Éléments de Géométrie Algébrique — English

§18. Complements on étale morphisms. Henselian local rings

In the present section, we study various properties special to étale morphisms. In addition, the notion of étale morphism allows one to develop in a very natural fashion Nagata's theory of Henselian rings, as well as that of strictly local rings. These rings play an important role in many recent developments, by appearing each time that one needs a "localization" procedure finer than the one provided by the Zariski topology (cf. for example [43], while awaiting the appearance of the chapter of our Treatise devoted to the study of the "étale topology").

18.1. A remarkable equivalence of categories

Proposition (18.1.1).

Let $S$ be a prescheme, S_0 a closed sub-prescheme of $S$, X_0 an S_0-prescheme smooth (resp. étale) over S_0, $x_0$ a point of X_0. Then there exist an open neighbourhood U_0 of $x_0$ in X_0, an $S$-prescheme $U$ smooth (resp. étale) over $S$, and an S_0-isomorphism $U{\times }_S{S}_0\stackrel{\sim }{\to }{U}_0$.

Note that if X_0 is étale over S_0 at the point $x_0$, then it is a fortiori unramified over S_0 at that point; if one has constructed an $S$-prescheme $U$ smooth over $S$ such that $U{\times }_S{S}_0$ is isomorphic to U_0, then since the fibres of the morphisms $U_0\to S_0$ and $U\to S$ containing $x_0$ are then isomorphic, it will follow that $U$ is unramified over $S$ at the point $x_0$ (17.4.1, d), hence also in a neighbourhood of $x_0$; replacing $U$ by this neighbourhood, one concludes that $U$ will be étale over $S$. It is therefore enough to prove the proposition when one assumes only X_0 smooth over S_0.

The question being local on $S$ and on X_0, one may suppose that $S=\mathrm{Spec}(A)$ and $X_0=\mathrm{Spec}(C_0)$ are affine, so that $S_0=\mathrm{Spec}(A_0)$, where A_0 is a quotient ring of $A$, $C_0=B_0/{\mathfrak{J}}_0$, where $B_0=A_0[T_1,\cdots ,T_n]$ and ${\mathfrak{J}}_0$ is an ideal of finite type of B_0; finally, C_0 is a formally smooth A_0-algebra for the discrete topologies. Let ${\mathfrak{p}}_0$ be the ideal $j_{x_0}$ in C_0; one has ${\mathfrak{p}}_0={\mathfrak{q}}_0/{\mathfrak{J}}_0$, where ${\mathfrak{q}}_0$ is a prime ideal of B_0. The Jacobian criterion (0, 22.6.4) combined with (0, 19.1.12) shows that there exist in ${\mathfrak{J}}_0$ a family of $r$ polynomials $u_i$ $(1⩽i⩽r)$ and $r$ indices $j_h$ $(1⩽h⩽r)$ such that the images of the $u_i$ in $({\mathfrak{J}}_0{)}_{{\mathfrak{q}}_0}/({\mathfrak{J}}_0^2{)}_{{\mathfrak{q}}_0}$ generate this $(B_0{)}_{{\mathfrak{q}}_0}$-module and such that

$$(\mathrm{18.1.1.1})det(\partial u_i/\partial T_{j_h})\notin {\mathfrak{q}}_0.$$

Since $(B_0{)}_{{\mathfrak{q}}_0}$ is a local ring, it follows from Nakayama's lemma that one may suppose the images of the $u_i$ in $({\mathfrak{J}}_0{)}_{{\mathfrak{q}}_0}$ generate this $(B_0{)}_{{\mathfrak{q}}_0}$-module, then, by replacing if necessary X_0 by an affine open neighbourhood of $x_0$, that the $u_i$ generate ${\mathfrak{J}}_0$ $(0_I,\mathrm{5.2.2})$. Put then $B=A[T_1,\cdots ,T_n]$; B_0 is thus a quotient ring

of $B$, ${\mathfrak{p}}_0$ is the image of a prime ideal $\mathfrak{q}$ of $B$, and $\mathfrak{q}$ is the inverse image of ${\mathfrak{q}}_0$. For each $i$, let $v_i\in B$ be an element whose image is $u_i$ in B_0, and let $\mathfrak{J}$ be the ideal of $B$ generated by the $v_i$, so that ${\mathfrak{J}}_0$ is the image of $\mathfrak{J}$ in B_0. The proposition will be established by taking for $U$ an open neighbourhood of the point of $\mathrm{Spec}(B/\mathfrak{J})$ corresponding to the prime ideal $\mathfrak{p}=\mathfrak{q}/\mathfrak{J}$, provided one proves that $B_{\mathfrak{q}}/{\mathfrak{J}}_{\mathfrak{q}}$ is a formally smooth $A$-algebra for the discrete topologies. Now, this follows from the Jacobian criterion, for the images of the $v_i$ in ${\mathfrak{J}}_{\mathfrak{q}}/{\mathfrak{J}}_{\mathfrak{q}}^2$ generate this $B_{\mathfrak{q}}$-module, and it follows from (18.1.1.1) that one has $det(\partial v_i/\partial T_{j_h})\notin \mathfrak{q}$.

Theorem (18.1.2).

Let $S$ be a prescheme, S_0 a closed sub-prescheme of $S$ whose underlying space is identical to that of $S$. Then the functor

  X ↦ X ×_S S_0

from the category of $S$-preschemes étale over $S$ to the category of S_0-preschemes étale over S_0 is an equivalence of categories.

Let us show first that this functor is fully faithful. Let $X$, $Y$ be two $S$-preschemes étale over $S$, and put $X_0=X{\times }_S{S}_0$, $Y_0=Y{\times }_S{S}_0$. If $Z=X{\times }_{S}Y$, the set ${\mathrm{Hom}}_S(X,Y)$ is in canonical bijective correspondence with the set of $X$-sections $\Gamma (Z/X)$, and similarly ${\mathrm{Hom}}_{S_0}(X_0,Y_0)$ is in canonical bijective correspondence with $\Gamma (Z_0/X_0)$, where $Z_0=Z{\times }_S{S}_0=X_0{\times }_{S_0}{Y}_0$. Now $Z$ is étale over $X$, Z_0 étale over X_0, and X_0 (resp. Z_0) is a closed sub-prescheme of $X$ (resp. $Z$) having the same underlying space. The open subsets of $Z$ such that the restriction of the morphism $Z\to X$ is surjective and radicial are therefore the same as the subsets of Z_0 having the corresponding properties, and our assertion follows accordingly from (17.9.3).

To complete the proof, it suffices to see that for every S_0-prescheme X_0 étale over S_0, there exist an $S$-prescheme $X$ étale over $S$ and an S_0-isomorphism $X_0\stackrel{\sim }{\to }X{\times }_S{S}_0$. By virtue of Prop. (18.1.1), there is an open cover $(U_{\alpha })$ of X_0 and, for each $\alpha$, an $S$-prescheme $V_{\alpha }$ which is étale over $S$, and finally an S_0-isomorphism ${\theta }_{\alpha }:U_{\alpha }\stackrel{\sim }{\to }{V}_{\alpha }{\times }_S{S}_0$. Moreover, by the first part of the proof, there exists a unique S_0-isomorphism ${\varphi }_{\alpha \beta }$ from ${\theta }_{\alpha }(U_{\alpha }\cap U_{\beta })$ onto ${\theta }_{\beta }(U_{\alpha }\cap U_{\beta })$, corresponding to the identity automorphism of $U_{\alpha }\cap U_{\beta }$, and it is immediate, for the same reason, that these isomorphisms satisfy the gluing condition $(0_I,\mathrm{4.1.7})$. There is consequently an $S$-prescheme $X$ such that the $V_{\alpha }$ are canonically identified with sub-preschemes induced on open sets of $X$, the ${\theta }_{\alpha }$ being identified with S_0-isomorphisms which coincide on the intersections $U_{\alpha }\cap U_{\beta }$ and therefore define an S_0-isomorphism $X_0\stackrel{\sim }{\to }X{\times }_S{S}_0$. It is clear that $X$ is étale over $S$ (17.3.2), which completes the proof.

Corollary (18.1.3).

Let $S$ be a prescheme, $X$ an $S$-prescheme étale over $S$, $S^{\prime }$ an $S$-prescheme, $S_0^{\prime }$ a closed sub-prescheme of $S^{\prime }$ having the same underlying space. Then the canonical map $X(S^{\prime }{)}_S\to X(S_0^{\prime }{)}_S$ (I, 3.4.3) is bijective.

Indeed, put $X^{\prime }=X{\times }_S{S}^{\prime }$, $X_0^{\prime }=X{\times }_S{S}_0^{\prime }$, so that $X(S^{\prime }{)}_S=\Gamma (X^{\prime }/S^{\prime })$ and $X(S_0^{\prime }{)}_S=\Gamma (X_0^{\prime }/S_0^{\prime })$; the corollary follows from the fact that the functor defined in (18.1.2) (with $S$ and S_0 replaced by $S^{\prime }$ and $S_0^{\prime }$ respectively) is fully faithful (or directly from (17.9.3)).

18.2. Étale covers

(18.2.1) Given a ring $A$ and a commutative $A$-algebra $B$ which is finite and is a free $A$-module, recall (Bourbaki, Alg., chap. VIII, §12, n° 2) that one defines on $B$ an $A$-linear form $T{r}_{B/A}$, the "trace form"; from this one deduces the definition of a symmetric $A$-bilinear form (also called the "trace form")

$$(\mathrm{18.2.1.1})(x,y)\mapsto T{r}_{B/A}(xy)$$

whose datum is equivalent to that of the associated $A$-linear map $ast{r}_{B/A}:B\to \check{B}$ of the $A$-module $B$ into its dual $\check{B}$, equal to its transpose. When $A$ is a field, it is equivalent to say that this bilinear form is nondegenerate or that $B$ is a separable $A$-algebra (Bourbaki, Alg., chap. IX, §2, prop. 5).

Let $f:A\to A^{\prime }$ be a homomorphism of rings; put $B^{\prime }=B{\otimes }_A{A}^{\prime }$, and let $g:B\to B^{\prime }$ be the canonical homomorphism; the image under $g$ of a basis of the $A$-module $B$ is then a basis of the $A^{\prime }$-module $B^{\prime }$, and it follows from the definitions that one has, for every $x\in B$,

$$(\mathrm{18.2.1.2})T{r}_{B^{\prime }/A^{\prime }}(g(x))=f(T{r}_{B/A}(x)).$$

(18.2.2) Consider now a ringed space $(X,{\mathcal{O}}_X)$ and let $\mathcal{B}$ be an ${\mathcal{O}}_X$-Algebra which, as an ${\mathcal{O}}_X$-Module, is locally free of finite rank; then, for every open $U\subset X$ such that $\mathcal{B}|U$ is (as an ${\mathcal{O}}_U$-Module) isomorphic to ${\mathcal{O}}_U^n$ (for an $n$ depending on $U$), $\Gamma (U,\mathcal{B})$ is a $\Gamma (U,{\mathcal{O}}_X)$-algebra which, as a $\Gamma (U,{\mathcal{O}}_X)$-module, is free of finite rank and therefore defines a $\Gamma (U,{\mathcal{O}}_X)$-linear form $T{r}_{\Gamma (U,\mathcal{B})/\Gamma (U,{\mathcal{O}}_X)}$, which we shall also denote $T{r}_{\mathcal{B}/{\mathcal{O}}_X,U}$; from this one deduces an associated linear map

  astr_{ℬ/𝒪_X, U} : Γ(U, ℬ) → Γ(U, ℬ)̌ = Γ(U, ℬ̌).

Moreover, it follows from (18.2.1.2) that these linear maps are compatible with the operations of restriction from $U$ to a smaller open set, and therefore define on the one hand a homomorphism of ${\mathcal{O}}_X$-Modules, also called the trace homomorphism:

$$(\mathrm{18.2.2.1})T{r}_{\mathcal{B}/{\mathcal{O}}_X}:\mathcal{B}\to {\mathcal{O}}_X$$

and on the other hand a homomorphism of ${\mathcal{O}}_X$-Modules

$$(\mathrm{18.2.2.2})ast{r}_{\mathcal{B}/{\mathcal{O}}_X}:\mathcal{B}\to \check{\mathcal{B}}$$

said to be associated with the trace, and equal to its own transpose. It also follows from (18.2.1.2) that for every $x\in X$ one has

$$(\mathrm{18.2.2.3})(T{r}_{\mathcal{B}/{\mathcal{O}}_X}{)}_x=T{r}_{{\mathcal{B}}_x/{\mathcal{O}}_{X,x}}(\mathrm{18.2.2.4})(ast{r}_{\mathcal{B}/{\mathcal{O}}_X}{)}_x=ast{r}_{{\mathcal{B}}_x/{\mathcal{O}}_{X,x}}.$$

Finally, under the conditions of (18.2.1), if one puts $X=\mathrm{Spec}(A)$ and if $\mathcal{B}=\tilde{B}$ is the ${\mathcal{O}}_X$-Algebra corresponding to $B$, the form $T{r}_{\mathcal{B}/{\mathcal{O}}_X}$ (resp. the homomorphism $ast{r}_{\mathcal{B}/{\mathcal{O}}_X}$) corresponds to the form $T{r}_{B/A}$ (resp. to the homomorphism of $A$-modules $ast{r}_{B/A}$), as also follows from (18.2.1.2).

Proposition (18.2.3).

Let $f:X\to Y$ be a finite morphism of preschemes and let $\mathcal{B}=f_{\ast }({\mathcal{O}}_X)$. The following conditions are equivalent:

a) $f$ is étale.

a') $f$ is a flat morphism of finite presentation and, for every $x\in X$, if one puts $y=f(x)$, then ${\mathcal{O}}_{X,x}/{\mathfrak{m}}_y\cdot {\mathcal{O}}_{X,x}$ is a field, a finite separable extension of $k(y)$.

b) $\mathcal{B}$ is a locally free ${\mathcal{O}}_Y$-Module and, for every $y\in Y$, ${\mathcal{B}}_y{\otimes }_{{\mathcal{O}}_{Y,y}}k(y)$ is a finite separable $k(y)$-algebra (hence the direct composite of a finite number of fields, finite separable extensions of $k(y)$).

c) $\mathcal{B}$ is a locally free ${\mathcal{O}}_Y$-Module and the homomorphism $ast{r}_{\mathcal{B}/{\mathcal{O}}_Y}:\mathcal{B}\to \check{\mathcal{B}}$ (18.2.2) is bijective.

Taking into account that $f$ is quasi-compact, the equivalence of a) and a') has already been proved (17.6.2). To prove the rest of the proposition, one may restrict to the case where $Y=\mathrm{Spec}(A)$ and $X=\mathrm{Spec}(B)$ are affine, $B$ being a finite $A$-algebra and $\mathcal{B}=\tilde{B}$. To say that $f$ is a morphism of finite presentation amounts then to saying that $B$ is an $A$-module of finite presentation (1.4.7). If in addition $f$ is flat, hence $B$ a flat $A$-module, one knows (Bourbaki, Alg. comm., chap. II, §5, n° 2, cor. 2 of th. 1) that $B$ is a projective $A$-module, hence $\mathcal{B}$ a locally free ${\mathcal{O}}_Y$-Module (loc. cit., n° 2, th. 1), and the converse is immediate. On the other hand, $f^{-1}(y)$ is none other than the spectrum of the $k(y)$-algebra $\mathcal{B}(y)={\mathcal{B}}_y{\otimes }_{{\mathcal{O}}_{Y,y}}k(y)$, which completes the proof of the equivalence of a') and b). To see that b) is equivalent to c), note that the second assertion of b) is equivalent to the fact that the homomorphism $ast{r}_{\mathcal{B}(y)/k(y)}:\mathcal{B}(y)\to \mathcal{B}\check{y}$ is bijective; since $\mathcal{B}(y)={\mathcal{B}}_y{\otimes }_{{\mathcal{O}}_{Y,y}}k(y)$ and $\mathcal{B}\check{y}={\check{\mathcal{B}}}_y{\otimes }_{{\mathcal{O}}_{Y,y}}k(y)$, and since ${\mathcal{B}}_y$ and ${\check{\mathcal{B}}}_y$ are free ${\mathcal{O}}_{Y,y}$-modules, it follows from (18.2.2.4) and from Bourbaki, Alg. comm., chap. II, §3, n° 3, cor. of prop. 6, that the homomorphism $ast{r}_{{\mathcal{B}}_y/{\mathcal{O}}_{Y,y}}:{\mathcal{B}}_y\to {\check{\mathcal{B}}}_y$ is also bijective; the converse being obvious, this completes the proof.

When an ${\mathcal{O}}_X$-Algebra $\mathcal{B}$ verifies the equivalent conditions b) and c) of (18.2.3), one says that $\mathcal{B}$ is a finite étale ${\mathcal{O}}_X$-Algebra. When $X=\mathrm{Spec}(A)$ is affine and one therefore has $\mathcal{B}=\tilde{B}$, where $B$ is an $A$-algebra, it amounts to the same, by virtue of (18.2.3), to say that $\mathcal{B}$ is a finite étale ${\mathcal{O}}_X$-Algebra or that $B$ is a finite étale $A$-algebra (in the sense of (17.3.2)).

Corollary (18.2.4).

Let $f:X\to Y$ be a finite morphism of finite presentation, and put $\mathcal{B}=f_{\ast }({\mathcal{O}}_X)$. Let $y$ be a point of $Y$. The following conditions are equivalent:

a) There exists an open neighbourhood $U$ of $y$ in $Y$ such that the restriction $f^{-1}(U)\to U$ of $f$ is an étale morphism.

b) ${\mathcal{B}}_y$ is a free ${\mathcal{O}}_{Y,y}$-module of finite type and ${\mathcal{B}}_y{\otimes }_{{\mathcal{O}}_{Y,y}}k(y)$ is a separable $k(y)$-algebra.

It is clear that a) implies b) by virtue of (18.2.3). On the other hand, $\mathcal{B}$ is an ${\mathcal{O}}_Y$-Module of finite presentation (1.6.3 and 1.4.7); hence, if ${\mathcal{B}}_y$ is a free ${\mathcal{O}}_{Y,y}$-module, there exists an open neighbourhood $U$ of $y$ in $Y$ such that $\mathcal{B}|U$ is a locally free ${\mathcal{O}}_U$-Module $(0_I,\mathrm{5.2.7})$; in addition, by hypothesis, the homomorphism $ast{r}_{{\mathcal{B}}_y/{\mathcal{O}}_{Y,y}}:{\mathcal{B}}_y\to {\check{\mathcal{B}}}_y$ being

bijective, it also follows from $(0_I,\mathrm{5.2.7})$ that one may suppose $U$ chosen so that the homomorphism $ast{r}_{\mathcal{B}|U/{\mathcal{O}}_U}$ is bijective. The fact that b) implies a) then follows from (18.2.3).

Corollary (18.2.5).

Let $Y$ be a quasi-compact or locally Noetherian prescheme, $f:X\to Y$ a finite morphism of finite presentation; put $\mathcal{B}=f_{\ast }({\mathcal{O}}_X)$. Suppose that for every closed point $y$ of $Y$, ${\mathcal{B}}_y$ is a free ${\mathcal{O}}_{Y,y}$-module and ${\mathcal{B}}_y{\otimes }_{{\mathcal{O}}_{Y,y}}k(y)$ is a separable $k(y)$-algebra. Then $f$ is étale.

Indeed, it follows from (18.2.4) that every closed point $y$ of $Y$ has an open neighbourhood $U$ such that the restriction $f^{-1}(U)\to U$ of $f$ is étale; the conclusion follows from the fact that in the two cases considered, every non-empty closed subset of $Y$ contains a closed point $(\mathrm{5.1.11}and{0}_I,\mathrm{2.1.3})$.

Corollary (18.2.6).

If $f:X\to Y$ is a finite étale morphism, and if one puts $\mathcal{B}=f_{\ast }({\mathcal{O}}_X)$, then the ${\mathcal{O}}_Y$-homomorphism $T{r}_{\mathcal{B}/{\mathcal{O}}_Y}:\mathcal{B}\to {\mathcal{O}}_Y$ (18.2.2) (also denoted $T{r}_f$) is surjective.

The question being local, one may, by virtue of (18.2.3), suppose that $Y=\mathrm{Spec}(A)$, $X=\mathrm{Spec}(B)$ with $\mathcal{B}=\tilde{B}$, $B$ being a free $A$-module; since by virtue of (18.2.3) the bilinear form (18.2.1.1) is nondegenerate, this entails in particular that the linear form $T{r}_{B/A}$ is surjective.

Remarks (18.2.7). — (i) When $f:X\to Y$ is a finite morphism such that $f_{\ast }({\mathcal{O}}_X)$ is a locally free ${\mathcal{O}}_Y$-Module (resp. locally free of rank $n$), one says further that $f$ is a finite locally free morphism (resp. locally free of rank $n$). This condition, by virtue of (18.2.3), is verified if $f$ is a finite étale morphism, but does not by itself imply that $f$ be étale, as is shown by the example where $X=\mathrm{Spec}(K)$ and $Y=\mathrm{Spec}(k)$ are spectra of fields, $K$ being a finite non-separable extension of $k$. When $f:X\to Y$ is a finite étale morphism, one also says that $X$ is an étale cover of $Y$. One will note that in that case, $f$ is universally open and universally closed, and in particular $f(X)$ is a subset of $Y$ both open and closed.

One says that an étale cover $X$ of $Y$ is trivial if $X$ is a sum of a finite number of preschemes isomorphic to $Y$. One says that an étale cover $X$ of $Y$ is locally trivial if the morphism $f:X\to Y$ is such that every point $y\in Y$ has an open neighbourhood $U$ for which the cover $f^{-1}(U)$ of $U$ is trivial.

(ii) Let $f:X\to Y$ be a finite morphism, locally free of rank $n$; put $\mathcal{B}=f_{\ast }({\mathcal{O}}_X)$ and let $u=ast{r}_{\mathcal{B}/{\mathcal{O}}_Y}:\mathcal{B}\to \check{\mathcal{B}}$; from this one deduces a homomorphism of $n$-th exterior power ${\Lambda }^{n}u:{\Lambda }^n\mathcal{B}\to {\Lambda }^n\check{\mathcal{B}}=\check{{\Lambda }^n\mathcal{B}}$ between invertible ${\mathcal{O}}_Y$-Modules, and consequently $(0_I,\mathrm{5.4.2})$ an element

  (18.2.7.1)    d_{X/Y} ∈ Γ(Y, (Λ^n ℬ̌) ⊗_{𝒪_Y} (Λ^n ℬ̌))

called the discriminant of $X$ over $Y$. Moreover, since $({\Lambda }^n\check{\mathcal{B}}){\otimes }_{{\mathcal{O}}_Y}({\Lambda }^n\check{\mathcal{B}})$ is the dual of $({\Lambda }^n\mathcal{B}){\otimes }_{{\mathcal{O}}_Y}({\Lambda }^n\mathcal{B})$, $d_{X/Y}$ may also be identified with a homomorphism

  (18.2.7.2)    (Λ^n ℬ) ⊗_{𝒪_Y} (Λ^n ℬ) → 𝒪_Y

and one denotes by ${\mathcal{D}}_{X/Y}$ the quasi-coherent Ideal of finite type of ${\mathcal{O}}_Y$, image of the homomorphism (18.2.7.2), also called the discriminant Ideal of $X$ over $Y$.

That being so, for the homomorphism $u$ to be bijective, it is necessary and sufficient that ${\Lambda }^{n}u$ be bijective, or again that the section $d_{X/Y}$ have an invertible germ at every point $y\in Y$, which one also writes $d_{X/Y}(y)\ne 0$ for every $y$ $(0_I,\mathrm{5.5.2})$. It also amounts to the same to say that the discriminant Ideal ${\mathcal{D}}_{X/Y}$ is equal to ${\mathcal{O}}_Y$.

The terminology of "cover" introduced in (18.2.7, (i)) is justified by the following proposition:

Proposition (18.2.8).

Let $f:X\to Y$ be a morphism étale, separated and of finite type, and for every $y\in Y$ let $n(y)$ be the geometric number of points of $f^{-1}(y)$. Then the function $y\mapsto n(y)$ is lower semi-continuous in $Y$. For it to be continuous at a point $y$ (hence constant in a neighbourhood of $y$), it is necessary and sufficient that there exist an open neighbourhood $U$ of $y$ such that the restriction $f^{-1}(U)\to U$ of $f$ be a finite (étale) morphism.

Since $f$ is quasi-finite (17.6.1) and locally of finite presentation, it amounts to the same to say that $f$ is finite or that $f$ is proper (8.11.1): in addition, each fibre $f^{-1}(y)$ is geometrically reduced over $k(y)$. The conclusions then follow from (15.5.9, (i) and (ii)) and from the fact that $f$ is flat.

Corollary (18.2.9).

Let $Y$ be a connected prescheme, $f:X\to Y$ a morphism étale, separated and of finite type. For $f$ to be finite (in other words, for $X$ to be an étale cover of $Y$), it is necessary and sufficient that all the fibres of $f$ have the same geometric number of points.

Remarks (18.2.10). — (i) The example of the "affine line with one point doubled" (I, 5.5.11) shows that an étale morphism of finite type of Noetherian preschemes can be non-separated; for this example, the first assertion of (18.2.8) no longer holds.

(ii) For a separated, étale morphism of finite type $f:X\to Y$ to make $X$ into a locally trivial cover, it is necessary and sufficient that for every $x\in X$ there exist an open neighbourhood $U$ of $y=f(x)$ and a $U$-section $g$ of $f^{-1}(U)$ such that $g(y)=x$. Indeed, the condition is obviously necessary; the fact that it is sufficient follows from the fact that every fibre $f^{-1}(y)$ is finite ((17.6.1) and (I, 6.2.2)), from the characterization of sections of an étale $Y$-scheme (17.9.3), and from prop. (18.2.8).

18.3. Finite étale algebras

Proposition (18.3.1).

Let $A$ be a ring, $B$ an $A$-algebra of finite presentation.

(i) For $B$ to be an unramified $A$-algebra, it is necessary and sufficient that $B$ be an $A$-module of finite presentation and that $B$ be a projective $(B{\otimes }_{A}B)$-module.

(ii) Suppose moreover that $B$ is a finite $A$-algebra. For $B$ to be an étale $A$-algebra, it is necessary and sufficient that $B$ be a projective $A$-module and a projective $(B{\otimes }_{A}B)$-module.

It is understood that the $(B{\otimes }_{A}B)$-module structure on $B$ is the one coming from the $(B{\otimes }_{A}B)$-algebra structure on $B$ corresponding to the canonical $A$-homomorphism of rings $B{\otimes }_{A}B\to B$, which is surjective and of kernel ${\mathfrak{J}}_{B/A}$ (0, 20.4.1).

(i) To say that the morphism $\mathrm{Spec}(B)\to \mathrm{Spec}(A)$ is locally of finite presentation is equivalent to saying that $B$ is an $A$-algebra of finite presentation (1.4.6). To say

that $B$ is an unramified $A$-algebra means then (17.4.2) that $\mathrm{Spec}((B{\otimes }_{A}B)/{\mathfrak{J}}_{B/A})$ is a sub-scheme induced on an open and closed subset of $\mathrm{Spec}(B{\otimes }_{A}B)$, and one knows that for this to be so, it is necessary and sufficient that ${\mathfrak{J}}_{B/A}$ be a direct factor ideal of $B{\otimes }_{A}B$ (Bourbaki, Alg. comm., chap. II, §4, n° 3, prop. 15); but it amounts to the same to say that the $(B{\otimes }_{A}B)$-module quotient $(B{\otimes }_{A}B)/{\mathfrak{J}}_{B/A}$ is projective (Bourbaki, Alg., chap. II, 3rd ed., §2, n° 2, prop. 4).

(ii) If one recalls that a flat $A$-module of finite presentation is projective and conversely (Bourbaki, Alg. comm., chap. II, §5, n° 2, cor. 2 of th. 1), the assertion of (ii) follows from that of (i) and from (17.6.2).

Proposition (18.3.2).

Let $A$ be a ring, $\mathfrak{J}$ an ideal of $A$ such that, for the $\mathfrak{J}$-preadic topology, $A$ is separated and complete; put $A_0=A/\mathfrak{J}$. Then the functor

  B ↦ B ⊗_A A_0

is an equivalence from the category of finite étale $A$-algebras to the category of finite étale A_0-algebras.

We shall first prove the following lemma:

Lemma (18.3.2.1).

Let $A$ be a ring, $\mathfrak{J}$ an ideal of $A$ such that, for the $\mathfrak{J}$-preadic topology, $A$ is separated and complete.

(i) Every projective $A$-module of finite type $M$ is separated and complete for the $\mathfrak{J}$-preadic topology, hence the projective limit of the projective $(A/{\mathfrak{J}}^{n+1})$-modules $M/{\mathfrak{J}}^{n+1}M$.

(ii) Conversely, put $A_n=A/{\mathfrak{J}}^{n+1}$, and let $(M_n)$ be a projective system of $A_n$-modules such that, for each $n$, the homomorphism $M_{n+1}{\otimes }_{A_{n+1}}{A}_n\to M_n$ deduced from the di-homomorphism of transition $M_{n+1}\to M_n$ is bijective. Suppose moreover that the $M_n$ are projective and M_0 of finite type. Then $M=\underleftarrow{\lim }{M}_n$ is a projective $A$-module of finite type such that the canonical homomorphism $M{\otimes }_A{A}_0\to M_0$ is bijective.

(i) There exists a free $A$-module of finite type $L$ such that $M$ is isomorphic to a direct factor of $L$; since $L$ is separated for the $\mathfrak{J}$-preadic topology, so is every submodule $N$ of $L$, since ${\mathfrak{J}}^{n+1}N\subset {\mathfrak{J}}^{n+1}L$; in particular $M$ is separated for this topology, and since the surjective homomorphism $f:L\to M$ is continuous for the $\mathfrak{J}$-preadic topology, its kernel $N$ is closed for the topology induced by that of $L$; since $L$ is complete and $f$ a strict morphism, one concludes that $M$ is complete (Bourbaki, Top. gén., chap. IX, 2nd ed., §3, n° 1, prop. 4).

(ii) It follows from Nakayama's lemma that if M_0 is generated by a finite family $(x_{i,0})$ of $r$ elements and if for each $n$, $x_{i,n}$ is an element of $M_n$ whose image in $M_{n-1}$ is $x_{i,n-1}$, then $(x_{i,n}{)}_{1⩽i⩽r}$ is a system of generators of $M_n$ (Bourbaki, Alg. comm., chap. II, §3, cor. 2 of prop. 4). That being so, for each $n$, put $L_n=A_n^r$; if $(e_{i,n}{)}_{1⩽i⩽r}$ is the canonical basis of $L_n$, let $u_n:L_n\to M_n$ be the $A$-linear map such that $u_n(e_{i,n})=x_{i,n}$ for each $i$. By hypothesis one has a split exact sequence

  0 → N_n →^{v_n} L_n →^{u_n} M_n → 0

and since $L_n=L_{n+1}/{\mathfrak{J}}^{n+1}{L}_{n+1}$ and $M_n=M_{n+1}/{\mathfrak{J}}^{n+1}{M}_{n+1}$, the vertical arrows in the commutative diagram

  0 → N_{n+1} →^{v_{n+1}} L_{n+1} →^{u_{n+1}} M_{n+1} → 0
              ↓                  ↓                    ↓
  0 → N_n     →^{v_n}     L_n    →^{u_n}     M_n      → 0

are all three surjective. Now one has $M=\underleftarrow{\lim }{M}_n$ and $L=A^r=\underleftarrow{\lim }{L}_n$; if one puts $N=\underleftarrow{\lim }{N}_n$, $u=\underleftarrow{\lim }{u}_n$, $v=\underleftarrow{\lim }{v}_n$, one has, by virtue of $(0_{III},\mathrm{13.2.2})$, the exact sequence

  (18.3.2.2)    0 → N →^v L →^u M → 0.

Moreover, since for each $n$, $v_n$ is left-invertible and $M_n$ is a projective $A_n$-module, it follows from (0, 19.1.8) that the exact sequence (18.3.2.2) is split, which proves the lemma.

That being so, let us show first that the functor in the statement of (18.3.2) is fully faithful. Put, as in the lemma, $A_n=A/{\mathfrak{J}}^{n+1}$; let $B$, $C$ be two finite étale $A$-algebras, and put, for each $n$, $B_n=B{\otimes }_A{A}_n$, $C_n=C{\otimes }_A{A}_n$; by virtue of (18.3.1) and (18.3.2.1), $B$ and $C$ are separated and complete for the $\mathfrak{J}$-preadic topology, and one has $B=\underleftarrow{\lim }{B}_n$, $C=\underleftarrow{\lim }{C}_n$; moreover every homomorphism of $A$-algebras $u:B\to C$ is continuous for the $\mathfrak{J}$-preadic topologies, and therefore gives a projective system of homomorphisms of $A_n$-algebras $u_n=u\otimes 1:B_n\to C_n$ of which it is the projective limit; the converse being obvious, one has therefore a canonical bijection

  Hom_{A-alg.}(B, C) ⥲ lim_← Hom_{A_n-alg.}(B_n, C_n).

But since $B$ and $C$ are étale $A$-algebras, it follows at once from (18.1.2) that the canonical map

  Hom_{A_{n+1}-alg.}(B_{n+1}, C_{n+1}) → Hom_{A_n-alg.}(B_n, C_n)

is bijective for $n⩾0$, which finishes proving that the canonical map Hom_{A-alg.}(B, C) → Hom_{A_0-alg.}(B_0, C_0) is bijective.

To complete the proof of (18.3.2), it suffices to see that for every finite étale A_0-algebra B_0, there exist a finite étale $A$-algebra $B$ and an A_0-isomorphism $B_0\stackrel{\sim }{\to }B{\otimes }_A{A}_0$. Now, it follows from (18.1.2) that there is a projective system $(B_n)$ such that $B_n$ is a finite étale $A_n$-algebra and the homomorphisms $B_{n+1}{\otimes }_{A_{n+1}}{A}_n\to B_n$ are bijective. It follows from (18.3.1) and (18.3.2.1) that the $A$-algebra $B=\underleftarrow{\lim }{B}_n$ is a projective $A$-module of finite type and that B_0 is isomorphic to $B{\otimes }_A{A}_0$. To prove that $B$ is an étale $A$-algebra, it suffices, by virtue of (18.2.5), to show that for every maximal ideal $\mathfrak{m}$ of $A$, $B_{\mathfrak{m}}/\mathfrak{m}{B}_{\mathfrak{m}}$ is a separable $(A/\mathfrak{m})$-algebra. Now, since $\mathfrak{J}$ is contained in the radical of $A$ $(0_I,\mathrm{7.1.10})$,

one has $\mathfrak{J}\subset \mathfrak{m}$, and if ${\mathfrak{m}}_0=\mathfrak{m}/\mathfrak{J}$, one has $A_0/{\mathfrak{m}}_0=A/\mathfrak{m}$ and $B_{\mathfrak{m}}/\mathfrak{m}{B}_{\mathfrak{m}}=(B_0{)}_{{\mathfrak{m}}_0}/{\mathfrak{m}}_0(B_0{)}_{{\mathfrak{m}}_0}$; the conclusion therefore follows from the fact that B_0 is a finite étale A_0-algebra (18.2.5).

Example (18.3.3). — Prop. (18.3.2) applies in particular when $A$ is a separated and complete local ring, $\mathfrak{J}$ being the maximal ideal of $A$, so that A_0 is a field and the category of finite étale A_0-algebras is identical to that of finite separable A_0-algebras, hence isomorphic to direct composites of fields, separable and finite extensions of A_0. In particular, if the field A_0 is separably closed, these extensions are all identical to A_0, and consequently every étale cover of $\mathrm{Spec}(A)$ is trivial (18.2.7) by virtue of (18.3.2).

Theorem (18.3.4).

Let $A$ be a Noetherian ring, $\mathfrak{J}$ an ideal of $A$ such that $A$ is separated and complete for the $\mathfrak{J}$-preadic topology, $A_0=A/\mathfrak{J}$. Put $S=\mathrm{Spec}(A)$, $S_0=\mathrm{Spec}(A_0)$. Let $X$ be an $S$-scheme proper over $S$, and put $X_0=X{\times }_S{S}_0$. Then the functor

  Z ↦ Z ×_X X_0

from the category of $X$-schemes finite and étale over $X$ to the category of X_0-schemes finite and étale over X_0 is an equivalence of categories.

Let us show first that this functor is fully faithful. Let $Z^{\prime }$ and Z'' be two $X$-schemes finite and étale over $X$. Put $S_n=\mathrm{Spec}(A/{\mathfrak{J}}^{n+1})$, $X_n=X{\times }_S{S}_n$, $Z_n^{\prime }=Z^{\prime }{\times }_S{S}_n$, $Z_n^{\prime \prime }=Z^{\prime \prime }{\times }_S{S}_n$ for each $n⩾0$. It follows from (III, 5.4.1) that one has a canonical bijection Hom_X(Z', Z'') ⥲ lim_← Hom_{X_n}(Z'_n, Z''_n). Now, by virtue of (18.1.2), the canonical map ${\mathrm{Hom}}_{X_{n+1}}(Z_{n+1}^{\prime },Z_{n+1}^{\prime \prime })\to {\mathrm{Hom}}_{X_n}(Z_n^{\prime },Z_n^{\prime \prime })$ is bijective, which completes the proof of our assertion.

It remains to prove that if ${\mathcal{B}}_0$ is a finite étale ${\mathcal{O}}_{X_0}$-Algebra, there exists an ${\mathcal{O}}_X$-Algebra $\mathcal{B}$ finite and étale and an isomorphism ${\mathcal{B}}_0\stackrel{\sim }{\to }\mathcal{B}{\otimes }_{{\mathcal{O}}_X}{\mathcal{O}}_{X_0}$. It follows from (18.1.2) that there is a projective system $({\mathcal{B}}_n)$, where ${\mathcal{B}}_n$ is a finite étale ${\mathcal{O}}_{X_n}$-Algebra, and the second comparison theorem (III, 5.1.4) proves the existence of a coherent ${\mathcal{O}}_X$-Module $\mathcal{B}$ and of a projective system of isomorphisms ${\mathcal{B}}_n\stackrel{\sim }{\to }\mathcal{B}{\otimes }_{{\mathcal{O}}_X}{\mathcal{O}}_{X_n}$. The datum of an Algebra structure on a Module $\mathcal{F}$ being equivalent to that of a homomorphism $\mathcal{F}\otimes \mathcal{F}\to \mathcal{F}$ making commutative diagrams in which only tensor powers of $\mathcal{F}$ intervene, it follows from (III, 5.1.3), (I, 10.11.4) and (I, 10.11.7) that $\mathcal{B}$ is naturally endowed with an ${\mathcal{O}}_X$-Algebra structure for which the isomorphism ${\mathcal{B}}_n\stackrel{\sim }{\to }\mathcal{B}{\otimes }_{{\mathcal{O}}_X}{\mathcal{O}}_{X_n}$ is an isomorphism of Algebras. Moreover, $\mathcal{B}$ is a locally free ${\mathcal{O}}_X$-Module; this also follows from (III, 5.1.3), (I, 10.11.4), (I, 10.11.7) and from the fact that, in the category of coherent ${\mathcal{O}}_X$-Modules, locally free ${\mathcal{O}}_X$-Modules $\mathcal{F}$ may be defined as those for which the functor $\mathcal{G}\mapsto \mathcal{H}o{m}_{{\mathcal{O}}_X}(\mathcal{F},\mathcal{G})$ is exact. Finally, to see that $\mathcal{B}$ is an étale ${\mathcal{O}}_X$-Algebra, it suffices (18.2.5) to show that for every closed point $x\in X$, ${\mathcal{B}}_x{\otimes }_{{\mathcal{O}}_{X,x}}k(x)$ is a separable $k(x)$-algebra. But since the structure morphism $f:X\to S$ is proper, $f(x)$ is a closed point of $S$, hence belongs to S_0, since $\mathfrak{J}$ is contained in the radical of $A$ $(0_I,\mathrm{7.1.10})$; the conclusion therefore follows from the fact that $X_0=f^{-1}(S_0)$ and that ${\mathcal{B}}_0$ is a finite étale ${\mathcal{O}}_{X_0}$-Algebra.

18.4. Local structure of unramified and étale morphisms

Lemma (18.4.1).

Let $A$ be a ring, $B$ a finite monogenic $A$-algebra, $u$ a generator of the $A$-algebra $B$, $F\in A[T]$ a polynomial such that $F(u)=0$, $F^{\prime }$ the derived polynomial; put $u^{\prime }=F^{\prime }(u)$. Then the ideal of $B$, annihilator of ${\Omega }_{B/A}^1$, contains u' B; it is equal to u' B if the ideal $\mathfrak{J}$ of A[T] formed by the polynomials $G$ such that $G(u)=0$ is generated by $F$, in other words if the canonical surjective homomorphism $\varphi :A[T]/F\cdot A[T]\to B$ transforming the image of $T$ into $u$ is bijective.

Put $C=A[T]$, so that $B=C/\mathfrak{J}$. One has the exact sequence (0, 20.5.12.1)

  𝔍/𝔍² → Ω^1_{C/A} ⊗_C B → Ω^1_{B/A} → 0

and ${\Omega }_{B/A}^1$ is therefore identified with the quotient $B/{\mathfrak{J}}^{\prime }$, ${\mathfrak{J}}^{\prime }$ being the ideal generated by the elements G'(u), where $G$ ranges over a system of generators of the ideal $\mathfrak{J}$ (0, 20.5.13); the lemma follows immediately.

Proposition (18.4.2).

With the notations of (18.4.1), let $\mathfrak{q}$ be a prime ideal of $B$. Then:

(i) If $\mathfrak{q}$ does not contain $u^{\prime }$, $B_{\mathfrak{q}}$ is a formally unramified $A_{\mathfrak{p}}$-algebra ($\mathfrak{p}$ being the inverse image of $\mathfrak{q}$ in $A$); in other words, $\mathrm{Spec}(B_{\mathfrak{q}})$ is formally unramified over $\mathrm{Spec}(A)$.

(ii) Suppose moreover that $F$ is unitary and generates $\mathfrak{J}$. Then, for $\mathrm{Spec}(B)$ to be étale over $\mathrm{Spec}(A)$ at the point $\mathfrak{q}$, it is necessary and sufficient that $u^{\prime }\notin \mathfrak{q}$.

The hypothesis that $u^{\prime }\notin \mathfrak{q}$ entails that ${\Omega }_{B_{\mathfrak{q}}/A_{\mathfrak{p}}}^1=0$ (0, 20.5.9), hence (i) follows from (17.2.1). Moreover, under the hypotheses of (ii), $B$ is a free $A$-module by virtue of Euclidean division; since the annihilator ${\mathfrak{J}}^{\prime }$ of ${\Omega }_{B/A}^1$ is then equal to u' B by virtue of (18.4.1), and ${\Omega }_{B/A}^1$ is a $B$-module of finite presentation (16.4.22), the annihilator of ${\Omega }_{B_{\mathfrak{q}}/A_{\mathfrak{p}}}^1$ is equal to $u^{\prime }{B}_{\mathfrak{q}}$ (Bourbaki, Alg. comm., chap. II, §2, n° 4, formula (9)), and (ii) follows therefore from (i) and from the implication c) ⇒ a) in (17.6.1).

Corollary (18.4.3).

With the notations of (18.4.2), suppose that $F$ is unitary and generates $\mathfrak{J}$. Then, for $B$ to be an étale $A$-algebra, it is necessary and sufficient that $u^{\prime }$ be invertible in $B$ (or, what amounts to the same, that the ideal of A[T] generated by $F$ and $F^{\prime }$ be equal to A[T]).

Taking (18.4.2, (ii)) into account, to say that $\mathrm{Spec}(B)$ is étale over $\mathrm{Spec}(A)$ means indeed that $u^{\prime }$ does not belong to any prime ideal of $B$, i.e. that it is invertible in $B$.

One says that a unitary polynomial $F\in A[T]$ such that the ideal of A[T] generated by $F$ and $F^{\prime }$ is equal to A[T] itself is separable; it is immediate that this definition coincides with the usual definition (Bourbaki, Alg., chap. V, §7, n° 6) when $A$ is a field.

Lemma (18.4.4).

Let $A$ be a local ring, $B$ a finite monogenic $A$-algebra, $u$ a generator of the $A$-algebra $B$. Let ${\mathfrak{n}}_i$ $(1⩽i⩽r)$ be maximal ideals of $B$ such that $\mathrm{Spec}(B)$ is formally unramified over $\mathrm{Spec}(A)$ at the points ${\mathfrak{n}}_i$. Then there exists a unitary polynomial $F\in A[T]$ such that $F(u)=0$ and $F^{\prime }(u)\notin {\mathfrak{n}}_i$ for every index $i$ $(1⩽i⩽r)$. Moreover, if $k$ is the residue field

of $A$, $f\in k[T]$ the minimal polynomial of the image of $u$ in $B{\otimes }_{A}k$, there exists an $F\in A[T]$ of which $f$ is the canonical image and such that $F(u)=0$; such a polynomial $F$ verifies the conditions $F^{\prime }(u)\notin {\mathfrak{n}}_i$ for $1⩽i⩽r$.

The maximal ideal $\mathfrak{m}$ of $A$ is the inverse image of each of the ${\mathfrak{n}}_i$ (II, 6.1.10); let $L=B{\otimes }_{A}k$, which is a finite $k$-algebra. Let $\zeta$ be the image of $u$ in $L$, and let $n$ be the rank of $L$ over $k$; the minimal polynomial $f\in k[T]$ of $\zeta$ over $k$ is therefore of degree $n$, and $L$ is isomorphic to $k[T]/f\cdot k[T]$. If ${\mathfrak{n}}_i^{\prime }={\mathfrak{n}}_i/\mathfrak{m}B$, the hypothesis entails that $\mathrm{Spec}(L)$ is étale over $k$ at the points ${\mathfrak{n}}_i^{\prime }$ $(1⩽i⩽r)$ by virtue of (17.6.1, c), hence $f^{\prime }(\zeta )\notin {\mathfrak{n}}_i^{\prime }$ by virtue of (18.4.2, (ii)). Note now that $\mathfrak{m}B$ is contained in the radical of $B$; since $1,\zeta ,\cdots ,{\zeta }^{n-1}$ form a basis of $L$ over $k$, it follows from Nakayama's lemma that $1,u,\cdots ,u^{n-1}$ generate the $A$-module $B$, and consequently there exists a unitary polynomial $F\in A[T]$ of degree $n$ such that $F(u)=0$; in addition, since $\zeta$ is a root of the canonical image of $F$ in k[T], this image is necessarily equal to $f$. But then the image of F'(u) in $L$ is $f^{\prime }(\zeta )$, and since $f^{\prime }(\zeta )\notin {\mathfrak{n}}_i^{\prime }$, one has $F^{\prime }(u)\notin {\mathfrak{n}}_i$ for each $i$.

Proposition (18.4.5).

Let $A$ be a local ring, $k$ its residue field, $B$ a finite (resp. finite and of finite presentation) $A$-algebra. Suppose moreover either that the field $k$ is infinite, or that $B$ is a local ring. Let $n$ be the rank of $L=B{\otimes }_{A}k$ over $k$. For $B$ to be a formally unramified (resp. étale) $A$-algebra, it is necessary and sufficient that there exist a unitary separable polynomial $F\in A[T]$ (18.4.3) such that $B$ is isomorphic to a quotient of $A[T]/F\cdot A[T]$ (resp. isomorphic to $A[T]/F\cdot A[T]$). Moreover, one may suppose $F$ of degree $n$ (resp. $F$ is necessarily of degree $n$).

The conditions are sufficient by virtue of (18.4.2), without assuming $k$ infinite or $B$ local. To see that the conditions are necessary, note that if $B$ is a formally unramified $A$-algebra, $L$ is a finite and separable algebra over $k$, hence the direct composite of a finite number of finite and separable extensions $k_j$ of $k$ $(1⩽j⩽r)$, $k_j$ being therefore generated by an element ${\xi }_j$ of minimal polynomial $f_j$ $(1⩽j⩽r)$ (Bourbaki, Alg., chap. V, §11, n° 4, prop. 4). Let us show that by virtue of the hypotheses made on $k$ or $B$, there exists an element $\zeta$ of $L$ generating the $k$-algebra $L$. This is immediate if $B$ is local, since then $r=1$. Otherwise, $k$ being supposed infinite, one may suppose that the irreducible polynomials $f_j\in k[T]$ are all distinct (by replacing if necessary each ${\xi }_j$ by ${\xi }_j+a_j$, for a suitable element $a_j\in k$); if one puts $f=f_1{f}_2\cdots f_r$, it is clear that $L$ is isomorphic to $k[T]/f\cdot k[T]$ in both cases considered, hence is generated by an element $\zeta$ of minimal polynomial $f\in k[T]$ of degree $n$. If $u\in B$ is an element whose image is $\zeta$ in $L$, Nakayama's lemma shows that the elements $1,u,\cdots ,u^{n-1}$ generate the $A$-module $B$; this already shows that there exists a unitary polynomial $F\in A[T]$ of degree $n$ such that $F(u)=0$, $u$ generating the $A$-algebra $B$, which is consequently isomorphic to a quotient algebra of $A[T]/F\cdot A[T]$; moreover $B$ is a semi-local ring, and at each of its maximal ideals ${\mathfrak{n}}_i$ one has $F^{\prime }(u)\notin {\mathfrak{n}}_i$ by (18.4.4), which proves that F'(u) is invertible in $B$, hence that $F$ is a separable polynomial. Finally, if $B$ is an étale $A$-algebra, $B$ being a flat $A$-module of finite presentation (1.4.7) is a free $A$-module, and $1,u,\cdots ,u^{n-1}$ form a basis of the $A$-module $B$ (Bourbaki, Alg. comm., chap. II, §3, n° 2, prop. 5), in other words the $A$-algebra $B$ is

isomorphic to $A[T]/F\cdot A[T]$, and for every other unitary polynomial $G\in A[T]$ such that $B$ is isomorphic to $A[T]/G\cdot A[T]$, $G$ is necessarily of degree $n$.

Theorem (18.4.6) (Chevalley).

(i) Let $f:X\to Y$ be a morphism locally of finite type, $x$ a point of $X$, $y=f(x)$, and put $A={\mathcal{O}}_{Y,y}$. For ${\mathcal{O}}_{X,x}$ to be a formally étale (resp. formally unramified) $A$-algebra, it is necessary and sufficient that there exist a unitary polynomial $F\in A[T]$ and a maximal ideal $\mathfrak{n}$ of $B=A[T]/F\cdot A[T]$ (resp. of a quotient algebra $B$ of $A[T]/F\cdot A[T]$) such that ${\mathcal{O}}_{X,x}$ is $A$-isomorphic to $B_{\mathfrak{n}}$ and such that, if $u$ is the image of $T$ in $B$, one has $F^{\prime }(u)\notin \mathfrak{n}$.

(ii) Suppose moreover that $f$ is locally of finite presentation. Then, for $f$ to be étale at the point $x$, it is necessary and sufficient that one can in addition take $B=A[T]/F\cdot A[T]$.

The conditions are sufficient by virtue of (18.4.2). To see that they are necessary, one may obviously restrict to the case where $X$ and $Y$ are affine, and, taking remark (17.4.1.2) into account, to the case where $f$ is formally unramified and quasi-finite. Since $f$ is affine, it follows from (8.12.8) that there exist a finite $A$-algebra $C$ and a maximal ideal $\mathfrak{r}$ of $C$ (necessarily above the maximal ideal $\mathfrak{m}$ of $A$) such that ${\mathcal{O}}_{X,x}$ is $A$-isomorphic to $C_{\mathfrak{r}}$. Moreover (17.4.1.2), the residue field $C/\mathfrak{r}=C_{\mathfrak{r}}/\mathfrak{r}{C}_{\mathfrak{r}}$ is a finite separable extension of $k=A/\mathfrak{m}$, hence of the form k[v], where $v$ is separable over $k$. Let ${\mathfrak{r}}_i$ $(1⩽i⩽h)$ be the maximal ideals of the semi-local ring $C$ other than $\mathfrak{r}$; there exists an element $u\in C$ belonging to all the ${\mathfrak{r}}_i$ and such that its image in $C/\mathfrak{r}$ is equal to $v$ (Bourbaki, Alg. comm., chap. II, §1, n° 2, prop. 5). We shall show that the sub-$A$-algebra $B=A[u]$ of $C$ and the ideal (necessarily maximal since $C$ is finite over $B$) $\mathfrak{n}=\mathfrak{r}\cap B$ of $B$ answer the question.

To handle the case where ${\mathcal{O}}_{X,x}$ is a formally unramified $A$-algebra, it will suffice to prove that $B_{\mathfrak{n}}$ is isomorphic to $C_{\mathfrak{r}}$; indeed, $B_{\mathfrak{n}}$ will then be formally unramified over $A$, and the existence of the polynomial $F\in A[T]$ having the properties of the statement will follow from (18.4.4). Note now that, since $f$ is formally unramified, $C/\mathfrak{r}$ is isomorphic to the $A$-algebra $C_{\mathfrak{r}}/\mathfrak{m}{C}_{\mathfrak{r}}$ (17.4.1.2). One is thus reduced to proving the following lemma:

Lemma (18.4.6.1).

Let $A$ be a local ring, $\mathfrak{m}$ its maximal ideal, $C$ a finite $A$-algebra, $\mathfrak{r}$ a maximal ideal of $C$. Let $u$ be an element of $C$ belonging to all the maximal ideals of $C$ distinct from $\mathfrak{r}$, not belonging to $\mathfrak{r}$, and such that $C_{\mathfrak{r}}/\mathfrak{m}{C}_{\mathfrak{r}}$ is a monogenic algebra over $k=A/\mathfrak{m}$, generated by the image of $u$ in $C_{\mathfrak{r}}/\mathfrak{m}{C}_{\mathfrak{r}}$. Put $B=A[u]$, $\mathfrak{n}=\mathfrak{r}\cap B$. Then the canonical homomorphism $B_{\mathfrak{n}}\to C_{\mathfrak{r}}$ is an isomorphism.

Put $R=B-\mathfrak{n}$, $S=C-\mathfrak{r}$, so that $B_{\mathfrak{n}}=R^{-1}B$ and $C_{\mathfrak{r}}=S^{-1}C$; the canonical homomorphism $B_{\mathfrak{n}}\to C_{\mathfrak{r}}$ may be written as the composite

  R⁻¹ B → R⁻¹ C → S⁻¹ C

and it suffices to show that each of these two homomorphisms is bijective.

Let us first show that $h:R^{-1}C\to S^{-1}C=C_{\mathfrak{r}}$ is bijective; it suffices to see that the images in $R^{-1}C$ of the elements of $S$ are invertible, or again that every maximal ideal $\mathfrak{p}$ of $R^{-1}C$ has an inverse image in $C$ not meeting $S$, hence necessarily equal to $\mathfrak{r}$. Now, since $R^{-1}C$ is a $(R^{-1}B)$-algebra finite, the inverse image of $\mathfrak{p}$ in

$R^{-1}B=B_{\mathfrak{n}}$ is the unique maximal ideal $\mathfrak{n}{B}_{\mathfrak{n}}$ of this ring, hence the inverse image of $\mathfrak{p}$ in $B$ is equal to $\mathfrak{n}$. But on the other hand, if $\mathfrak{q}$ is the inverse image of $\mathfrak{p}$ in $C$, one has $\mathfrak{q}\cap B=\mathfrak{n}$, and since $\mathfrak{n}$ is maximal in $B$ and $C$ is a $B$-algebra finite, $\mathfrak{q}$ is necessarily one of the maximal ideals of $C$; in addition, one has $u\notin \mathfrak{q}$ since $u\notin \mathfrak{n}$ and $u\in B$, hence by hypothesis one has necessarily $\mathfrak{q}=\mathfrak{r}$.

On the other hand, since $B\subset C$, the homomorphism $g:R^{-1}B\to R^{-1}C$ is injective $(0_I,\mathrm{1.3.2})$; to see that it is surjective, note that $R^{-1}C$ is a $(R^{-1}B)$-module of finite type, and on the other hand that $\mathfrak{m}{R}^{-1}B$ is contained in the maximal ideal of the local ring $R^{-1}B$; by virtue of Nakayama's lemma, it suffices to prove that the homomorphism $R^{-1}B/\mathfrak{m}{R}^{-1}B\to R^{-1}C/\mathfrak{m}{R}^{-1}C$ is surjective. But by virtue of the first part of the proof, $R^{-1}C/\mathfrak{m}{R}^{-1}C$ is identified with $C_{\mathfrak{r}}/\mathfrak{m}{C}_{\mathfrak{r}}$; by hypothesis this $k$-algebra is generated by the image of $u$, and a fortiori it is equal to the image of $R^{-1}B/\mathfrak{m}{R}^{-1}B$.

Consider in the second place the case where $f$ is étale at the point $x$. Replacing $X$ by a neighbourhood of $x$, one may suppose that $X$ is a neighbourhood of $\mathfrak{n}$ in $\mathrm{Spec}(B)$ (1.7.2). Put $B^{\prime }=\mathrm{Spec}(A[T]/F\cdot A[T])$ and let ${\mathfrak{n}}^{\prime }$ be the inverse image of $\mathfrak{n}$ in $B^{\prime }$; since the image of F'(T) in $B^{\prime }$ does not belong to ${\mathfrak{n}}^{\prime }$ by hypothesis, the morphism $\mathrm{Spec}(B^{\prime })\to \mathrm{Spec}(A)$ is étale at the point ${\mathfrak{n}}^{\prime }$ by (18.4.2, (ii)). Since by hypothesis $\mathrm{Spec}(B)\to \mathrm{Spec}(A)$ is étale at the point $\mathfrak{n}$, one concludes (17.3.4) that $\mathrm{Spec}(B)\to \mathrm{Spec}(B^{\prime })$ is étale at the point $\mathfrak{n}$; but since this morphism is an immersion, it can be étale at a point only if it is a local isomorphism at that point (17.9.1), hence $B_{\mathfrak{n}}$ and $B_{{\mathfrak{n}}^{\prime }}^{\prime }$ are isomorphic.

Finally, suppose that ${\mathcal{O}}_{X,x}$ is a formally étale $A$-algebra; with the previous notations, it follows from (17.1.5) that $B_{\mathfrak{n}}$ is a formally étale $B_{\mathfrak{n}}$-algebra; but since the homomorphism $B_{\mathfrak{n}}\to {\mathcal{O}}_{X,x}$ is surjective, this can hold only if this homomorphism is bijective (0, 19.10.3, (i)). This completes the proof of (18.4.6).

Corollary (18.4.7).

Let $f:X\to Y$ be a morphism locally of finite type, $x$ a point of $X$. For $f$ to be formally unramified at the point $x$, it is necessary and sufficient that there exist an open neighbourhood $U$ of $x$ such that $f|U$ factors as $U{\to }^j{X}^{\prime }{\to }^{h}Y$, where $h$ is an étale morphism and $j$ a closed immersion.

One may obviously restrict to the case where $Y=\mathrm{Spec}(R)$ is affine and $f$ of finite type. If $A={\mathcal{O}}_{Y,y}$ with $y=f(x)$, the condition for $f$ to be formally unramified at the point $x$ is equivalent, by virtue of (17.4.1.2), to saying that ${\mathcal{O}}_{X,x}$ is a formally unramified $A$-algebra. If this is so, one may apply (18.4.6, (i)); replacing if necessary $Y$ by an affine neighbourhood of $y$, one may suppose (with the notations of (18.4.6)) that the polynomial $F$ is the image in A[T] of a unitary polynomial $G\in R[T]$. One then puts $X^{\prime }=\mathrm{Spec}(R[T]/G\cdot R[T])$; let $x^{\prime }$ be the image of the point $\mathfrak{n}$ of $\mathrm{Spec}(B)$ under the morphism corresponding to the composite homomorphism $R[T]/G\cdot R[T]\to A[T]/F\cdot A[T]\to B$. It follows from (18.4.2) that the morphism $h:X^{\prime }\to Y$ corresponding to the canonical homomorphism $R\to R[T]/G\cdot R[T]$ is étale at the point $x^{\prime }$, hence, by restricting if necessary $X^{\prime }$ and $Y$ to open neighbourhoods of $x^{\prime }$ and $y$ respectively, one may suppose $h$ étale. On the other hand, by virtue of (I, 6.5.1, (ii)) and of (1.7.2), it follows from the fact that one has a local homomorphism $\varphi :{\mathcal{O}}_{X^{\prime },x^{\prime }}\to {\mathcal{O}}_{X,x}$ that this homomorphism corresponds to a

morphism $j:U\to X^{\prime }$ of an open neighbourhood $U$ of $x$ in $X^{\prime }$; by restricting $U$ if necessary, one may, by applying (I, 6.5.1, (i)), suppose that $h\circ j=f|U$, hence $j$ is of finite type (I, 6.3.4, (v)); finally, the homomorphism $\varphi$ is surjective by (18.4.6, (i)), hence it follows from (I, 6.5.4, (i)) that one may, by restricting $U$ and $X^{\prime }$ again, suppose that $j$ is a closed immersion. This proves therefore the necessity of the stated condition; its sufficiency is immediate (17.1.3, (i) and (ii)).

Corollary (18.4.8).

Let $f:X\to Y$ be a morphism locally of finite presentation. For $f$ to be unramified (resp. étale), it is necessary and sufficient that there exist a family of flat morphisms $g_{\alpha }:Y_{\alpha }\to Y$ and, for each $\alpha$, an open set $U_{\alpha }$ in $X_{\alpha }=X{\times }_Y{Y}_{\alpha }$, such that, if $g_{\alpha }^{\prime }:X_{\alpha }\to X$ and $f_{\alpha }:X_{\alpha }\to Y_{\alpha }$ are the canonical projections, the $g_{\alpha }^{\prime }(U_{\alpha })$ form a cover of $X$, and each of the composite morphisms $U_{\alpha }\to X_{\alpha }\to Y_{\alpha }$ is a closed immersion (resp. an open immersion). One may moreover then take the $g_{\alpha }$ étale.

The necessity of the condition for étale morphisms is trivial, by taking a single $Y_{\alpha }$ equal to $X$, the morphism $g_{\alpha }$ being equal to $f$, and the open set $U\subset X{\times }_{Y}X$ being the diagonal. When $f$ is unramified, the necessity of the condition follows from (18.4.7); one takes an open cover $(V_{\alpha })$ of $X$ such that for each $\alpha$, $f|V_{\alpha }$ factors as $V_{\alpha }{\to }^{j_{\alpha }}{Y}_{\alpha }\to Y$, where $j_{\alpha }$ is a closed immersion and $g_{\alpha }$ an étale morphism. Then $j_{\alpha }:V_{\alpha }\to Y_{\alpha }$ factors as $V_{\alpha }{\to }^{s_{\alpha }}{X}_{\alpha }{\to }^{f_{\alpha }}{Y}_{\alpha }$, where $s_{\alpha }$ is a $V_{\alpha }$-section of $X_{\alpha }$, and since the morphism $g_{\alpha }^{\prime }:X_{\alpha }\to X$ is étale, $s_{\alpha }$ is an open immersion (17.4.1), and it suffices to take $U_{\alpha }=s_{\alpha }(V_{\alpha })$ to answer the question.

The sufficiency of the conditions follows from (17.7.1); one concludes from it that $f$ is unramified (resp. étale) at each point of $g_{\alpha }^{\prime }(U_{\alpha })$, hence in $X$ entirely since the $g_{\alpha }^{\prime }(U_{\alpha })$ cover $X$.

Proposition (18.4.9).

Let $S$ be a prescheme, $f:X\to S$ a morphism, $h:Y\to S$ a morphism locally of finite presentation, $g:X\to Y$ an $S$-morphism, $x$ a point of $X$, $y=g(x)$. The following conditions are equivalent:

a) $h$ is étale at the point $y$ and $g$ is flat at the point $x$.

b) $h$ is unramified at the point $y$ and $f$ is flat at the point $x$.

Since $f=h\circ g$, a) entails that $f$ is flat at the point $x$ (2.1.6), and obviously that $h$ is unramified at the point $y$, hence a) implies b).

To prove that b) entails a), one may first suppose that $h$ is unramified (by replacing $Y$ by a neighbourhood of $y$); then, by replacing $S$ by an open neighbourhood of $s=h(y)=f(x)$, one may suppose that there exist an étale morphism $u:S^{\prime }\to S$, a point $y^{\prime }$ of $Y^{\prime }=Y{\times }_S{S}^{\prime }$ above $y$, and an open neighbourhood $V$ of $y^{\prime }$ in $Y^{\prime }$ such that, if $h^{\prime }=h_{(S^{\prime })}:Y^{\prime }\to S^{\prime }$, the restriction of $h^{\prime }$ to $V$ is a closed immersion (18.4.8). If one then proves that $h^{\prime }$ is étale at the point $y^{\prime }$, it will follow that $h$ is étale at the point $y$ (17.7.1, (ii)); in addition, $f^{\prime }=f_{(S^{\prime })}:X{\times }_S{S}^{\prime }\to S^{\prime }$ is flat at every point $x^{\prime }$ above $x$. Since the projections $v:Y^{\prime }\to Y$, $w:X^{\prime }\to X$ are étale morphisms (hence flat), if one proves that $g^{\prime }=g_{(S^{\prime })}:X_{(S^{\prime })}\to Y_{(S^{\prime })}$ is flat at the point $x^{\prime }$, it will follow from (2.2.11, (iv)) that $g$ will be flat at the point $x$. One may therefore restrict to the case where $h$ is a closed immersion