ ) and suppose given a monomorphism $K_{1} \to L_{1}$ such that the diagram

  K₁ ────→ L₁
  ↑         ↑
  K ─────→ L

commutes; if $K_{1}$ (resp. $L_{1}$ ) is considered as a simple sheaf on $Y$ (resp. $X$ ), hence as an $R (Y)$ -algebra (resp. an $R (X)$ -algebra), this means that $K_{1} \to L_{1}$ is an $f$ -morphism. With this, if $X^{'}$ (resp. $Y^{'}$ ) is the integral closure of $X$ (resp. $Y$ ) relative to $L_{1}$ (resp. $K_{1}$ ), then $X^{'}$ (resp. $Y^{'}$ ) is an integral normal prescheme (6.3.6) whose field of rational functions is canonically identified with the algebraic closure $L^{'}$ (resp. $K^{'}$ )

of $L$ (resp. $K$ ) in $L_{1}$ (resp. $K_{1}$ ), and there exists a canonical (necessarily dominant) morphism $f^{'} : X^{'} \to Y^{'}$ making the diagram (6.3.5.1) commute.

The most important case is that in which one takes $L_{1} = L$ , so $K_{1}$ is then an extension of $K$ contained in $L$ , and one supposes $X$ integral and normal, hence $X^{'} = X$ . What precedes then shows that when $X$ is normal, and $Y^{'}$ is the integral closure of $Y$ relative to a field $K_{1} \subset L = R (X)$ , every dominant morphism $f : X \to Y$ factors as

  f : X →^{f'} Y' → Y

where $f^{'}$ is dominant; moreover, when the monomorphism $K_{1} \to L$ is given, $f^{'}$ is necessarily unique, as one sees by reducing to the case where $X$ and $Y$ are affine. We thus see that, for the data of $Y$ , $L$ , and a $K$ -monomorphism $K_{1} \to L$ , the integral closure $Y^{'}$ of $Y$ relative to $K_{1}$ is the solution of a universal problem.

Remark.

Let us take up the hypotheses of (6.3.6), supposing further that each of the algebras $A_{i}$ is of finite rank over $κ (ξ_{i})$ (which implies that $A_{i}$ is a direct sum of a finite number of fields); we can assert in certain cases that the structure morphism $X^{'} \to X$ is not only integral, but even finite. Let us restrict to the case where $X$ is reduced; since the question is local on $X$ , we may further assume that $X$ is affine of ring $C$ , and that $C$ has only a finite number of minimal ideals $p_{i}$ ( $1 \leq i \leq r$ ) such that the $C_{i} = C / p_{i}$ are integral; then $X^{'}$ will be finite over $X$ if the integral closure of each $C_{i}$ in every extension of finite degree of its field of fractions is a $C$ -module of finite type (6.3.4). We know that this condition is always satisfied when $C$ is an algebra of finite type over a field ([13], t. I, p. 267, th. 9), or over $Z$ ([9, I, p. 93, th. 3]), or over a complete local Noetherian ring ([25], p. 298). We conclude that $X^{'} \to X$ will be a finite morphism whenever $X$ is a scheme of finite type over a field, over $Z$ , or over a complete local Noetherian ring.

6.4. Determinant of an endomorphism of an $O_{X}$ -module

(6.4.1)

Let $A$ be a (commutative) ring, $E$ a free $A$ -module of rank $n$ , $u$ an endomorphism of $E$ ; we recall that to define the characteristic polynomial of $u$ , one considers the endomorphism $u \otimes 1$ of the free A[T]-module of rank $n$ , $E \otimes_{A} A [T]$ ( $T$ an indeterminate), and one sets

  P(u, T) = det(T · I − (u ⊗ 1))                                            (6.4.1.1)

( $I$ the identity automorphism of $E \otimes_{A} A [T]$ ). We have

  P(u, T) = T^n − σ₁(u) T^{n−1} + … + (−1)^n σ_n(u)                         (6.4.1.2)

where $σ_{i} (u)$ is an element of $A$ , equal to a homogeneous polynomial of degree $i$ (with integer coefficients) in the entries of the matrix of $u$ relative to an arbitrary basis of $E$ ; we say that the $σ_{i} (u)$ are the elementary symmetric functions of $u$ , and in particular we have $σ_{1} (u) = T r u$ and $σ_{n} (u) = d e t u$ . Recall that, by the Hamilton–Cayley theorem, we have

  P(u, u) = u^n − σ₁(u) u^{n−1} + … + (−1)^n σ_n(u) = 0                     (6.4.1.3)

which can also be written as

  (det u) · I_E = u · Q(u)                                                  (6.4.1.4)

(I_E the identity automorphism of $E$ ), with

  Q(u) = (−1)^{n+1} (u^{n−1} − σ₁(u) u^{n−2} + … + (−1)^{n−1} σ_{n−1}(u)).  (6.4.1.5)

Let $ϕ : A \to B$ be a ring homomorphism, making $B$ an $A$ -algebra; consider the $B$ -module $E_{(B)} = E \otimes_{A} B$ , which is free of rank $n$ , and the extension $u \otimes 1$ of $u$ to an endomorphism of $E_{(B)}$ ; it is immediate that one has $σ_{i} (u \otimes 1) = ϕ (σ_{i} (u))$ for every index $i$ .

(6.4.2)

Now suppose that $A$ is an integral ring, with field of fractions $K$ , and let $E$ be an $A$ -module of finite type (but not necessarily free). Let $n$ be the rank of $E$ , i.e. the rank of the free $K$ -module $E \otimes_{A} K$ ; to every endomorphism $u$ of $E$ canonically corresponds the endomorphism $u \otimes 1$ of $E \otimes_{A} K$ ; by abuse of language, we still call characteristic polynomial of $u$ , and denote by $P (u, T)$ , the polynomial $P (u \otimes 1, T)$ , whose coefficients $σ_{i} (u \otimes 1)$ (which belong to $K$ ) we also denote by $σ_{i} (u)$ and call the elementary symmetric functions of $u$ ; in particular $d e t u = d e t (u \otimes 1)$ by definition. With this notation, formulas (6.4.1.3) to (6.4.1.5) make sense and remain valid, provided one interprets $u^{j}$ as the homomorphism $E \to E \otimes_{A} K$ obtained by composing the endomorphism $u^{j} \otimes 1 = (u \otimes 1)^{j}$ of $E \otimes_{A} K$ with the canonical homomorphism $x \mapsto x \otimes 1$ .

If $F$ is the torsion submodule of $E$ and if $E_{0} = E / F$ , then $u (F) \subset F$ , hence, by passage to quotients, $u$ gives an endomorphism $u_{0}$ of $E_{0}$ ; moreover, $E \otimes_{A} K$ is identified with $E_{0} \otimes_{A} K$ and $u \otimes 1$ with $u_{0} \otimes 1$ , hence $σ_{i} (u) = σ_{i} (u_{0})$ for $1 \leq i \leq n$ .

When $E$ is torsion-free, $E$ is canonically identified with a sub- $A$ -module of $E \otimes_{A} K$ , and the relation $u \otimes 1 = 0$ is equivalent to $u = 0$ . When $E$ is a free $A$ -module, the two definitions of $σ_{i} (u)$ given in (6.4.1) and (6.4.2) coincide by the above, which justifies the notation adopted.

Note also that when $E$ is a torsion module — in other words, $E_{0} = 0$ — the exterior algebra of $E_{0}$ reduces to $K$ and the determinant of the unique endomorphism $u_{0}$ of $E_{0}$ is equal to 1.

Proposition.

Let $A$ be an integral ring, $E$ an $A$ -module of finite type, $u$ an endomorphism of $E$ ; then the elementary symmetric functions $σ_{i} (u)$ of $u$ (and in particular det u) are elements of $K$ integral over $A$ .

Proof. Let $A^{'}$ be the integral closure of $A$ ; since A'[T] is an integrally closed ring ([24], p. 99), it is the integral closure of A[T] in its field of fractions $K (T)$ . Replacing $u$ by $T \cdot I - u \otimes 1$ and $A$ by A[T], we see that we are reduced to proving that det u is integral over $A$ . If $n$ is the rank of $E$ , we have $d e t u = d e t (⋀^{n} u)$ and $(⋀^{n} u) \otimes 1 = ⋀^{n} (u \otimes 1)$ , so we may assume $n = 1$ . But then the map $u \mapsto d e t u$ is a homomorphism from the $A$ -module $Hom_{A} (E, E)$ to $K$ ; since $E$ is of finite type, $Hom_{A} (E, E)$ is isomorphic to a sub- $A$ -module of the $A$ -module of finite type $E^{n}$ (if $E$ admits a system of $n$ generators), so the elements det u belong to a sub- $A$ -module of $K$ of finite type, and are consequently integral over $A$ .

Corollary.

Under the hypotheses of (6.4.3), if we further suppose $A$ normal, then the $σ_{i} (u)$ (in particular Tr u and det u) belong to $A$ .

Proposition.

Let $A$ be an integral ring, $E$ an $A$ -module of finite type of rank $n$ , $u$ an endomorphism of $E$ such that the $σ_{i} (u)$ belong to $A$ . For $u$ to be an automorphism of $E$ , it is necessary that det u be invertible in $A$ ; this condition is sufficient when $E$ is torsion-free.

Proof. The condition is sufficient, since the hypothesis and the formulas (6.4.1.4) and (6.4.1.5) (valid in $E$ , and not only in $E \otimes_{A} K$ , since $E$ is torsion-free) prove that $(d e t u)^{- 1} Q (u)$ is the inverse of $u$ .

The condition is necessary, since if $u$ is invertible, it follows from (6.4.3) that $d e t (u^{- 1})$ belongs to the integral closure $A^{'}$ of $A$ in its field of fractions $K$ , and is evidently the inverse of det u in $A^{'}$ . Our assertion then follows from:

Lemma.

Let $A$ be a subring of a ring $A^{'}$ such that $A^{'}$ is integral over $A$ . If an element $x \in A$ is invertible in $A^{'}$ , it is also invertible in $A$ .

Proof. In the contrary case, $x$ would belong to a maximal ideal $m$ of $A$ , and it follows from the first theorem of Cohen–Seidenberg ([13], t. I, p. 257, th. 3) that there would be a maximal ideal $m^{'}$ of $A^{'}$ such that $m = m^{'} \cap A$ ; we would then have $x \in m^{'}$ , which is absurd.

Corollary.

Let $A$ be an integral and integrally closed ring, $E$ a torsion-free $A$ -module of finite type, $u$ an endomorphism of $E$ . For $u$ to be an automorphism of $E$ , it is necessary and sufficient that det u be invertible in $A$ .

Proof. This follows from (6.4.4) and (6.4.5).

Remark.

We will need below a generalisation of the preceding results. Consider a reduced Noetherian ring $A$ ; let $p_{α}$ ( $1 \leq α \leq r$ ) be its minimal ideals, $K_{α}$ the field of fractions of the integral ring $A / p_{α}$ , $K$ the total ring of fractions of $A$ , the direct sum of the fields $K_{α}$ . Let $E$ be an $A$ -module of finite type, and suppose that $E \otimes_{A} K$ is a free $K$ -module of rank $n$ (which here is no longer a consequence of the other hypotheses); it amounts to the same to say that all the $K_{α}$ -vector spaces $E \otimes_{A} K_{α} = E_{α}$ have the same dimension $n$ ; if then $u$ is an endomorphism of $E$ , we again set $P (u, T) = P (u \otimes 1, T)$ and $σ_{j} (u) = σ_{j} (u \otimes 1)$ , and in particular $d e t u = d e t (u \otimes 1)$ ; the $σ_{j} (u)$ are thus elements of $K$ . It is immediate that $E \otimes_{A} K$ is the direct sum of the $E_{α}$ and that the latter are stable under $u \otimes 1$ ; the restriction of $u \otimes 1$ to $E_{α}$ is in fact the extension $u_{α}$ of $u$ to this $K_{α}$ -vector space; we deduce that $σ_{j} (u)$ is the element of $K$ whose components in the $K_{α}$ are the $σ_{j} (u_{α})$ . Since the integral closure of $A$ in $K$ is the direct sum of the integral closures of $A$ in the $K_{α}$ , the $σ_{j} (u)$ are integral over $A$ .

Lemma.

The sub- $A$ -algebra of $K$ generated by all the elements $σ_{j} (u)$ ( $1 \leq j \leq n$ ), as $u$ runs over $Hom_{A} (E, E)$ , is an $A$ -module of finite type.

Proof. It suffices to prove that the sub-A[T]-algebra of K[T] generated by the $P (u, T)$ is an A[T]-module of finite type, since if the $F_{i} (T)$ ( $1 \leq i \leq m$ ) form a system of generators of this A[T]-module, then the coefficients of the $F_{i} (T)$ are integral over $A$ , hence generate an $A$ -algebra which is an $A$ -module of finite type ([13], t. I, p. 255, th. 1). We may therefore

replace $A$ by A[T] (which is Noetherian) and $E$ by $E \otimes_{A} A [T] = E^{'}$ , which is such that $E^{'} \otimes_{A [T]} K [T] = E \otimes_{A} K [T]$ is a free K[T]-module of rank $n$ . Returning to the initial notation, we therefore see that it suffices to prove that the $A$ -module generated by the elements det u, as $u$ runs over $Hom_{A} (E, E)$ , is of finite type; a fortiori (since every submodule of an $A$ -module of finite type is of finite type) it suffices to prove that as $v$ runs over the set of endomorphisms of $⋀^{n} E$ , the $A$ -module generated by the det v is of finite type; in other words, we are again reduced to the case $n = 1$ . But then the proposition follows from the fact that $Hom_{A} (E, E)$ is an $A$ -module of finite type, and that $v \mapsto d e t v$ is a homomorphism from this $A$ -module to $K$ .

Let $F$ be the kernel of the canonical homomorphism $E \to E \otimes_{A} K$ and let $E_{0} = E / F$ ; we see as above that $E \otimes_{A} K$ is identified with $E_{0} \otimes_{A} K$ , that $u (F) \subset F$ , and that if $u_{0}$ is the endomorphism of $E_{0}$ deduced from $u$ by passage to quotients, then $u \otimes 1$ is identified with $u_{0} \otimes 1$ , and $σ_{j} (u) = σ_{j} (u_{0})$ for every $j$ . If we have $F = 0$ , then formulas (6.4.1.3) and (6.4.1.5) make sense and remain valid when $E$ is identified with a submodule of $E \otimes_{A} K$ and the $u^{j}$ with homomorphisms $E \to E \otimes_{A} K$ ; consequently, Proposition (6.4.5) also extends to this case, with the same proof.

(6.4.8)

Let $(X, A)$ be a ringed space, $E$ a locally free $A$ -module (of finite rank), $u$ an endomorphism of $E$ . There exists by hypothesis a base $B$ of the topology of $X$ such that for every $V \in B$ , $E ∣ V$ is isomorphic to $A^{n} ∣ V$ (for some $n$ that may vary with $V$ ). Let $u$ be an endomorphism of $E$ ; for every $V \in B$ , $u_{V}$ is thus an endomorphism of the $Γ (V, A)$ -module $Γ (V, E)$ , which is free by hypothesis; the determinant $d e t u_{V}$ is therefore defined and belongs to $Γ (V, A)$ . Moreover, if $e_{1}, \dots, e_{n}$ form a basis of $Γ (V, E)$ , their restrictions to any open $W \subset V$ form a basis of $Γ (W, E)$ over $Γ (W, A)$ , so $d e t u_{W}$ is the restriction of $d e t u_{V}$ to $W$ . There exists therefore a unique section of $A$ over $X$ , which we denote by det u and call the determinant of $u$ , such that the restriction of det u to every $V \in B$ is $d e t u_{V}$ . It is clear that for every $x \in X$ we have $(d e t u)_{x} = d e t u_{x}$ ; for two endomorphisms $u$ , $v$ of $E$ , we have

  det(v ∘ u) = (det v)(det u)                                               (6.4.8.1)

as well as

$d e t (1_{E}) = 1_{A} (6.4.8.2)$

and, if the rank of $E$ is constant (which will be the case (0, 5.4.1) if $X$ is connected) and equal to $n$ ,

  det(s · u) = s^n det u                                                    (6.4.8.3)

for every $s \in Γ (X, A)$ (note that $d e t (0) = 0_{A}$ if $n \geq 1$ , but $d e t (0) = 1_{A}$ for $n = 0$ ). Moreover, for $u$ to be an automorphism of $E$ , it is necessary and sufficient that det u be invertible in $Γ (X, A)$ .

If the rank of $E$ is constant, one can define in the same way the elementary symmetric functions $σ_{i} (u)$ , which are elements of $Γ (X, A)$ ; the relations (6.4.1.3) to (6.4.1.5) still hold.

We have thus defined a homomorphism $d e t : Hom_{A} (E, E) \to Γ (X, A)$ of multiplicative monoids; if we note that $Hom_{A} (E, E) = Γ (X, Hom_{A} (E, E))$ by definition, we see that we can replace $X$ in this definition by an arbitrary open $U$ of $X$ , which immediately defines a homomorphism $d e t : Hom_{A} (E, E) \to A$ of sheaves of multiplicative monoids. When $E$ has constant rank, one similarly defines homomorphisms $σ_{i} : Hom_{A} (E, E) \to A$ of sheaves of sets; for $i = 1$ , the homomorphism $σ_{1} = T r$ is a homomorphism of $A$ -modules.

Let $(Y, B)$ be a second ringed space, and let $f : (X, A) \to (Y, B)$ be a morphism of ringed spaces; if $F$ is a locally free $B$ -module, then $f * (F)$ is a locally free $A$ -module (of the same rank as $F$ if the latter is of constant rank) (0, 5.4.5). For every endomorphism $v$ of $F$ , $f * (v)$ is an endomorphism of $f * (F)$ , and it follows immediately from the definitions that $d e t f * (v)$ is the section of $A = f * (B)$ over $X$ which canonically corresponds to $d e t v \in Γ (Y, B)$ . We can also say that the homomorphism $f * (d e t) : f * (Hom_{B} (F, F)) \to f * (B) = A$ is the composite

  f*(𝓗𝓸𝓶_ℬ(ℱ, ℱ)) →^{γ^♯} 𝓗𝓸𝓶_𝒜(f*(ℱ), f*(ℱ)) →^{det} 𝒜

(0, 4.4.6). We have analogous results for the $σ_{i}$ .

(6.4.9)

Suppose now that $X$ is a locally integral prescheme, so that the sheaf $R (X)$ of rational functions on $X$ is a locally simple sheaf of fields (I, 7.3.4), quasi-coherent as an $O_{X}$ -module. If $E$ is a quasi-coherent $O_{X}$ -module of finite type, then $E^{'} = E \otimes_{O_{X}} R (X)$ is a locally free $R (X)$ -module (I, 7.3.6); for every endomorphism $u$ of $E$ , $u \otimes 1_{R (X)}$ is then an endomorphism of $E^{'}$ , and $d e t (u \otimes 1)$ is a section of $R (X)$ over $X$ , which we still call determinant of $u$ and still denote by det u. It follows from (6.4.3) that det u is a section of the integral closure of $O_{X}$ in $R (X)$ (6.3.2); if, further, $X$ is normal, then det u is a section of $O_{X}$ over $X$ , and if we further suppose that $E$ is torsion-free, for $u$ to be an automorphism of $E$ , it is necessary and sufficient that det u be invertible, by virtue of (6.4.6). The formulas (6.4.8.1) to (6.4.8.3) still hold; from the homomorphism $u \mapsto d e t u$ , applied to the modules of sections of $Hom_{O_{X}} (E, E)$ , one deduces a homomorphism of sheaves $d e t : Hom_{O_{X}} (E, E) \to R (X)$ , which takes its values in $O_{X}$ when $X$ is normal. We have analogous definitions and results for the other elementary symmetric functions $σ_{j} (u)$ when $E^{'}$ is of constant rank; if moreover $X$ is normal, the $σ_{j} (u)$ are sections of $O_{X}$ over $X$ .

Finally, let $X$ , $Y$ be two integral preschemes, and let $f : X \to Y$ be a dominant morphism. We know that there then exists a canonical homomorphism $f * (R (Y)) \to R (X)$ (I, 7.3.8), whence we obtain, for every quasi-coherent $O_{Y}$ -module $F$ of finite type, a canonical homomorphism $θ : f * (F \otimes_{O_{Y}} R (Y)) \to f * (F) \otimes_{O_{X}} R (X)$ . If $v$ is an

endomorphism of $F$ , then $f * (v \otimes 1_{R (Y)})$ is an endomorphism of $f * (F \otimes_{O_{Y}} R (Y))$ , and one has a commutative diagram

  f*(ℱ ⊗_{𝒪_Y} ℛ(Y)) ─────f*(v⊗1)────→ f*(ℱ ⊗_{𝒪_Y} ℛ(Y))
        │                                       │
       θ↓                                      θ↓
  f*(ℱ) ⊗_{𝒪_X} ℛ(X) ───f*(v)⊗1────→ f*(ℱ) ⊗_{𝒪_X} ℛ(X)

One then concludes easily that $d e t f * (v)$ is the canonical image, under the homomorphism $f * (R (Y)) \to R (X)$ , of the section det v of $R (Y)$ ; indeed, one reduces immediately to the case where $X = Spec (A)$ and $Y = Spec (B)$ are affine, $A$ and $B$ being integral rings with fields of fractions $K$ and $L$ respectively, the homomorphism $B \to A$ being injective and extending therefore to a monomorphism $L \to K$ ; if $F = \tilde{M}$ , where $M$ is a $B$ -module of finite type, the rank over $L$ of $M \otimes_{B} L$ is equal to that of $(M \otimes_{B} A) \otimes_{A} K$ over $K$ , and $d e t ((u \otimes 1) \otimes 1)$ is the image in $K$ of $d e t (u \otimes 1)$ for every endomorphism $u$ of $M$ , whence the conclusion.

(6.4.10)

Suppose finally that $X$ is a locally Noetherian reduced prescheme, so that the sheaf $R (X)$ of rational functions on $X$ is again a quasi-coherent $O_{X}$ -module (I, 7.3.4); let furthermore $E$ be a coherent $O_{X}$ -module such that $E^{'} = E \otimes_{O_{X}} R (X)$ is locally free (of finite rank). By virtue of (6.4.7), if $E^{'}$ is of constant rank, we can, for every endomorphism $u$ of $E$ , define the $σ_{j} (u)$ , which are sections of $R (X)$ over $X$ . When $E^{'}$ is no longer assumed of constant rank, we can still define the homomorphism $d e t : Hom_{O_{X}} (E, E) \to R (X)$ .

6.5. Norm of an invertible sheaf

(6.5.1)

Let $(X, A)$ be a ringed space and let $B$ be a (commutative) $A$ -algebra. The $A$ -module $B$ is canonically identified with a sub- $A$ -module of $Hom_{A} (B, B)$ , a section $f$ of $B$ over an open $U$ of $X$ being identified with multiplication by this section. If $(X, A)$ and $B$ satisfy one of the conditions enumerated in (6.4.8), (6.4.9), or (6.4.10), we can therefore define $d e t (f)$ (and in certain cases the $σ_{j} (f)$ ), which are sections of $A$ or of $R (X)$ over $U$ , which we shall call the norm of $f$ (resp. the elementary symmetric functions of $f$ ) and denote by $N_{B / A} (f)$ . We shall suppose that one of the following conditions is satisfied:

(I) $B$ is a locally free $A$ -module (of finite rank).
(II) $(X, A)$ is a locally Noetherian reduced prescheme, $B$ is a coherent $A$ -module such that $B \otimes_{A} R (X)$ is a locally free $R (X)$ -module, and, for every section $f \in Γ (U, B)$ over an open $U \subset X$ , $N_{B / A} (f)$ is a section of $A$ over $U$ .

Note that this last condition is automatically satisfied when the locally Noetherian prescheme $X$ is normal (6.4.9).

On the other hand, the hypothesis that $B \otimes_{A} R (X)$ is locally free can also be expressed as follows: denote by $X_{α}$ the closed reduced subpreschemes of $X$ having as their underlying spaces the irreducible components of $X$ (I, 5.2.1), which are therefore locally Noetherian integral preschemes. Every $x \in X$ belongs to a finite number of the subspaces $X_{α}$ ; on the other hand, $B \otimes_{A} R (X_{α})$ is a locally free $R (X_{α})$ -module of constant rank $k_{α}$ (I, 7.3.6); to say that $B \otimes_{A} R (X)$ is a locally free $R (X)$ -module means that, for every $x \in X$ , the ranks $k_{α}$ corresponding to the indices such that $x \in X_{α}$ are equal. The question is in fact local, and one reduces to the case $X = Spec (C)$ , where $C$ is a reduced Noetherian ring, and $B = \tilde{D}$ , where $D$ is a $C$ -algebra that is a $C$ -module of finite type; if $p_{i}$ ( $1 \leq i \leq m$ ) are the minimal prime ideals of $C$ , the total ring of fractions $L$ of $C$ is the direct sum of the fields of fractions $K_{i}$ of the integral rings $C / p_{i}$ , and $D \otimes_{C} L$ is the direct sum of the $D \otimes_{C} K_{i}$ , whence the conclusion.

It is clear that under hypotheses (I) or (II), one thus defines a homomorphism of sheaves of multiplicative monoids $N_{B / A} : B \to A$ , also denoted by $N$ if no confusion may arise, and called the norm homomorphism. For two sections $f$ , $g$ of $B$ over the same open $U$ , one therefore has

$N_{B / A} (f g) = N_{B / A} (f) \cdot N_{B / A} (g) (6.5.1.1)$

for the corresponding sections of $A$ over $U$ ;

$N_{B / A} (1_{B}) = 1_{A}; (6.5.1.2)$

finally, for every section $s$ of $A$ over $U$ ,

$N_{B / A} (s \cdot 1_{B}) = s^{n} (6.5.1.3)$

if the rank of $B$ is constant and equal to $n$ (for $s = 0_{A}$ , this formula gives $N (0_{B}) = 0_{A}$ if $n \geq 1$ , and $N (0_{B}) = N (1_{B}) = 1_{A}$ if $n = 0$ ).

Under hypothesis (I), for $f \in Γ (U, B)$ to be invertible it is necessary and sufficient that $N (f) \in Γ (U, A)$ be so. Under hypothesis (II), this condition is necessary; it is sufficient (by (6.4.7)) when one supposes that $B \to B \otimes_{A} R (X)$ is injective and that the following more restrictive hypothesis is satisfied:

(II bis) $(X, A)$ is a locally Noetherian reduced prescheme, $B$ is a coherent $A$ -module such that $B \otimes_{A} R (X)$ is a locally free $R (X)$ -module, and, for every section $f \in Γ (U, B)$ over an open $U$ such that $B \otimes_{A} R (X) ∣ U$ is of constant rank $n$ over $R (X) ∣ U$ , the $σ_{j} (f)$ ( $1 \leq j \leq n$ ) are sections of $A$ over $U$ .

(Again note that this condition is satisfied when $X$ is normal.)

(6.5.2)

Suppose that one of the hypotheses (I), (II) of (6.5.1) is satisfied, and let $L^{'}$ be an invertible $B$ -module. We shall associate to it canonically (up to unique isomorphism) an invertible $A$ -module $L$ in the following way. Denote by $A *$ (resp. $B *$ ) the subsheaf of $A$ (resp. $B$ ) such that $Γ (U, A *)$ (resp. $Γ (U, B *)$ ) is the set of invertible

elements of $Γ (U, A)$ (resp. $Γ (U, B)$ ) for every open $U \subset X$ ; these are sheaves of multiplicative groups, and $N_{B / A}$ , restricted to $B *$ , is a homomorphism $B * \to A *$ of sheaves of groups (6.5.1). Let $L$ be the set of pairs $(U_{λ}, η_{λ})$ having the following property: $U_{λ}$ is an open of $X$ and $η_{λ}$ is an isomorphism $L^{'} ∣ U_{λ} \sim B ∣ U_{λ}$ of $(B ∣ U_{λ})$ -modules. By hypothesis, the $U_{λ}$ form a cover of $X$ ; for any two indices $λ$ , $μ$ , one sets $ω_{λ μ} = (η_{λ} ∣ U_{λ} \cap U_{μ}) \circ (η_{μ} ∣ U_{λ} \cap U_{μ})^{- 1}$ , which is an automorphism of $B ∣ U_{λ} \cap U_{μ}$ , canonically identified with a section of $B *$ over $U_{λ} \cap U_{μ}$ , and $(ω_{λ μ})$ is a 1-cocycle of the cover $U = (U_{λ})$ with values in $B *$ (0, 5.4.7). The fact that $N_{B / A} : B * \to A *$ is a homomorphism then implies that $(N_{B / A} ω_{λ μ})$ is a 1-cocycle of $U$ with values in $A *$ , and therefore corresponds (up to unique isomorphism) to an invertible $A$ -module $L$ (0, 5.4.7) which we shall denote by $N_{B / A} (L^{'})$ and call the norm of the invertible $B$ -module $L^{'}$ .

Let $M$ be a subset of $L$ such that the corresponding $U_{λ}$ still form a cover of $X$ , and let $B$ be this cover; the restriction of the cocycle $(ω_{λ μ})$ to $B$ defines a 1-cocycle $(N_{B / A} ω_{λ μ})$ of $B$ with values in $A *$ , the restriction of the 1-cocycle $(N_{B / A} ω_{λ μ})$ of $U$ ; it is clear that there is a canonical isomorphism from the invertible $A$ -module defined by this 1-cocycle of $B$ onto $N_{B / A} (L^{'})$ , which allows the invertible $A$ -module to be defined by any sub-cover of $U$ . This possibility immediately shows that, if $L_{1}^{'}$ , $L_{2}^{'}$ are two invertible $B$ -modules, then, by virtue of (6.5.1.1) and (6.5.1.2),

  N(ℒ'₁ ⊗_ℬ ℒ'₂) = N(ℒ'₁) ⊗_𝒜 N(ℒ'₂)                                       (6.5.2.1)

and

$N_{B / A} (B) = A (6.5.2.2)$

as well as

$N (L^{' - 1}) = (N (L^{'}))^{- 1} (6.5.2.3)$

up to canonical isomorphisms. Moreover, it follows from (6.5.1.3) that if $L$ is an invertible $A$ -module and if $B$ is of constant rank $n$ over $A$ in case (I) (resp. $B \otimes_{A} R (X)$ of constant rank $n$ over $R (X)$ in case (II)), one has, up to canonical isomorphism,

  N_{ℬ/𝒜}(ℒ ⊗_𝒜 ℬ) = ℒ^{⊗n}.                                               (6.5.2.4)

(6.5.3)

We now show that $N_{B / A} (L^{'})$ is a covariant functor in the category of invertible $B$ -modules. Indeed, let $h^{'} : L_{1}^{'} \to L_{2}^{'}$ be a homomorphism of invertible $B$ -modules, and let $B = (U_{λ})$ be an open cover of $X$ such that for every $λ$ one has $(B ∣ U_{λ})$ -isomorphisms $η_{λ}^{(1)} : L_{1}^{'} ∣ U_{λ} \sim B ∣ U_{λ}$ and $η_{λ}^{(2)} : L_{2}^{'} ∣ U_{λ} \sim B ∣ U_{λ}$ ; there is then for each $λ$ an endomorphism $h_{λ}^{'}$ of $B ∣ U_{λ}$ such that $h_{λ}^{'} \circ η_{λ}^{(1)} = η_{λ}^{(2)} \circ (h^{'} ∣ U_{λ})$ , and one can evidently identify $h_{λ}^{'}$ with a section of $B$ over $U_{λ}$ (0, 5.1.1). Hence, for every pair $(λ, μ)$ of indices, the restrictions to $U_{λ} \cap U_{μ}$ of $(η_{λ}^{(2)})^{- 1} \circ h_{λ}^{'} \circ η_{λ}^{(1)}$ and $(η_{μ}^{(2)})^{- 1} \circ h_{μ}^{'} \circ η_{μ}^{(1)}$ coincide. We deduce that for the 1-cocycles $(ω_{λ μ}^{(1)})$ and $(ω_{λ μ}^{(2)})$ with values in $B *$ corresponding to $L_{1}^{'}$ and $L_{2}^{'}$ , the relation

  ω_{λμ}^{(2)} h'_μ = h'_λ ω_{λμ}^{(1)}.

If we set $h_{λ} = N (h_{λ}^{'})$ , we therefore have the analogous relations

  N(ω_{λμ}^{(2)}) h_μ = h_λ N(ω_{λμ}^{(1)})

and consequently the $h_{λ}$ define a homomorphism $N (L_{1}^{'}) \to N (L_{2}^{'})$ which we shall denote by $N_{B / A} (h^{'})$ or $N (h^{'})$ . Under hypothesis (I), for $h^{'}$ to be an isomorphism, it is necessary and sufficient (since this is a local question) that $N_{B / A} (h^{'})$ be one. Under hypothesis (II), this condition is again necessary; it is sufficient if hypothesis (II bis) is satisfied and $B \to B \otimes_{A} R (X)$ is injective.

Taking in particular $L_{1}^{'} = B$ , the homomorphisms $B \to L^{'}$ are identified (0, 5.1.1) with the sections of $L^{'}$ over $X$ , whence a canonical map

  N_{ℬ/𝒜} : Γ(X, ℒ') → Γ(X, N_{ℬ/𝒜}(ℒ')).

It also follows from (6.5.1.1) that if $f_{1}^{'} \in Γ (X, L_{1}^{'})$ , $f_{2}^{'} \in Γ (X, L_{2}^{'})$ , one has

  N(f'₁ ⊗ f'₂) = N(f'₁) ⊗ N(f'₂).                                          (6.5.3.1)

For every invertible $A$ -module $L$ and every section $f \in Γ (X, L)$ , one has, taking (6.5.2.4) into account,

$N_{B / A} (f \otimes 1_{B}) = f^{\otimes n} (6.5.3.2)$

when $B$ is of constant rank $n$ under hypothesis (I) (resp. when $B \otimes_{A} R (X)$ is of constant rank $n$ under hypothesis (II)). Finally, for the homomorphism $B \to L^{'}$ corresponding to a section $f^{'}$ of $L^{'}$ over $X$ to be an isomorphism, it is necessary and sufficient that $f_{x}^{'}$ be a basis of $L_{x}^{'}$ for every $x \in X$ ; under hypothesis (I), this condition is therefore equivalent to saying that $(N (f^{'}))_{x}$ is a basis of $(N (L^{'}))_{x}$ for every $x$ ; under hypothesis (II), this condition is again necessary, and it is sufficient when $B$ satisfies (II bis) and $B \to B \otimes_{A} R (X)$ is injective.

(6.5.4)

Let $(X, A)$ , $(X^{'}, A^{'})$ be two ringed spaces,

$f : X^{'} \to X$

a morphism, $B$ an $A$ -algebra, $B^{'} = f * (B)$ the inverse image $A^{'}$ -algebra. One supposes that one of the following hypotheses is satisfied:

1° $B$ satisfies hypothesis (I) of (6.5.1).
2° $(X, A)$ and $B$ satisfy hypothesis (II) of (6.5.1), $(X^{'}, A^{'})$ is a locally Noetherian reduced prescheme, and, if one denotes by $X_{α}$ and $X_{β}^{'}$ the closed reduced subpreschemes of $X$ and $X^{'}$ respectively having as underlying spaces the irreducible components of these spaces, the restriction of $f$ to each $X_{β}^{'}$ is a dominant morphism from $X_{β}^{'}$ into some $X_{α}$ .

Under these conditions, $B^{'}$ satisfies hypothesis (I) or hypothesis (II) of (6.5.1) respectively; the first assertion is immediate. To establish the second, it suffices to see that for every $x^{'} \in X^{'}$ , the ranks of the $B^{'} \otimes_{O_{X^{'}}} R (X_{β}^{'})$ for the $β$ such that $x^{'} \in X_{β}^{'}$ are the same. But if the restriction of $f$ to $X_{β}^{'}$ is a dominant morphism into $X_{α}$ , the rank of $B^{'} \otimes_{O_{X^{'}}} R (X_{β}^{'})$ is equal to that of $B \otimes_{O_{X}} R (X_{α})$ (as one sees at once by reducing to the affine case as in (6.4.9)), whence our assertion by virtue of hypothesis (II) and (6.5.1).

This being so, it follows from (6.4.8), (6.4.9), and (6.4.10) that if $s$ is a section of $B$ over an open $U \subset X$ , $s^{'}$ the corresponding section of $B^{'}$ over $f^{- 1} (U)$ , then $N_{B^{'} / A^{'}} (s^{'})$ is the section of $A^{'}$ over $f^{- 1} (U)$ which corresponds to the section $N_{B / A} (s)$ of $A$ over $U$ .

If $M$ is an invertible $B$ -module, one deduces from the preceding that if $M^{'} = f * (M)$ (which is an invertible $B^{'}$ -module), one has $N_{B^{'} / A^{'}} (M^{'}) = f * (N_{B / A} (M))$ up to canonical isomorphism.

(6.5.5)

Suppose henceforth that $(X, A)$ is a prescheme. The data of a quasi-coherent $A$ -algebra $B$ that is an $A$ -module of finite type is then equivalent, as we know, to that of a finite morphism $g : X^{'} \to X$ such that $g_{*} (O_{X^{'}}) = B$ , defined up to $X$ -isomorphism (6.1.2 and 1.3.1). Furthermore, to give a quasi-coherent $O_{X^{'}}$ -module $F^{'}$ is equivalent to giving a quasi-coherent $B$ -module $F$ such that $g_{*} (F^{'}) = F$ (1.4.3), and for $F^{'}$ to be invertible it is necessary and sufficient that $F$ be so (6.1.12). To translate the preceding results in terms of the finite morphism $g$ , it will be necessary to suppose either that $g_{*} (O_{X^{'}})$ is a locally free $O_{X}$ -module (of finite type), or that $(X, O_{X})$ and $g_{*} (O_{X^{'}})$ satisfy hypothesis (II). For every invertible $O_{X^{'}}$ -module $L^{'}$ , one then sets

$N_{X^{'} / X} (L^{'}) = N_{g_{*} (O_{X^{'}}) / O_{X}} (g_{*} (L^{'})) (6.5.5.1)$

and one calls this the norm (relative to $g$ ) of $L^{'}$ . Similarly, if $h^{'} : L_{1}^{'} \to L_{2}^{'}$ is a homomorphism of invertible $O_{X^{'}}$ -modules, one sets

$N_{X^{'} / X} (h^{'}) = N_{g_{*} (O_{X^{'}}) / O_{X}} (g_{*} (h^{'})) : N_{X^{'} / X} (L_{1}^{'}) \to N_{X^{'} / X} (L_{2}^{'}) . (6.5.5.2)$

In particular, for $L_{1}^{'} = O_{X^{'}}$ , one thus obtains a canonical map

  N_{X'/X} : Γ(X', ℒ') → Γ(X, N_{X'/X}(ℒ')).                                (6.5.5.3)

We leave to the reader most of these translations, and confine ourselves to spelling out the following:

Proposition.

Let $g : X^{'} \to X$ be a finite morphism, and suppose that either $g_{*} (O_{X^{'}})$ is a locally free $O_{X}$ -module or that $(X, O_{X})$ and $g_{*} (O_{X^{'}})$ satisfy (II bis) (which will be the case in particular when $X$ is locally Noetherian and normal). For a homomorphism $h^{'} : L_{1}^{'} \to L_{2}^{'}$ of invertible $O_{X^{'}}$ -modules to be an isomorphism, it is necessary and sufficient, under the first hypothesis, that $N_{X^{'} / X} (h^{'})$ be one; under the second hypothesis, this condition is again necessary, and it is sufficient when the homomorphism $g_{*} (O_{X^{'}}) \to g_{*} (O_{X^{'}}) \otimes_{O_{X}} R (X)$ is injective.

Proof. Note that here we use the fact that, for $g_{*} (h^{'})$ to be an isomorphism, it is necessary and sufficient that $h^{'}$ be one (1.4.2).

Corollary.

Let $g : X^{'} \to X$ be a finite morphism, and suppose that either $g_{*} (O_{X^{'}})$ is a locally free $O_{X}$ -module or that $(X, O_{X})$ and $g_{*} (O_{X^{'}})$ satisfy (II bis) and $g_{*} (O_{X^{'}}) \to g_{*} (O_{X^{'}}) \otimes_{O_{X}} R (X)$ is injective. Let $L^{'}$ be an invertible $O_{X^{'}}$ -module, $f^{'}$ a

section of $L^{'}$ over $X^{'}$ , and $f = N_{X^{'} / X} (f^{'})$ the corresponding section of $L = N_{X^{'} / X} (L^{'})$ over $X$ (6.5.5.1). Then $g (X^{'} - X_{f^{'}}^{'}) = X - X_{f}$ , and $X_{f}$ is the largest open subset $U$ of $X$ such that $g^{- 1} (U) \subset X_{f^{'}}^{'}$ .

Proof. Indeed, $g (X^{'} - X_{f^{'}}^{'})$ is closed in $X$ (6.1.10); it therefore suffices to prove the last assertion. Now the relation $U \subset X_{f}$ is equivalent to the fact that the homomorphism $O_{X} ∣ U \to L ∣ U$ defined by $f ∣ U$ is an isomorphism. By (6.5.6), this is equivalent to saying that the homomorphism $O_{X^{'}} ∣ g^{- 1} (U) \to L^{'} ∣ g^{- 1} (U)$ defined by $f^{'} ∣ g^{- 1} (U)$ is an isomorphism, that is, to the relation $g^{- 1} (U) \subset X_{f^{'}}^{'}$ .

Proposition.

Let $g : X^{'} \to X$ be a finite morphism, $f : Y \to X$ a morphism; let $Y^{'} = X_{(Y)}^{'}$ , $g^{'} = g_{(Y)}$ , $f^{'} = f_{(X^{'})}$ , so that one has the commutative diagram

  X' ←─f'── Y'
  │          │
  g↓         ↓g'
  X ←──f──── Y

Suppose that either $g_{*} (O_{X^{'}})$ is locally free, or that $(X, O_{X})$ and $g_{*} (O_{X^{'}})$ satisfy (II), that $Y$ is a locally Noetherian reduced prescheme, and that the restriction of $f$ to every irreducible component of $Y$ is a dominant morphism into an irreducible component of $X$ . Then, for every invertible $O_{X^{'}}$ -module $L^{'}$ , one has

$N_{Y^{'} / Y} (f^{'} * (L^{'})) = f * (N_{X^{'} / X} (L^{'}))$

up to canonical isomorphism.

Proof. Note that one has $f * (g_{*} (L^{'})) = g_{*}^{'} (f^{'} * (L^{'}))$ by virtue of (1.5.2), and in particular $g_{*}^{'} (O_{Y^{'}}) = f * (g_{*} (O_{X^{'}}))$ ; if $g_{*} (O_{X^{'}})$ is locally free, then so is $g_{*}^{'} (O_{Y^{'}})$ . The conclusion then follows from the definitions and (6.5.4).

Remark.

We shall later generalise the notion of norm developed above, and relate it to the notion of direct image of a divisor.

6.6. Application: criteria for ampleness

Proposition.

Let $Y$ be a prescheme, $f : X \to Y$ a quasi-compact morphism, $g : X^{'} \to X$ a finite and surjective morphism. Suppose that either $g_{*} (O_{X^{'}})$ is a locally free $O_{X}$ -module, or that $(X, O_{X})$ and $g_{*} (O_{X^{'}})$ satisfy (II bis). Then, for every invertible $O_{X^{'}}$ -module $L^{'}$ that is ample for $f \circ g$ , $L = N_{X^{'} / X} (L^{'})$ is ample for $f$ .

Proof. We may assume $Y$ affine (4.6.4), and then by virtue of (4.6.6) the statement is equivalent to:

Corollary.

Let $X$ be a quasi-compact prescheme, $g : X^{'} \to X$ a finite surjective morphism such that either $g_{*} (O_{X^{'}})$ is a locally free $O_{X}$ -module or that $(X, O_{X})$ and $g_{*} (O_{X^{'}})$ satisfy (II bis). Then, for every ample $O_{X^{'}}$ -module $L^{'}$ , $L = N_{X^{'} / X} (L^{'})$ is ample.

Proof. Under the second hypothesis, we may further assume that the canonical homomorphism $g_{*} (O_{X^{'}}) \to g_{*} (O_{X^{'}}) \otimes_{O_{X}} R (X)$ is injective. Indeed, if not, let

$T$ be the kernel of this homomorphism, which is a coherent ideal of $B = g_{*} (O_{X^{'}})$ (I, 6.1.1), and set $X^{''} = Spec (B / T)$ ; we then have a commutative diagram

  X'' ──h──→ X'
   \         /
   g'\      / g
       ↘  ↙
         X

where $h$ is a closed immersion (1.4.10). Furthermore, we know that the support of $T$ is a closed set (0, 5.2.2) rare in $X$ (I, 7.4.6), whence we conclude that for the generic point $x$ of an irreducible component of $X$ , there is an affine open neighbourhood $U$ of $x$ such that $B ∣ U = (B / T) ∣ U$ . Since $g$ is by hypothesis surjective, we thus conclude that $x \in g^{'} (X^{''})$ ; $g^{'}$ is therefore dominant, and being a finite morphism, it is surjective (6.1.10); by definition, $g_{*}^{'} (O_{X^{''}}) \otimes_{O_{X}} R (X) = (B / T) \otimes_{O_{X}} R (X) = g_{*} (O_{X^{'}}) \otimes_{O_{X}} R (X)$ , so $(X, O_{X})$ and $g_{*}^{'} (O_{X^{''}})$ satisfy (II bis), and moreover $g_{*}^{'} (O_{X^{''}}) \to g_{*}^{'} (O_{X^{''}}) \otimes_{O_{X}} R (X)$ is injective. Finally, $h * (L^{'}) = L^{''}$ is an ample $O_{X^{''}}$ -module (4.6.13, (i bis)), and we have $N_{X^{''} / X} (L^{''}) = N_{X^{'} / X} (L^{'})$ . Indeed, to define these two invertible $O_{X}$ -modules one can use a single affine open cover $(U_{λ})$ of $X$ such that the restrictions of $g_{*} (L^{'})$ and $g_{*}^{'} (L^{''})$ to $U_{λ}$ are respectively isomorphic to $B ∣ U_{λ}$ and $(B / T) ∣ U_{λ}$ . One sees at once that to every isomorphism $η_{λ} : g_{*} (L^{'}) ∣ U_{λ} \to B ∣ U_{λ}$ there canonically corresponds an isomorphism

$η_{λ}^{'} : g_{*}^{'} (L^{''}) ∣ U_{λ} \to (B / T) ∣ U_{λ}$

so that, if $(ω_{λ μ})$ and $(ω_{λ μ}^{'})$ are the 1-cocycles corresponding to the systems of isomorphisms $(η_{λ})$ and $(η_{λ}^{'})$ (6.5.2), $ω_{λ μ}^{'}$ is the canonical image in $Γ (U_{λ} \cap U_{μ}, B / T)$ of $ω_{λ μ} \in Γ (U_{λ} \cap U_{μ}, B)$ . By virtue of the definition of $T$ , one therefore concludes that

$N_{B / A} (ω_{λ μ}) = N_{(B / T) / A} (ω_{λ μ}^{'})$

(with $A = O_{X}$ ), whence the asserted equality.

Suppose therefore that the homomorphism $g_{*} (O_{X^{'}}) \to g_{*} (O_{X^{'}}) \otimes_{O_{X}} R (X)$ is injective when we are under hypothesis (II bis). It suffices to prove that, as $f$ runs over the sections of $L^{\otimes n}$ (for $n > 0$ ) over $X$ , the $X_{f}$ form a base of the topology of $X$ (4.5.2). Now let $x \in X$ and let $U$ be an arbitrary neighbourhood of $x$ ; since $g^{- 1} (x)$ is finite (6.1.7) and $L^{'}$ is ample, there exists an integer $n > 0$ and a section $f^{'}$ of $L^{' \otimes n}$ over $X^{'}$ such that $X_{f^{'}}^{'}$ is a neighbourhood of $g^{- 1} (x)$ contained in $g^{- 1} (U)$ (4.5.4). Since

$L^{\otimes n} = N_{X^{'} / X} (L^{' \otimes n})$

it suffices to take $f = N_{X^{'} / X} (f^{'})$ ; indeed, then $X - X_{f} = g (X^{'} - X_{f^{'}}^{'})$ (6.5.7), so $x \in X_{f} \subset U$ .

Corollary.

Under the hypotheses of (6.6.1), for an invertible $O_{X}$ -module $L$ to be ample for $f$ , it is necessary and sufficient that $L^{'} = g * (L)$ be ample for $f \circ g$ .

Proof. The condition is necessary, since $g$ is affine (5.1.12). To prove that the condition is sufficient, we may assume $Y$ affine (4.6.4), hence $X$ and $X^{'}$ quasi-compact and $L^{'}$ ample (4.6.6), and it remains to show that $L$ is ample. Now, the set of points

$x \in X$ in a neighbourhood of which $g_{*} (O_{X^{'}})$ (resp. $g_{*} (O_{X^{'}}) \otimes_{O_{X}} R (X)$ ) has a given rank $n$ in the first (resp. second) hypothesis is both open and closed in $X$ , so $X$ is the prescheme sum of a finite number of these opens, and we may therefore assume that it is equal to one of them (4.6.17). But then $N_{X^{'} / X} (L^{'}) = L^{\otimes n}$ , hence $L^{\otimes n}$ is ample by virtue of (6.6.2), and so is $L$ (4.5.6).

Corollary.

Suppose that the hypotheses of (6.6.1) are satisfied and suppose further that $f : X \to Y$ is of finite type. Then, for $f$ to be quasi-projective, it is necessary and sufficient that $f \circ g$ be so. If, moreover, we suppose that $Y$ is a quasi-compact scheme, or a prescheme whose underlying space is Noetherian, then, for $f$ to be projective, it is necessary and sufficient that $f \circ g$ be so.

Proof. The hypothesis implies that $f \circ g$ is of finite type. Taking into account the definition of quasi-projective morphisms (5.3.1), the first assertion follows from (6.6.1) and (6.6.3). Taking this result and (5.5.3, (ii)) into account, it remains to prove that when $f$ is assumed quasi-projective, for $f$ to be proper, it is necessary and sufficient that $f \circ g$ be proper. Now $f$ is then separated (5.3.1) and of finite type; since $g$ is surjective, our assertion follows from (5.4.2, (ii)) and (5.4.3, (ii)).

In particular:

Corollary.

Let $X$ be a prescheme of finite type over a field $K$ , $K^{'}$ a finite-degree extension of $K$ . For $X$ to be projective (resp. quasi-projective) over $K$ , it is necessary and sufficient that $X^{'} = X \otimes_{K} K^{'}$ be projective (resp. quasi-projective) over $K^{'}$ .

Proof. The condition is indeed necessary (5.3.4, (iii) and 5.5.5, (iii)). Conversely, suppose it is satisfied, and let $g : X^{'} \to X$ be the canonical projection. It is clear that $g$ is a finite morphism (6.1.5, (iii)) and surjective (I, 3.5.2, (ii)). Moreover, $g_{*} (O_{X^{'}})$ is a locally free $O_{X}$ -module, being isomorphic to $O_{X} \otimes_{K} K^{'}$ (1.5.2). It then follows from the hypothesis and from (6.1.11) and (5.5.5, (ii)) that $X^{'}$ is projective (resp. quasi-projective) over $K$ ; we then deduce from (6.6.4) that $X$ is projective (resp. quasi-projective) over $K$ .

In Chapter V, we will show that the statement of (6.6.5) remains valid when $K^{'}$ is an arbitrary extension of $K$ .

The end of this section is devoted to the proof of the criterion (6.6.11), which is a rather technical refinement of (6.6.1); it may be omitted on a first reading.

Lemma.

Let $X$ be a reduced Noetherian prescheme, $E$ a coherent $O_{X}$ -module such that $E \otimes_{O_{X}} R (X)$ is a locally free $R (X)$ -module of constant rank $n$ . Then there exists a reduced Noetherian prescheme $Z$ and a finite birational morphism $h : Z \to X$ having the following property: the morphisms of sheaves of sets $σ_{i} : Hom_{O_{X}} (E, E) \to R (X)$ ( $1 \leq i \leq n$ ) (cf. (6.4.10)) send $Hom_{O_{X}} (E, E)$ into the coherent $O_{X}$ -algebra $h_{*} (O_{Z})$ .

Proof. Consider an affine open $U$ of $X$ of ring $A (U) = A$ ; let $E = Γ (U, E)$ , and let C_U be the sub-algebra of $R (U)$ generated by the $σ_{i} (u)$ as $u$ runs over $Hom_{A} (E, E)$ ; we have seen (6.4.7.1) that this $A$ -algebra is of finite rank. Moreover, it is clear that the formation of the algebras C_U commutes with the restriction operations from an affine open $U$ to an affine open $U^{'} \subset U$ . We have thus defined a finite sub- $O_{X}$ -algebra $C$ of $R (X)$ such that $Γ (U, C) = C_{U}$ for every affine open $U$ of $X$ . We

shall take $Z = Spec (C)$ , and for $h$ the structure morphism, which is therefore finite (6.1.2); since $C$ is reduced, $Z$ is a reduced Noetherian prescheme (1.3.8). Finally, the total ring of fractions of C_U is $R (U)$ by definition, and since C_U is contained in the integral closure of $A (U)$ in $R (U)$ , there is a bijective correspondence between minimal prime ideals of $A (U)$ and minimal prime ideals of C_U ([13], t. I, p. 259), which proves that $h$ is birational and completes the proof.

Corollary.

Under the hypotheses of (6.6.6), let $W$ be an open of $X$ such that, for every $x \in W$ , either $X$ is normal at the point $x$ , or $E_{x}$ is a free $O_{x}$ -module. Then we may suppose $h$ defined so that the restriction of $h$ to $h^{- 1} (W)$ is an isomorphism of $h^{- 1} (W)$ onto $W$ .

Proof. Indeed, either hypothesis implies that if $U \subset W$ is an affine open, then, with the notation of (6.6.6), $(σ_{i} (u))_{x} \in A_{x}$ for every $x \in U$ (6.4.3), so $σ_{i} (u) \in A$ , and the conclusion follows from the definition of $h$ given in (6.6.6).

(6.6.8)

Let $X$ be a reduced Noetherian prescheme, $g : X^{'} \to X$ a finite surjective morphism, so that $B = g_{*} (O_{X^{'}})$ is a coherent $O_{X}$ -algebra, and suppose moreover that $B \otimes_{O_{X}} R (X)$ is a locally free $R (X)$ -module of constant rank $n$ . We can then apply the lemma (6.6.6) taking $E = B$ , whence, with the notation of (6.6.6), a homomorphism of sheaves of multiplicative monoids $σ_{n} : Hom_{O_{X}} (B, B) \to h_{*} (O_{Z})$ , and composing this homomorphism with the canonical homomorphism $B \to Hom_{O_{X}} (B, B)$ (6.5.1), one therefore obtains a homomorphism of sheaves of multiplicative monoids:

  N' : ℬ = g_*(𝒪_{X'}) → h_*(𝒪_Z) = 𝒞.                                     (6.6.8.1)

This being so, for every invertible $O_{X^{'}}$ -module $L^{'}$ , $g_{*} (L^{'})$ is an invertible $B$ -module (6.1.12), and the method of (6.5.2) allows us to associate to $L^{'}$ functorially an invertible $C$ -module, which we shall denote by $N^{'} (g_{*} (L^{'}))$ .

Lemma.

Let $X$ be a reduced Noetherian prescheme, $g : X^{'} \to X$ a finite surjective morphism such that $g_{*} (O_{X^{'}}) \otimes_{O_{X}} R (X)$ is a locally free $R (X)$ -module of constant rank. Then there exists a reduced Noetherian prescheme $Z$ and a finite birational morphism $h : Z \to X$ having the following property: for every ample $O_{X^{'}}$ -module $L^{'}$ , the invertible $O_{Z}$ -module $M$ such that $h_{*} (M) = N^{'} (g_{*} (L^{'}))$ (notation of (6.6.8)) is ample.

Proof. Suppose first that the homomorphism $B \to B \otimes_{O_{X}} R (X)$ is injective. Define $Z$ and $h$ as in (6.6.6) (with $E = g_{*} (O_{X^{'}})$ ). Let $z \in Z$ ; we have to show that there exists an integer $m > 0$ and a section $t$ of $M^{\otimes m}$ over $Z$ such that $Z_{t}$ is an affine open containing $z$ (4.5.2). Let $x = h (z)$ , and let $U$ be an affine open of $X$ containing $x$ ; then $h^{- 1} (U)$ is an affine open neighbourhood of $z$ , and it suffices to find $t$ such that $z \in Z_{t} \subset h^{- 1} (U)$ , since $Z_{t}$ will then necessarily be affine (5.5.8). There exists by hypothesis an integer $n > 0$ and a section $s^{'}$ of $L^{' \otimes n}$ over $X^{'}$ such that

$g^{- 1} (x) \subset X_{s^{'}}^{'} \subset g^{- 1} (U) (6.6.9.1)$

by virtue of (4.5.4). By definition, $s^{'}$ is also a section of $g_{*} (L^{'})$ over $X$ , and there corresponds to it, as in (6.5.2), a section $s = N^{'} (s^{'})$ of $N^{'} (g_{*} (L^{'}))$ over $X$ .

We shall see that if $t$ is the section $s$ considered as a section of $M$ over $Z$ , then $t$ answers the question. Set

  V = X − g(X' − X'_{s'})                                                  (6.6.9.2)

which is an open of $X$ containing $x$ and contained in $U$ , by virtue of (6.6.9.1) and (6.1.10). We shall show that

$h^{- 1} (V) \subset Z_{t} \subset h^{- 1} (U), (6.6.9.3)$

which will complete the proof. It amounts to saying that the set $T$ of $y \in X$ such that $s_{y}$ is invertible contains $V$ and is contained in $U$ . For this, consider first an affine open $W$ contained in $V$ ; $g^{- 1} (W)$ is an affine open in $X^{'}$ , and by virtue of (6.6.9.2), $s_{y^{'}}^{'}$ is invertible for every $y^{'} \in g^{- 1} (W)$ ; by virtue of the hypotheses on $X$ and $B$ , one can apply the results of (6.4.7) and one sees that if $y = g (y^{'})$ , then $s_{y}$ is invertible; in other words, $V \subset T$ . On the other hand, it also follows from (6.4.7) that, conversely, if $s_{y}$ is invertible, so is $s_{y^{'}}^{'}$ , which implies $y^{'} \in g^{- 1} (U)$ by (6.6.9.1), and therefore $y \in U$ , whence $T \subset U$ in this case.

One passes from this to the general case by the same reasoning as in (6.6.2), replacing $X^{'}$ by X'' such that $g_{*}^{'} (O_{X^{''}}) \to g_{*}^{'} (O_{X^{''}}) \otimes_{O_{X}} R (X)$ is injective, and $L^{'}$ by an ample $O_{X^{''}}$ -module $L^{''}$ such that $N^{'} (g_{*} (L^{'})) = N^{'} (g_{*}^{'} (L^{''}))$ . The lemma (6.6.9) is therefore proven in all cases (with a suitable choice of $h$ ).

Corollary.

Suppose that the hypotheses of (6.6.9) are satisfied; for every invertible $O_{X}$ -module $L$ such that $g * (L)$ is ample, $h * (L)$ is ample.

Proof. Indeed, if we set $L^{'} = g * (L)$ , then $g_{*} (L^{'}) = L \otimes_{O_{X}} B$ (0, 5.4.10), so

  N'(g_*(ℒ')) = (ℒ ⊗_{𝒪_X} 𝒞)^{⊗n}

(by the same reasoning as for (6.5.2.4)). We conclude that $M = (h * (L))^{\otimes n}$ , and since $M$ is ample, so is $h * (L)$ (4.5.6).

Proposition.

Let $Y$ be an affine scheme, $X$ a reduced Noetherian prescheme, $f : X \to Y$ a quasi-compact morphism, $g : X^{'} \to X$ a finite surjective morphism. Let $W$ be an open part of $X$ such that, for every $x \in W$ , either $X$ is normal at the point $x$ , or there exists an open neighbourhood $T \subset W$ of $x$ such that $(g_{*} (O_{X^{'}})) ∣ T$ is a locally free $(O_{X} ∣ T)$ -module. Then there exists a reduced $Y$ -prescheme $Z$ and a finite and birational $Y$ -morphism $h : Z \to X$ such that the restriction of $h$ to $h^{- 1} (W)$ is an isomorphism $h^{- 1} (W) \sim W$ and which possesses the following property: for every invertible $O_{X}$ -module $L$ such that $g * (L)$ is ample relative to $f \circ g$ , $h * (L)$ is ample relative to $f \circ h$ .

Proof. Since $Y$ is affine, $g * (L)$ is ample, and it is then a matter of proving that for a suitable choice of $h$ , $h * (L)$ is ample (4.6.6). We shall see that we can replace $g$ by a finite surjective morphism $g^{'} : X^{''} \to X$ such that $g^{'} * (L)$ is ample and $g_{*}^{'} (O_{X^{''}}) \otimes_{O_{X}} R (X)$ is a locally free $R (X)$ -module of constant rank; we will then be reduced to the conditions of (6.6.10) and the proposition will be proven.

For this, let $B = g_{*} (O_{X^{'}})$ ; denote by $X_{i}$ ( $1 \leq i \leq n$ ) the closed reduced subpreschemes of $X$ having as their underlying spaces the irreducible components of $X$ (I, 5.2.1); they are integral by hypothesis. Let $X_{i}^{'}$ be the closed

subprescheme $g^{- 1} (X_{i})$ of $X^{'}$ , $g_{i} : X_{i}^{'} \to X_{i}$ the morphism $g$ restricted to $X_{i}^{'}$ , which is finite (6.1.5, (iii)) and surjective; let $k_{i}$ be the rank of the $O_{X_{i}}$ -algebra $B_{i} = (g_{i})_{*} (O_{X_{i}^{'}})$ . Since $B_{i} \otimes_{O_{X_{i}}} R (X_{i})$ is a constant presheaf (I, 7.3.5), the rank $k_{i}$ is also the rank of the $(O_{X} ∣ U)$ -algebra $B ∣ U$ for every open $U$ of $X$ meeting only the single irreducible component $X_{i}$ . If $T$ is an open set in $X$ such that $B ∣ T$ is isomorphic to $O_{X}^{m} ∣ T$ , it follows from the preceding remark that the numbers $k_{i}$ are equal to $m$ for all the indices $i$ such that $T \cap X_{i} \neq = \emptyset$ . Let then $U$ be the open of $X$ formed by the points in a neighbourhood of which $B$ is a locally free $O_{X}$ -module, and let $U_{j}$ ( $1 \leq j \leq s$ ) be its connected components, which are open in $X$ and finite in number (since $U$ is Noetherian); denote by $V_{j}$ the closed subprescheme of $X^{'}$ , closure of the subprescheme induced on the open $g^{- 1} (U_{j})$ (I, 9.5.11). By the above, for all indices $i$ such that $X_{i} \cap U_{j} \neq = \emptyset$ , the ranks $k_{i}$ are all equal to a single integer $m_{j}$ ; note moreover that a single $X_{i}$ cannot meet two $U_{j}$ of distinct indices. Let $i_{λ}$ be the indices $i$ such that $X_{i} \cap U = \emptyset$ . Consider the product $k$ of all the $k_{i}$ , set $n_{i} = k / k_{i}$ , and let X'' be the prescheme defined as follows. For each $j$ ( $1 \leq j \leq s$ ), one considers $k / m_{j}$ preschemes isomorphic to $V_{j}$ , and for each $λ$ , $k / k_{i_{λ}} = n_{i_{λ}}$ preschemes isomorphic to $X_{i_{λ}}^{'}$ ; X'' is the sum of all these preschemes. One defines a morphism $g^{''} : X^{''} \to X^{'}$ reducing to the canonical injection on each of the summands of X''; it is clear that g'' is a finite dominant morphism, hence surjective (since a finite morphism is closed (6.1.10)); set $g^{'} = g \circ g^{''}$ , which is a finite surjective morphism $X^{''} \to X$ ; one has $g^{'} * (L) = g^{''} * (g * (L))$ , so $g^{'} * (L)$ is an ample $O_{X^{''}}$ -module (5.1.12). It is clear then that for this new prescheme X'', the ranks defined like the $k_{i}$ for $X^{'}$ are all equal to $k$ ; taking (I, 7.3.3) into account, one immediately concludes that for every affine open $T$ of $X$ , $(g_{*}^{'} (O_{X^{''}}) \otimes_{O_{X}} R (X)) ∣ T$ is an $(R (X) ∣ T)$ -module isomorphic to $(R (X) ∣ T)^{k}$ .

Corollary.

If, in the statement of (6.6.11), we have $W = X$ , then for an invertible $O_{X}$ -module $L$ to be ample relative to $f$ , it is necessary and sufficient that $g * (L)$ be ample relative to $f \circ g$ .

Remark.

In Chapter III, we will see that, if $Y$ is Noetherian, $f$ of finite type, and if the restriction of $f$ to the closed reduced subprescheme of $X$ having $X - W$ as its underlying space is proper, then the conclusion of (6.6.12) is still valid. But we shall give in Chapter V examples of algebraic schemes $X$ over a field $K$ (the structure morphism $X \to Spec (K)$ not being proper) whose normalisation $X^{'}$ is quasi-affine, but which is not quasi-affine (so that $O_{X}$ is not ample, even though $O_{X^{'}}$ is (5.1.12) and the morphism $X^{'} \to X$ is finite and surjective (6.3.10)). We shall see in the next section that this circumstance cannot occur when we replace "quasi-affine" by "affine".

6.7. Chevalley's theorem

We shall establish (with the help of Serre's criterion (5.2.1)) the following theorem, which was proven by C. Chevalley by other methods, in the case of algebraic schemes.

Theorem.

Let $X$ be an affine scheme, $Y$ a Noetherian prescheme, and $f : X \to Y$ a finite surjective morphism. Then $Y$ is an affine scheme.

Proof. It is clear that f_red : X_red → Y_red is finite (6.1.5, (vi)); since $X_{re d}$ is an affine scheme, and saying that $Y$ is affine is equivalent to saying that $Y_{re d}$ is affine (since $Y$ is Noetherian (I, 6.1.7)), we see that we may suppose $Y$ reduced. For every closed part $Y^{'}$ of $Y$ , there is then a single reduced subprescheme of $Y$ having $Y^{'}$ as its underlying space (I, 5.1.2); its inverse image $f^{- 1} (Y^{'})$ , canonically isomorphic to $X \times_{Y} Y^{'}$ (I, 4.4.1), is affine as a closed subprescheme of $X$ , and the restriction of $f$ to $f^{- 1} (Y^{'})$ , which is identified with $f \times_{Y} 1_{Y^{'}}$ , is a finite surjective morphism (6.1.5, (iii)). By virtue of the principle of Noetherian induction (0, 2.2.2), we may therefore (taking (I, 6.1.7) into account) reduce to proving the theorem under the hypothesis that for every closed part $Y^{'} \neq = Y$ , every closed subprescheme of $Y$ having $Y^{'}$ as its underlying space is affine. We thus conclude that, for every coherent $O_{Y}$ -module $F$ whose (closed) support $Z$ is distinct from $Y$ , one has $H^{1} (Y, F) = 0$ . Indeed, there exists a closed subprescheme $Y^{'}$ of $Y$ having $Z$ as its underlying space and such that, if $j : Y^{'} \to Y$ is the canonical injection, one has $F = j_{*} (j * (F))$ (I, 9.3.5); consequently (5.2.3) $H^{1} (Y, F) = H^{1} (Y^{'}, j * (F)) = 0$ by (I, 5.1.9.2).

Suppose first that $Y$ is not irreducible, and let $Y^{'}$ be an irreducible component of $Y$ and $Y^{''} = Y - Y^{'}$ ; we again denote by $Y^{'}$ the closed reduced subprescheme of $Y$ having $Y^{'}$ as its underlying space, and by $j$ the canonical injection $Y^{'} \to Y$ . Let $F$ be a coherent $O_{Y}$ -module, and consider the canonical homomorphism

  ρ : ℱ → ℱ' = j_*(j*(ℱ))

(0, 4.4.3); $F^{'}$ is a coherent $O_{Y}$ -module by virtue of (0, 5.3.10) and (0, 5.3.12), since $j_{*} (O_{Y^{'}}) = O_{Y} / J$ , denoting by $J$ the sheaf of ideals of $O_{Y}$ defining the subprescheme $Y^{'}$ . Consequently $G = Ker ρ$ and $K = I m ρ$ are also coherent $O_{Y}$ -modules (0, 5.3.4); now by definition the fibre $F_{y}^{'}$ of $F^{'}$ at the generic point $y$ of $Y^{'}$ is equal to $F_{y}$ , since $y$ is interior to $Y^{'}$ and therefore $J_{y} = 0$ , since $Y$ is reduced. We conclude that $y$ is not contained in the (closed) support of $G$ ; moreover, the support of $F^{'}$ (and a fortiori that of $K$ ) is contained in $Y^{'}$ ; in other words, the supports of $G$ and $K$ are distinct from $Y$ . We deduce that $H^{1} (Y, G) = H^{1} (Y, K) = 0$ , and the exact sequence of cohomology applied to the exact sequence $0 \to G \to F \to K \to 0$ shows that $H^{1} (Y, F) = 0$ . We conclude by Serre's criterion (5.2.1).

So suppose $Y$ irreducible, and hence integral. We may also suppose $X$ integral: indeed, if we denote by $X_{i}$ the closed reduced subpreschemes of $X$ having as their underlying spaces the irreducible components of $X$ (I, 5.2.1) and by $g_{i}$ the restriction of $g$ to $X_{i}$ , then at least one of the $g_{i}$ is dominant, and since it is a finite morphism (6.1.5), it is surjective (6.1.10); since $X_{i}$ is also an affine scheme, we see that we may replace $X$ by $X_{i}$ in the statement.

Lemma.

Let $X$ , $Y$ be two integral Noetherian preschemes, $x$ (resp. $y$ ) the generic point of $X$ (resp. $Y$ ), $f : X \to Y$ a finite surjective morphism. Let $L$ be an invertible $O_{X}$ -module such that there exists an affine open neighbourhood $U$ of $y$ and a section $g \in Γ (X, L)$ for which

$x \in X_{g} \subset f^{- 1} (U)$ . Then there exist two integers $m > 0$ , $n > 0$ , a homomorphism $u : O_{Y}^{m} \to f_{*} (L^{\otimes n})$ , and an open neighbourhood $V$ of $y$ such that the restriction $u ∣ V$ is an isomorphism of $O_{Y}^{m} ∣ V$ onto $f_{*} (L^{\otimes n}) ∣ V$ .

Proof. Let $C$ be the (integral) ring of $U$ , $k = O_{y}$ its field of fractions: since $f$ is finite, $U^{'} = f^{- 1} (U)$ is affine (1.3.2); let $D$ be its (integral) ring of fractions $K = O_{x}$ ; by hypothesis $D$ is a $C$ -module of finite type (6.1.4), so $K$ is an extension of finite rank of $k$ . The fibre $f^{- 1} (y) = X \times_{Y} Spec (κ (y)) = X \times_{Y} Spec (O_{y})$ is identified with $Spec (K)$ (I, 3.6.5); let $s_{i}$ ( $1 \leq i \leq m$ ) be elements of $D$ forming a basis of $K$ over $k$ . There exists $n > 0$ such that the sections $(s_{i} ∣ X_{g}) g^{\otimes n}$ of $L^{\otimes n}$ over $X_{g}$ extend to sections $b_{i}$ ( $1 \leq i \leq m$ ) of $L^{\otimes n}$ over $X$ (I, 9.3.1). The $b_{i}$ are also, by definition, sections of $f_{*} (L^{\otimes n})$ over $Y$ and therefore define a homomorphism $u : O_{Y}^{m} \to f_{*} (L^{\otimes n})$ (0, 5.1.1); we shall see that $u$ answers the question. One has $L^{\otimes n} ∣ U^{'} = \tilde{M}$ , where $M$ is a $D$ -module of finite type, so if $ϕ$ is the injection $C \to D$ corresponding to the morphism $f^{- 1} (U) \to U$ given by the restriction of $f$ , then $M_{[ϕ]}$ is a $C$ -module of finite type; consequently

$f_{*} (L^{\otimes n}) ∣ U = \tilde{M_{[ϕ]}}$

(I, 1.6.3) is coherent, and since $U$ is an arbitrary affine open of $Y$ , $f_{*} (L^{\otimes n})$ is coherent; moreover, $u ∣ U = \tilde{θ}$ , where $θ$ is a $C$ -homomorphism $C^{m} \to M_{[ϕ]}$ , and $u_{y} = θ_{y}$ is the homomorphism $θ \otimes 1 : k^{m} = C^{m} \otimes K \to M_{[ϕ]} \otimes K$ ; but the latter is by definition an isomorphism, since the $(b_{i})_{x}$ form a basis of $M_{[ϕ]} \otimes K$ over $k$ , $g_{x}$ being by hypothesis a generator of $(L^{\otimes n})_{x}$ . We conclude that the supports of the $O_{Y}$ -modules Ker u and Coker u do not contain $y$ ; since these $O_{Y}$ -modules are coherent (0, 5.3.3), their supports are closed (0, 5.2.2), whence the lemma.

This being so, the hypotheses of the lemma (6.7.1.1) are fulfilled in the case we are considering, by taking $L = O_{X}$ , since $X$ is affine (I, 1.1.10); we shall set $A = O_{Y}$ and $B = f_{*} (O_{X})$ . By virtue of Serre's criterion (5.2.1.1), it suffices to prove that for every coherent $O_{Y}$ -module $F$ , one has $H^{1} (Y, F) = 0$ ; it even suffices to prove this when $F \subset O_{Y}$ , which entails that $F$ is torsion-free since $Y$ is integral; in fact we shall show that $H^{1} (Y, F) = 0$ for every coherent torsion-free $O_{Y}$ -module $F$ . Now the homomorphism $u : A^{m} \to B$ defines a homomorphism

  v : 𝒢 = 𝓗𝓸𝓶_𝒜(ℬ, ℱ) → 𝓗𝓸𝓶_𝒜(𝒜^m, ℱ) = ℱ^m.

We first show that $v$ is injective: by hypothesis $T = C o k er u$ has a support not meeting $V$ , hence is a torsion $O_{Y}$ -module (I, 7.4.6); the exact sequence

$A^{m} \to B \to T \to 0$

gives, by left exactness of the functor $Hom_{A}$ , the exact sequence

  0 → 𝓗𝓸𝓶_𝒜(𝒯, ℱ) → 𝒢 →^v ℱ^m.

But since $F$ is torsion-free, one has $Hom_{A} (T, F) = 0$ (0, 5.2.6), whence our assertion. We therefore have the exact sequence

  0 → 𝒢 → ℱ^m → Coker v → 0

where $G$ and Coker v are coherent $O_{Y}$ -modules ((0, 5.3.4) and (0, 5.3.5)); by virtue of the exact sequence of cohomology, it will suffice to show that $H^{1} (Y, G) = H^{1} (Y, C o k er v) = 0$ in order to deduce $H^{1} (Y, F^{m}) = (H^{1} (Y, F))^{m} = 0$ , and therefore $H^{1} (Y, F) = 0$ . Now the restriction $v ∣ V$ is an isomorphism, hence the support of Coker v is distinct from $Y$ , whence $H^{1} (Y, C o k er v) = 0$ by the hypothesis at the start. On the other hand, $G$ is a coherent $B$ -module (I, 9.6.4); since $X$ is affine over $Y$ , there exists a quasi-coherent $O_{X}$ -module $K$ such that $G$ is isomorphic to $f_{*} (K)$ (1.4.3); since $X$ is affine, one has $H^{1} (X, K) = 0$ (I, 5.1.9.2), hence also $H^{1} (Y, G) = 0$ by (5.2.3), which completes the proof of theorem (6.7.1).

Corollary.

Let $X$ be a Noetherian prescheme, $(X_{i})_{1 \leq i \leq n}$ a finite cover of the space $X$ consisting of closed sets. For $X$ to be affine, it is necessary and sufficient that, for each $i$ , there exist a closed subprescheme of $X$ having $X_{i}$ as its underlying space and which is affine.

Proof. Indeed, if this is so, let $X^{'}$ be the scheme sum of the $X_{i}$ ; it is clear that $X^{'}$ is affine, and one defines a surjective morphism $f : X^{'} \to X$ by taking as the restriction of $f$ to $X_{i}$ the canonical injection. Everything comes down to verifying that $f$ is finite, by virtue of (6.7.1), and this has been seen in (6.1.5).

Corollary.

For a Noetherian prescheme $X$ to be affine, it is necessary and sufficient that the closed reduced subpreschemes having as their underlying spaces the irreducible components of $X$ be affine.

Éléments de Géométrie Algébrique — English