# I. Étale morphisms

To simplify the exposition, we assume that all preschemes in the following are locally Noetherian (at least, starting from §2).

## 1. Basics of differential calculus

Let X be a prescheme over Y, and \Delta_{X/Y} the diagonal morphism X\to X\times_Y X. This is an immersion, and thus a closed immersion of X into an open subset V of X\times_Y X. Let {\mathscr{I}}_X be the ideal of the closed sub-prescheme corresponding to the diagonal in V (N.B. if one really wishes to do things intrinsically, without assuming that X is separated over Y — a misleading hypothesis — then one should consider the set-theoretic inverse image of {\mathcal{O}}_{X\times X} in X and denote by {\mathscr{I}}_X the augmentation ideal in the above …). The sheaf {\mathscr{I}}_X/{\mathscr{I}}_X^2 can be thought of as a quasi-coherent sheaf on X, which we denote by \Omega_{X/Y}^1. This sheaf is of finite type if X\to Y is of finite type, and it behaves well with respect to a base change Y'\to Y. We also introduce the sheaves {\mathcal{O}}_{X\times_Y X}/{\mathscr{I}}_X^{n+1}={\mathscr{P}}^n_{X/Y}, which are sheaves of rings on X, giving us preschemes denoted by \Delta_{X/Y}^n and called the n-th infinitesimal neighbourhood of X/Y. The polysyllogism is entirely trivial, even if rather long10; it seems wise to not discuss it until we use it for something helpful, with smooth morphisms.

## 2. Quasi-finite morphisms

Let A\to B be a local homomorphism (N.B. all rings are now Noetherian) and {\mathfrak{m}} the maximal ideal of A. Then the following conditions are equivalent:

1. B/{\mathfrak{m}}B is of finite dimension over k=A/{\mathfrak{m}}.
2. {\mathfrak{m}}B is an ideal of definition, and B/{\mathfrak{r}}(B)=k(B) is an extension of k=k(A).
3. The completion \widehat{B} of B is finite over the completion \widehat{A} of A.

We then say that B is quasi-finite over A. A morphism f\colon X\to Y is said to be quasi-finite at x (or the Y-prescheme f is said to be quasi-finite at x) if {\mathcal{O}}_x is quasi-finite over {\mathcal{O}}_{f(x)}. This is equivalent to saying that x is isolated in its fibre f^{-1}(x). A morphism is said to be quasi-finite if it is quasi-finite at each point11.

If A is complete, then quasi-finiteness is equivalent to finiteness.

We could also give the usual polysyllogism (i), (ii), (iii), (iv), (v) for quasi-finite morphisms, but that doesn’t seem necessary here.

## 3. Unramified morphisms

Let f\colon X\to Y be a morphism of finite type, x\in X, and y=f(x). Then the following conditions are equivalent:

1. {\mathcal{O}}_x/{\mathfrak{m}}_y{\mathcal{O}}_x is a finite separable extension of k(y).
2. \Omega_{X/Y}^1 is zero at x.
3. The diagonal morphism \Delta_{X/Y} is an open immersion on a neighbourhood of x.

Proof. For the implication (i) \implies (ii), we can use Nakayama to reduce to the case where Y=\operatorname{Spec}(k) and X=\operatorname{Spec}(k'), where it is well known, and also trivial by the definition of separable; (ii) \implies (iii) comes from a nice and easy characterisation of open immersions, using Krull; (iii) \implies (i) follows as well from reducing to the case where Y=\operatorname{Spec}(k) and the diagonal morphism is everywhere an open immersion. We must then prove that X is finite with separable ring over k, and this leads us to consider the case where k is algebraically closed. But then every closed point of X is isolated (since it is identical to the inverse image of the diagonal by the morphism X\to X\times_k X defined by x), whence X is finite. We can thus suppose that X consists of a single point, with ring A, and so A\otimes_k A\to A is an isomorphism, hence A=k.

1. Let f satisfy one of the equivalent conditions of 3.1. We then say that f is unramified at x, or that X is unramified at x on Y.
2. Let A\to B be a local homomorphism. We say that it is unramified, or that B is a local unramified algebra on A, if B/{\mathfrak{r}}(A)B is a finite separable extension of A/{\mathfrak{r}}(A), i.e. if {\mathfrak{r}}(A)B={\mathfrak{r}}(B) and k(B) is a separable extension of k(A).12

The fact that B is unramified over A can be seen at the level of the completions of A and B. Unramified implies quasi-finite.

The set of points where f is unramified is open.

Let X' and X be preschemes of finite type over Y, and g\colon X'\to X a Y-morphism. If X is unramified over Y, then the graph morphism \Gamma_g\colon X'\to X\times_Y X is an open immersion.

Indeed, this is the inverse image of the diagonal morphism X\to X\times_Y X by g\times_Y \mathrm{id}_{X'}\colon X'\times_Y X\to X\times_Y X. One can also introduce the annihilator ideal {\mathfrak{d}}_{X/Y} of \Omega_{X/Y}^1, called the different ideal of X/Y; it defines a closed sub-prescheme of X which, set-theoretically, is the set of point where X/Y is ramified, i.e. not unramified.

1. An immersion is ramified.
2. The composition of two ramified morphisms is also ramified.
3. Base extension of a ramified morphisms is also ramified.

We are rather indifferent about (ii) and (iii) (the second seems more interesting to me). We can, of course, also be more precise, by giving some one-off statements; this is more general only in appearance (except for in the case of definition (b)), and is boring. We obtain, as per usual, the corollaries:

1. The cartesian product of two unramified morphisms is unramified.
2. If gf is unramified then so too is f.
3. If f is unramified then so too is f_\text{red}.

Let A\to B be a local homomorphism, and suppose that the residue extension k(B)/k(A) is trivial, with k(A) algebraically closed. In order for B/A to be unramified, it is necessary and sufficient that \widehat{B} be (as an \widehat{A}-algebra) a quotient of \widehat{A}.

• In the case where we don’t suppose that the residue extension is trivial, we can reduce to the case where it is by taking a suitable finite flat extension of A which destroys the aforementioned extension.
• Consider the example where A is the local ring of an ordinary double point of a curve, and B a point of its normalisation: then A\subset B, B is unramified over A with trivial residue extension, and \widehat{A}\to\widehat{B} is surjective but not injective. We are thus going to strengthen the notion of unramified-ness.

## 4. Étale morphisms. Étale covers

We are going to suppose that everything concerning flat morphisms that we need to be true is indeed true; these facts will be proved later, if there is time13.

1. Let f\colon X\to Y be a morphism of finite type. We say that f is étale at x if f is both flat and unramified at x. We say that f is étale if it is étale at all points. We say that X is étale at x over Y, or that it is a Y-prescheme which is étale at x etc.
2. Let f\colon A\to B be a local homomorphism. We say that f is étale, or that B is étale over A, if B is flat and unramified over A.14

For B/A to be étale, it is necessary and sufficient that \widehat{B}/\widehat{A} be étale.

Proof. This is true individually for both “unramified” and “flat”.

Let f\colon X\to Y be of finite type, and x\in X. The property of f being étale at x depends only on the local homomorphism {\mathcal{O}}_{f(x)}\to{\mathcal{O}}_x, and in fact only on the corresponding homomorphism for the completions.

Suppose that the residue extension k(A)\to k(B) is trivial, or that k(A) is algebraically closed. Then B/A is étale if and only if \widehat{A}\to\widehat{B} is an isomorphism.

We can combine flatness with 3.7.

Let f\colon X\to Y be a morphism of finite type. Then the set of points where f is étale is open.

Proof. Again, this is true individually for both “unramified” and “flat”.

This proposition shows that we can forget about the “one-off” comments in the study of morphisms of finite type that are somewhere étale.

1. An open immersion is étale.
2. The composition of two étale morphisms is étale.
3. The base extension of an étale morphism is étale.

Proof. Indeed, (i) is trivial, and for (ii) and (iii) it suffices to note that it is true for “unramified” and “flat”.

As a matter of fact, there are also corresponding comments to make about local homomorphisms (without the finiteness condition), which in any case should appear in the multiplodoque (starting with the case of unramified).

[Trans.] Grothendieck’s “multiplodoque d’algèbre homologique” was the final version of his Tohoku paper — see (2.1) in ‘Life and work of Alexander Grothendieck’ by Ching-Li Chan and Frans Oort for more information.

The cartesian product of two étale morphisms is étale.

Let X and X' be of finite type over Y, and g\colon X\to X' a Y-morphism. If X' is unramified over Y and X is étale over Y, then g is étale.

Proof. Indeed, g is the composition of the graph morphism \Gamma_g\colon X\to X\times_Y X' (which is an open immersion by 3.4 and the projection morphism, which is étale since it is just a “change of base” by X'\to Y of the étale morphism X\to Y.

We say that a cover of Y is étale (resp. unramified) if it is a Y-scheme X that is finite over Y and étale (resp. unramified) over Y.

The first condition means that X is defined by a coherent sheaf of algebras {\mathscr{B}} over Y. The second means that {\mathscr{B}} is locally free over Y (resp. means absolutely nothing) and, further, that, for all y\in Y, the fibre {\mathscr{B}}(y)={\mathscr{B}}_y\otimes_{{\mathcal{O}}_y}k(y) is a separable algebra (i.e. a finite composition of finite separable extensions) over k(y).

Let X be a flat cover of Y of degree n (the definition of this term deserved to figure in 4.9) defined by a locally free coherent sheaf {\mathscr{B}} of algebras. We define, as usual, the trace homomorphism {\mathscr{B}}\to{\mathscr{A}} (that is a homomorphism of {\mathscr{A}}-modules, where {\mathscr{A}}={\mathcal{O}}_Y). For X to be étale it is necessary and sufficient that the corresponding bilinear form \operatorname{tr}_{{\mathscr{B}}/{\mathscr{A}}}xy define an isomorphism of {\mathscr{B}} over {\mathscr{B}}, or, equivalently, that the discriminant section d_{X/Y} = d_{{\mathscr{B}}/{\mathscr{A}}} \in \Gamma\big(Y,\wedge^n\check{{\mathscr{B}}}\otimes_{\mathscr{A}}\wedge^n\check{{\mathscr{B}}}\big) is invertible, or that the discriminant ideal defined by this section is the unit ring.

Proof. We can reduce to the case where Y=\operatorname{Spec}(k), and then it is a well-known criterion of separability, and thus trivial by passing to the algebraic closure of k.

We will have a less trivial statement to make later on, when we do not suppose a priori that X is flat over Y, but instead require some normality hypothesis.

## 5. Fundamental property of étale morphisms

Let f\colon X\to Y be a morphism of finite type. For f to be an open immersion, it is necessary and sufficient for it to be an étale and radicial morphism.

Proof. Recall what “radicial” means: injective, with radicial residual extensions (and we also recall that it means that the morphism remains injective for any base extension). The necessity is trivial, and the sufficiency remains to be shown. We are going to give two different proofs: the first is shorter, the second is more elementary.

1. A flat morphism is open, and so we can suppose (by replacing Y with f(X)) that f is an onto homeomorphism. For any base extension, it remains true that f is flat, radicial, surjective, thus a homeomorphism, and a fortiori closed. Thus f is proper. Thus f is finite (reference: Chevalley’s theorem), defined by a coherent sheaf {\mathscr{B}} of algebras. Now {\mathscr{B}} is locally free, and further, by hypothesis, of rank 1 everywhere, and so X=Y.

2. We can suppose that Y and X are affine. We can further easily reduce to proving the following: if Y=\operatorname{Spec}(A), with A local, and if f^{-1}(y) is non-empty (where y is the closed point of Y), then X=Y (indeed, this would imply that every y\in f(X) has an open neighbourhood U such that X|U=U). We will then have that X=\operatorname{Spec}(B), and wish to prove that A=B. But, for this, we can reduce to proving the analogous claim where we replace A by \widehat{A}, and B by B\otimes_A\widehat{A} (taking into account the fact that \widehat{A} is faithfully flat over A). We can thus suppose that A is complete. Let x be the point over y. By 2.2, {\mathcal{O}}_x is finite over A, and is thus (being flat and radicial over A) identical to A. So X=Y\coprod X' (disjoint sum). But since X is radicial over Y, X' is empty.

Let f\colon X\to Y be a morphism that is both a closed immersion and étale. If X is connected, then f is an isomorphism from X to a connected component of Y.

Proof. Indeed, f is also an open immersion.

We thus deduce:

Let X be an unramified Y-scheme, with Y connected. Then every section of X over Y is an isomorphism from Y to a connected component of X. There is thus a bijective correspondence between the set of such sections and the set of connected components X_i of X such that the projection X_i\to Y is an isomorphism (or, equivalently, by 5.1, surjective and radicial). In particular, a section is determined by its value at a point.

Proof. Only the first claim demands a proof; by 5.2, it suffices to note that a section is a closed immersion (since X is separated over Y) and also étale, by 4.8.

Let X and Y be preschemes over S, with X unramified and separated over S, and Y connected. Let f and g be S-morphisms from Y to X, and suppose that y is a point of Y such that f(y)=g(y)=x, and such that the residue homomorphisms k(x)\to k(y) defined by f and g are identical (“f and g agree geometrically at y”). Then f and g are identical.

Proof. This follows from 5.3 by reducing to the case where Y=S, and by replacing X with X\times_S Y.

Here is a particularly important variant of 5.3:

Let S be a prescheme, and let X and Y be S-preschemes. Let S_0 be a closed sub-prescheme of S having the same underlying space as S, and let X_0=X\times_S S_0 and Y_0=Y\times_S S_0 be the “restrictions” of X and Y to S_0. Suppose that X is étale over S. Then the natural map \operatorname{Hom}_S(Y,X) \to \operatorname{Hom}_{S_0}(X_0,Y_0) is bijective.

Proof. We can again reduce to the case where Y=S, and then this follows from the “topological” description of sections of X/Y given in 5.3.

This result contains a claim of uniqueness and of existence of morphisms. It can also be expressed (if X and Y are both étale over S) by saying that the functor X\mapsto X_0 from the category of étale S-schemes to the category of étale S_0-schemes is fully faithful, i.e. establishes an equivalence between the first and a full subcategory of the second. We will see below that it is in fact an equivalence between the first and the second (which will be a theorem of existence for étale S-schemes).

The following form (which looks more general) of 5.5 is often useful:

(Extension of liftings). Consider a commutative diagram of morphisms \begin{CD} X @<<< Y_0 \\@VVV @VVV \\S @<<< Y \end{CD} where X\to S is étale, and Y_0\to Y is a bijective closed immersion. Then we can find a unique morphism Y\to X such that the two corresponding triangles commute.

Proof. By replacing S with Y, and X with X\times_S Y, we can reduce to the case where Y=S, and this is then a particular case of 5.5 for Y=S.

We also note the following immediate consequence of 5.1 (which we did not give as a corollary, in order to not interrupt the line of ideas developed following 5.1:

Let X and X' be preschemes that are of finite type and flat over Y, and let g\colon X\to X' be a Y-morphism. For g to be an open immersion (resp. an isomorphism), it is necessary and sufficient for the induced morphism on the fibres g\otimes_Y k(y)\colon X\otimes_Y k(y) \to X'\otimes_Y k(y) to be an open immersion (resp. an isomorphism) for all y\in Y.

Proof. It suffices to prove sufficiency; since it is true for the property of being a surjection, we can reduce to the case of an open immersion. By 5.1, we have to show that g is radicial (which is trivial) and étale (which follows from 5.9 below).

Let X and X' be Y-preschemes, g\colon X\to X' a Y-morphism, x a point of X, and y the projection of x in Y. For g to be quasi-finite (resp. unramified) at x, it is necessary and sufficient for g\otimes_Y k(y) to be so.

Proof. The two algebras over k(g(x)) that we have to study in order to see whether or not we do indeed have a morphism which is quasi-finite (resp. unramified) at x are the same for g and g\otimes_Y k(y).

With the notation of 5.8, suppose that X and X' are flat and of finite type over Y. For g to be flat (resp. étale) at x, it is necessary and sufficient for g\otimes_Y k(y) to be so.

Proof. For “flat”, the statement only serves as a reminder, since this is one of the fundamental criteria of flatness15. For “étale”, this follows by taking 5.8 into account.

## 6. Application to étale extensions of complete local rings

This section is a particular case of the results on formal preschemes, which should appear in the multiplodoque. Nevertheless, here we get away without much difficulty, i.e. without the explicit local determination of the étale morphisms in §7 (using the Main Theorem). This is perhaps sufficient reason to keep this current section (even in the multiplodoque) where it is.

Let A be a complete local ring (Noetherian, of course), with residue field k. For any A-algebra B, let R(B)=B\otimes_Ak be thought of as a k-algebra; this depends functorially on B. Then R defines an equivalence between the category of A-algebras that are finite and étale over A and the category of algebras that are finite rank and separable over k.

Firstly, the functor in question is fully faithful, as follows from the more general fact:

Let B and B' be A-algebras that are finite over A. If B is étale over A, then the canonical map \operatorname{Hom}_{A\mathsf{-alg}}(B,B') \to \operatorname{Hom}_{k\mathsf{-alg}}(R(B),R(B')) is bijective.

Proof. We can reduce to the case where A is Artinian (by replacing A by A/{\mathfrak{m}}^n), and then this is a particular case of 5.5.

It remains to prove that, for every finite and separable k algebra (or we can simply say “étale”, for brevity) L, there exists some B étale over A such that R(B) is isomorphic to L. We can suppose that L is a separable extension of k, and, as such, it admits a generator x, i.e. it is isomorphic to an algebra k[t]/Fk[t], where F\in k[t] is a monic polynomial. We can lift F to a monic polynomial F_1 in A[t], and we take B=A[t]/F_1A[t].

## 7. Local construction of unramified and étale morphisms

Let A be a Noetherian ring, B an algebra which is finite over A, and u a generator of B over A. Let F\in A[t] be such that F(u)=0 (we do not assume F to be monic), and u'=F'(u) (where F' is the differentiated polynomial). Let {\mathfrak{q}} be a prime ideal of B not containing u', and {\mathfrak{p}} its intersection with A. Then B_{\mathfrak{q}} is unramified over A_{\mathfrak{p}}.

In other words, taking Y=\operatorname{Spec}(A), X=\operatorname{Spec}(B), and X_{u'}=\operatorname{Spec}(B_{u'}), we claim that X_{u'} is unramified over Y. This statement follows from the following, more precise:

The different ideal of B/A contains u'B, and is equal to u'B if the natural homomorphism A[t]/FA[t]\to B (sending t to u) is an isomorphism.

Proof. Let J be the kernel of the homomorphism C=A[t]\to B, so that J contains FA[t], and is equal to it in the second case described in 7.2. Since the homomorphism C\to B is surjective, \Omega_{B/A}^1 can be identified with the quotient of \Omega_{C/A}^1 by the sub-module generated by J\Omega_{C/A}^1 and \mathrm{d}(J) (we should have explicitly described the definition of the homomorphism d and the calculation of \Omega^1 for an algebra of polynomials in §1). Identifying \Omega_{C/A}^1 with C, via the basis \mathrm{d}t, we obtain B/B\cdot J', and so the different ideal is generated by the set J' of images in B of derivatives of G\in J (and it suffices to take G that generate J). Since F\in J (resp. F is a generator of J), we are done.

We thus find:

Under the conditions of 7.1, and supposing that F is monic and that A[t]/FA[t]\to B is an isomorphism, in order for B_{\mathfrak{q}} to be étale over A_{\mathfrak{p}}, it is necessary and sufficient for {\mathfrak{q}} to not contain u'.

Proof. Since B is flat over A, being étale is equivalent to being unramified, and we can apply 7.2.

Under the conditions of 7.3, in order for B to be étale over A, it is necessary and sufficient for u' to be invertible, or for the ideal F' generated by F in A[t] to be the unit ideal.

Proof. The second claim follows from the first along with Nakayama (in B).

A monic polynomial F\in A[t] that has the property stated in 7.4 is said to be a separable polynomial (if F is not monic, we must at least require that the coefficient of its leading term be invertible; in the case where A is a field, we recover the usual definition).

Let B be an algebra which is finite over the local ring A. Suppose that K(A) is infinite, or that B is local. Let n be the rank of L=B\otimes_A K(A) over K(A)=k. For B to be unramified (resp. étale) over A, it is necessary and sufficient for B to be isomorphic to a quotient of (resp. isomorphic to) A[t]/FA[t], where F is a separable monic polynomial, which we can assume to be (resp. which is necessarily of) degree n.

Proof. We only have to prove necessity. Suppose that B is unramified over A, and thus that L is separable over k. It then follows from the hypotheses that L/k admits a generator \xi, and thus the \xi^i (for 0\leqslant i<n) form a basis for L over k. Let u\in B be a lift of \xi; by Nakayama, the u^i (for 0\leqslant i<n) generate (resp. form a basis of) the A-module B, and, in particular, we obtain a monic polynomial F\in A[t] such that F(u)=0; B is then isomorphic to a quotient of (resp. isomorphic to) A[t]/FA[t]. Finally, by applying 7.4 to L/k, we see that F and F' generate A[t] modulo {\mathfrak{m}}A[t], and so (by Nakayama in A[t]/FA[t]) F and F' generate A[t], and we are done.

Let A be a local ring, and A\to{\mathcal{O}} a local homomorphism such that {\mathcal{O}} is isomorphic to the localisation of an algebra of finite type over A. Suppose that {\mathcal{O}} is unramified over A. Then we can find an A-algebra B which is integral over A, a maximal ideal {\mathfrak{n}} of B, a generator u of B over A, and a monic polynomial F\in A[t] such that {\mathfrak{n}}\not\ni F'(u) and such that {\mathcal{O}} is isomorphic (as an A-algebra) to B_{\mathfrak{n}}. If {\mathcal{O}} is étale over A, then we can take B=A[t]/FA[t].

(Of course, these conditions are more than sufficient …)

Before proving 7.6, we first state some nice corollaries:

For {\mathcal{O}} to be unramified over A, it is necessary and sufficient for {\mathcal{O}} to be isomorphic to the quotient of an algebra which is unramified and étale over A.

Proof. We can take {\mathcal{O}}'=B'_{{\mathfrak{n}}'}, where B'=A[t]/FA[t] and where {\mathfrak{n}}' is the inverse image of {\mathfrak{n}} in B'.

Let f\colon X\to Y be a morphism of finite type, and x\in X. For f to be unramified at x, it is necessary and sufficient for there to exist an open neighbourhood U of x such that f|U factors as U\to X'\to Y, where the first arrow is a closed immersion, and the second is an étale morphism.

Proof. This is a simple translation of 7.7.

We will now show how the jargon of 7.6 follows from the main theorem: there exists, by 7.7, an epimorphism {\mathcal{O}}'\to{\mathcal{O}}, where {\mathcal{O}} has all the desired properties; but since {\mathcal{O}}' and {\mathcal{O}} are étale over A, the morphism {\mathcal{O}}'\to{\mathcal{O}} is étale by 4.8, and thus an isomorphism.

Proof (of Theorem 7.6). This repeats a proof from the Séminaire Chevalley. By the “Main Theorem”, we have that {\mathcal{O}}=B_{\mathfrak{n}}, where B is an algebra which is finite over A, and {\mathfrak{n}} is a maximal ideal. Then B/{\mathfrak{n}}=B({\mathcal{O}}) is a separable, and thus monogenous, extension of k; if {\mathfrak{n}}_i (for 1\leqslant i\leqslant r) are maximal ideas of B that are distinct from {\mathfrak{n}}, then there thus exists an element u of B that belongs to all the {\mathfrak{n}}_i, and thus whose image in B/{\mathfrak{n}} is a generator. But B/{\mathfrak{n}}=B_{\mathfrak{n}}/{\mathfrak{n}}B_{\mathfrak{n}}=B_{\mathfrak{n}}/{\mathfrak{m}}B_{\mathfrak{n}} (where {\mathfrak{m}} is the maximal ideal of A).

Suppose, for the moment, that we have both 7.9 and 7.10. Let n be the rank of the k-algebra L=B\otimes_A k. By Nakayama, there exists a monic polynomial of degree n in A[t] such that F(u)=0. Let f be the polynomial induced from F by reduction \mod{\mathfrak{m}}. Then L is k-isomorphic to k[t]/fk[t], and so, by 7.3, f'(\xi) is not contained in the maximal ideal of L that corresponds to {\mathfrak{n}} (where \xi denotes the image of t in L, i.e. the image of u in L). Since f'(\xi) is the image of F'(u), we are done.

Let A be a local ring, B an algebra which is finite over A, {\mathfrak{n}} a maximal ideal of B, and u an element of B whose image in B_{\mathfrak{n}}/{\mathfrak{m}}B_{\mathfrak{n}} is a generator as an algebra over k=A/{\mathfrak{m}}, and such that u is contained in every maximal ideal of B that is distinct from {\mathfrak{n}}. Let B'=B[u] and {\mathfrak{n}}'={\mathfrak{n}}B'. Then the canonical homomorphism B'_{{\mathfrak{n}}'}\to B_{\mathfrak{n}} is an isomorphism.

Let B be a algebra which is finite over A and generated by a single element u, and let {\mathfrak{n}} be a maximal ideal of B such that B_{\mathfrak{n}} is unramified over A. Then there exists a monic polynomial F\in A[t] such that F(u)=0 and F'(u)\not\in{\mathfrak{n}}.

N.B. 7.10 should have appeared as a corollary to 7.1, and before 7.5 (which it implies).

So 7.6 now follows from the combination of 7.9 and 7.10; it remains only to prove 7.9.

Proof (of Lemma 7.9). Let S'=B'\setminus{\mathfrak{n}}', so that B'S'^{-1}=B'_{\mathfrak{n'}}. Similarly, let S=B\setminus{\mathfrak{n}}, so that BS^{-1}=B_{\mathfrak{n}}. We then have a natural homomorphism BS'^{-1}\to BS^{-1}=B_{\mathfrak{n}}; we will show that this is an isomorphism, i.e. that the elements of S are invertible in BS'^{-1}, i.e. that every maximal ideal {\mathfrak{p}} of BS'^{-1} does not meet S, i.e. that every maximal ideal of BS'^{-1} induces {\mathfrak{n}} on B. \begin{CD} B @>>> BS'^{-1} @>>> BS^{-1} = B_{\mathfrak{n}} \\@AAA @AAA \\B' @>>> B'S'^{-1} = B'_{{\mathfrak{n}}'} \end{CD} Since BS'^{-1} is finite over B'S'^{-1}=B'_{{\mathfrak{n}}'}, {\mathfrak{p}} induces the unique maximal ideal {\mathfrak{n}}'B_{{\mathfrak{n}}'} of B'_{{\mathfrak{n}}'}, and thus induces the maximal ideal {\mathfrak{n}}' of B'; since B is finite over B', the ideal {\mathfrak{q}} of B induced by {\mathfrak{p}}, which lives over {\mathfrak{n}}', is necessarily maximal, and does not contain u, and is thus identical to {\mathfrak{n}}. (We have just used the fact that u belongs to every maximal ideal of B that is distinct from {\mathfrak{n}}). We now prove that BS'^{-1} is equal to B'S'^{-1}: since the former is finite over the latter, we can reduce, by Nakayama, to proving equality modulo {\mathfrak{n}}'BS'^{-1}, and, a fortiori, it suffices to prove equality modulo {\mathfrak{m}}BS'^{-1}; but BS'^{-1}/{\mathfrak{m}}BS'^{-1}=B_{\mathfrak{n}}/{\mathfrak{m}}B_{\mathfrak{n}} is generated, over k, by u (here we use the other property of u), and so the image of B' (and, a fortiori, of B'S'^{-1}) inside is everything (as a sub-ring that contains k and the image of u.)

We must be able to state 7.6 for a ring {\mathcal{O}} which is only semi-local, so that we also cover 7.5: we make the hypothesis that {\mathcal{O}}/{\mathfrak{m}}{\mathcal{O}} is a monogenous k-algebra; we can thus find some u\in B whose image in B/{\mathfrak{m}}B is a generator, and belongs to every maximal ideal of B that doesn’t come from {\mathcal{O}}. Both 7.9 and 7.10 should be able to be adapted without difficulty. More generally, …

## 8. Infinitesimal lifting of étale schemes. Applications to formal schemes

Let Y be a prescheme, Y_0 a sub-prescheme, X_0 an étale Y_0-scheme, and x a point of X_0. Then there exists an étale Y-scheme X, a neighbourhood U_0 of x in X_0, and a Y_0-isomorphism U_0\xrightarrow{\sim}X\times_Y Y_0.

Proof. Let y be the projection of x in Y_0; applying 7.6 to the étale local homomorphism A_0\to B_0 of local rings of y and x in Y_0 and X_0, we obtain an isomorphism \begin{aligned} B_0 &= (C_0)_{{\mathfrak{n}}_0} \\C_0 &= A_0[t]/F_0A_0[t] \end{aligned} where F_0 is a monic polynomial, and {\mathfrak{n}}_0 is a maximal ideal of C_0 not containing the class of F'_0(t) in C_0. Let A be the local ring of y in Y, let F be a monic polynomial in A[t] that gives F_0 under the surjective homomorphism A\to A_0 (we lift the coefficients of F_0), and let C=A[t]/FA[t], with {\mathfrak{n}} the maximal ideal of C given by the inverse image of {\mathfrak{n}}_0 under the natural epimorphism C\to C\otimes_A A_0=C_0. Let B = C_{\mathfrak{n}}. It is immediate, by construction and by 7.1, that B is étale over A, and that we have an isomorphism B\otimes_A A_0=A_0.

We know that there exists a Y-scheme X of finite type, along with a point z of X over y such that {\mathcal{O}}_z is A-isomorphic to C; since the latter is étale over A={\mathcal{O}}_y, we can (by taking X to be small enough) assume that X is étale over Y. Let X'_0=X\times_Y Y_0. Then the local ring of z in X'_0 can be identified with {\mathcal{O}}_z\otimes_A A_0=B\otimes_A A_0, and is thus isomorphic to B_0. This isomorphism is defined by an isomorphism from a neighbourhood U_0 of x in X to a neighbourhood of z in X'_0, and we can assume this to be identical to X'_0 by taking X to be small enough.

The analogous claim holds for étale covers, if we suppose the residue field k(y) to be infinite.

Proof. The proof is the same, just replacing 7.5 by 7.6.

The functor described in 5.5 is an equivalence of categories.

Proof. By 5.5, it remains only to show that every étale S_0-scheme X_0 is isomorphic to an S_0-scheme X\times_S S_0, where X is an étale S-scheme. The underlying topological space of X must necessarily be identical to the one of X_0, and with X_0 being identified with a closed sub-prescheme of X. The problem is thus equivalent to the following: find, on the underlying topological space |X_0| of X_0, a sheaf of algebras {\mathcal{O}}_X over f_0^*({\mathcal{O}}_S) (where f_0 is the projection X_0\to S_0, thought of here as a continuous map of the underlying spaces) that makes |X_0| an étale S-prescheme X, as well as an algebra homomorphism {\mathcal{O}}_X\to{\mathcal{O}}_{X_0} that is compatible with the homomorphism f_0^*({\mathcal{O}}_S)\to f_0^*({\mathcal{O}}_{S_0}) on the sheaves of scalars, and that induces an isomorphism {\mathcal{O}}_X\otimes_{f_0^*({\mathcal{O}}_S)*}f_0^*({\mathcal{O}}_{S_0})\xrightarrow{\sim}{\mathcal{O}}_{X_0}. (Then X will be an étale S-prescheme that is reduced along X_0, and thus separated over S, since X_0 is separated over S_0, and X satisfies all the desired properties).

If (U_i) is an open cover of X_0, and if we find a solution to the problem on each of the U_i, then it follows from the uniqueness theorem 5.5 that these solutions glue (i.e. the sheaves of algebras that they define, endowed with their augmentation homomorphisms, glue), and we claim that the ringed space thus constructed over S is an étale S-prescheme X endowed with an isomorphism X\times_S S_0\xleftarrow{\sim}X_0. It thus suffices to find a solution locally, which we know is possible by 8.1.

Let S be a locally Noetherian formal prescheme, endowed with an ideal of definition {\mathscr{J}}, and let S_0=(|S|,{\mathcal{O}}_S/{\mathscr{J}}) be the corresponding ordinary prescheme. Then the functor {\mathfrak{X}}\mapsto{\mathfrak{X}}\times_S S_0 from the category of étale covers of S to the category of étale covers of S_0 is an equivalence of categories.

Proof. Of course, we define an étale cover of a formal prescheme S to be a cover of S (i.e a formal prescheme over S defined by a coherent sheaf of algebras {\mathscr{B}}) such that {\mathscr{B}} is locally free, and such that the residue fibres {\mathscr{B}}_s\otimes_{{\mathcal{O}}_s}k(s) of {\mathscr{B}} are separable algebras over k(s). If we denote by S_n the ordinary prescheme (|S|,{\mathcal{O}}_S/{\mathscr{J}}^{n+1}), then the data of a coherent sheaf of algebras {\mathscr{B}} on S is equivalent to the data of a sequence of coherent sheaves of algebras {\mathscr{B}}_n on the S_n, endowed with a transitive system of homomorphisms {\mathscr{B}}_m\to{\mathscr{B}}_n (for m\geqslant n) defining the isomorphisms {\mathscr{B}}_m\otimes_{{\mathcal{O}}_{S_m}}{\mathcal{O}}_{S_n}\xrightarrow{\sim}{\mathscr{B}}_n. It is immediate that {\mathscr{B}} is locally free if and only if the {\mathscr{B}}_n are locally free over the S_n, and that the separability condition is satisfied if and only if it is satisfied for {\mathscr{B}}_0, or for all the {\mathscr{B}}_n. Thus {\mathscr{B}} is étale over S if and only if the {\mathscr{B}}_n are étale over the S_n. Taking this into account, 8.4 follows immediately from 8.3.

It was not necessary to restrict ourselves to the case of covers in 8.4, but this is the only case that we will use for the moment.

## 9. Invariance properties

Let A\to B be a local and étale morphism; we study here some cases where a certain property of A implies the same property for B, or vice versa. A certain number of such propositions are already consequences of the simple fact that B is quasi-finite and flat over A, and we content ourselves with “recalling” some of them: A and B have the same Krull dimension, and the same depth (Serre’s “cohomological codimension”, in the more modern language). It also follows, for example, that A is Cohen–Macaulay if and only if B is. Also, for any prime ideal {\mathfrak{q}} of B (inducing some {\mathfrak{p}} of A), B_{\mathfrak{q}} is again quasi-finite and flat over A_{\mathfrak{p}}, as long as we suppose that B is the localisation of an algebra of finite type over A (this follows from the fact that the set of points where a morphism of finite type is quasi-finite (resp. flat) is open); further, every prime ideal {\mathfrak{p}} of A is induced by a prime ideal {\mathfrak{q}} of B (since B is faithfully flat over A). It thus follows, for example, that {\mathfrak{q}} and {\mathfrak{p}} have the same rank; also, A has no embedded prime ideals if and only if B has none.

We will thus content ourselves with more specific propositions dealing with the case of étale morphisms.

Let A\to B be an étale local homomorphism. For A to be regular, it is necessary and sufficient for B to be regular.

Proof. Let k be the residue field of A, and L the residue field of B. Since B is flat over A, and since L=B\otimes_A k (i.e. {\mathfrak{n}}={\mathfrak{m}}B, where {\mathfrak{m}} and {\mathfrak{n}} are the maximal ideals of A and B, respectively), the {\mathfrak{m}}-adic filtration on B is identical to the {\mathfrak{n}}-adic filtration, and \operatorname{gr}^\bullet(B) = \operatorname{gr}^\bullet(A)\otimes_k L. It follows that \operatorname{gr}^\bullet(B) is a polynomial algebra over L if and only if \operatorname{gr}^\bullet(A) is a polynomial algebra over K. (N.B. we have not used the fact that L/k is separable.)

Let f\colon X\to Y be an étale morphism. If Y is regular, then X is regular; the converse is true if f is surjective.

Let f\colon X\to Y be an étale morphism. If Y is reduced, then X is reduced; the converse is true if f is surjective.

This is equivalent to the following:

Let f\colon A\to B be an étale local homomorphism, with B isomorphic to the localisation of an A-algebra of finite type over A. For A to be reduced, it is necessary and sufficient for B to be reduced.

Proof. The necessity is trivial, since A\to B is injective (since B is faithfully flat over A). For the sufficiency, let {\mathfrak{p}}_i be the minimal prime ideals of A. By hypothesis, the natural map A\to\prod_i A/{\mathfrak{p}}_i is injective, and so tensoring with the flat A-module B gives that B\to\prod_i B/{\mathfrak{p}}_iB is injective, and we can thus reduce to proving that the B/{\mathfrak{p}}_iB are reduced. Since B/{\mathfrak{p}}_iB is étale over A/{\mathfrak{p}}_i, we can reduce to the case where A is integral. Let K be the field of fractions of A, so that A\to K is injective, and thus so too is B\to B\otimes_A K (since B is A-flat), and we can thus reduce to proving that B\otimes_A K is reduced. But B is the localisation of an A-algebra of finite type over A, and thus is the local ring of a point x of a scheme of finite type X=\operatorname{Spec}(C) over Y=\operatorname{Spec}(A) that is also étale over Y, so B\otimes_A K is a localised ring (with respect to some suitable multiplicatively stable set) of the ring C\otimes_A K of X\otimes_A K. Since X\otimes_A K is étale over K, its ring is a finite product of fields (that are separable extensions of K), and thus so too is B\otimes_A K.

Let f\colon A\to B be an étale local homomorphism, with A analytically reduced (i.e. such that the completion \widehat{A} of A has no nilpotent elements). Then B is analytically reduced, and a fortiori reduced.

Proof. Indeed, \widehat{B} is finite and étale over \widehat{A}; we can apply 9.3.

Let f\colon A\to B be a local homomorphism, with B isomorphic to the localisation of an A-algebra of finite type over A.

1. If f is étale, then A is normal if and only if B is normal.
2. If A is normal, then f is étale if and only if f is injective and unramified (and then B is normal, by (i)).

We will give two different proofs of (i): the first using certain properties of quasi-finite flat morphisms (stated at the start of this section) and without using 7.6 (and thus the Main Theorem); the second proof does the opposite. For (ii), it seems like we do indeed need the Main Theorem, no matter what.

Proof. (First proof). We use the following necessary and sufficient condition for a local Noetherian ring A of dimension \neq0 to be normal.

1. For every rank-1 prime ideal {\mathfrak{p}} of A, A_{\mathfrak{p}} is normal (or, equivalently, regular);
2. For every rank-\geqslant 2 prime ideal {\mathfrak{p}} of A, the depth of A_{\mathfrak{p}} is \geqslant 2.16

We assume this criterion here, but it should also appear in the section on flatness. Its main advantage is that it does not suppose a priori that A is reduced, nor a fortiori that A is integral. Here, we can already suppose that \dim A=\dim B\neq0.

By the statements at the start of this section, the rank-1 (resp. rank-\geqslant 2) prime ideals {\mathfrak{p}} of A are exactly the intersections of A with the rank-1 (resp. rank-\geqslant 2) prime ideals {\mathfrak{q}} of B. Finally, if {\mathfrak{p}} and {\mathfrak{q}} correspond to one another, then B_{\mathfrak{q}} is étale over A_{\mathfrak{p}}, and thus of the same depth as A_{\mathfrak{p}}, and is regular if and only if A_{\mathfrak{p}} is (by 9.1). Applying Serre’s criterion, we see that A is normal if and only if B is.

Proof. (Second proof). Suppose that B is normal, with field of fractions L; let K be the field of fractions of A (and note that A is integral, since B is integral). We have already seen, in the proof of 9.3, that B\otimes_A K is a finite product of fields; since it is contained in L, it is a field; since it contains B, it is equal to L itself. An element of K which is integral over A is integral over B, and is thus in B, since B is normal, and thus also in A, since B\cap K=A (as follows from the fact that B is faithfully flat over A).

Now suppose that A is normal; we will prove that B is also normal. By 7.6, we have that B=B'_{\mathfrak{n}}, where B'=A[t]/FA[t] (with F and {\mathfrak{n}} being as in 7.6). Thus L=B\otimes_A K is a localisation of B'\otimes_A K=K[t]/FK[t], and also a product of fields (finite separable extensions of K). This latter product (B'\otimes_A K) is a direct factor of B'_K (since each time we localise an Artinian ring (here B'_K) with respect to a multiplicatively stable set), and thus corresponds to a decomposition F=F_1F_2 in K[t], with the generator of L corresponding to t being annihilated by F_1. But, since A is normal, the F_i are in A[t] (supposing that they are monic). Note that B\to L=B\otimes_A K is injective (since A\to K is, since B is flat over A), and so F_1(u)=0, with u being the class of t in L. Suppose that F were of minimal degree; then it would follows that F_2=1. (N.B. we would have F'(u)=F'_1(u)F_2(u)+F_1(u)F'_2(u)=F'_1(u)F_2(u), since F_1(u)=0, whence F'_1(u)\neq 0 since F'(u)\neq0.) Thus L = B\otimes_A K = K[t]/FK[t] \tag{$*$} and so F is a separable polynomial in K[T] (but evidently not necessarily in A[t]). (N.B. for now, we have only shown, essentially, that we can choose F and {\mathfrak{n}} in 7.6 such that, with the above notation, B'\to B'_{\mathfrak{n}}=B is injective; for this, we have used the fact that A is normal; I do not know if this remains true without this normality hypothesis).

Now recall the well-known lemma, taken from Serre’s lectures last year:

Let K be a ring, F\in K[t] a separable monic polynomial, L=K[t]/FK[t], and u the class of t in L (so that F'(u) is an invertible element of L). Then \operatorname{tr}_{L/K} u^i/F'(u) = \begin{cases} 0 &\text{if }0\leqslant i<n-1\text{;} \\1 &\text{if }i=n-1 \end{cases} where n=\deg F.

The determinant of the matrix (u^j\cdot u^i/F'(u))_{0\leqslant i,j\leqslant n-1} is equal to (-1)^{n(n-1)/2}, and thus invertible in every sub-ring A of K.

Let A be a sub-ring of K, V the A-module generated by the u^i (for 0\leqslant i\leqslant n-1), and V' the sub-A-module of L consisting of the x\in L such that \operatorname{tr}_{L/K}(xy)\in A for all y\in V (i.e. for y of the form u^i, for 0\leqslant i\leqslant n-1). Then V' is the A-module given by the basis u^i/F'(u) (for 0\leqslant i\leqslant n-1).

Suppose that K is the field of fractions of an integral normal ring A, with the coefficients of F lying in A. Then, with the notation of 9.8, V' contains the normal closure A' of A in L, which is thus contained in A[u]/F'(u), and a fortiori in A[u][F'(u)^{-1}].

We can apply the above corollary to the situation that we have obtained in the proof: since F'(u) is invertible in B, and since B contains A[u], B contains A'. By the Main Theorem (or by the fact that B=A[u]_{\mathfrak{n}}), B is a localisation of A'. Since A' is normal, so too is B.

Proof. (of ii). We proceed as in the above proof to show that we can choose F in 7.6 such that we again have (*). The only obstacle a priori is that we can no longer prove that B\to L is injective, since B is no longer assumed to be flat over A, and so we can only apply the same argument a priori to the image B_1 of B under the aforementioned homomorphism. It immediately follows that B_1 is flat over A (since it is the localisation of a free A-algebra). By 4.8, the morphism B\to B_1 is étale, and thus an isomorphism, which finishes the proof.

(From an editorial point of view, we should perform the two proofs above, and place the formal calculations of the lemma and of its corollaries in a separate section).

Let f\colon X\to Y be an étale morphism. If Y is normal, then X is normal; the converse is true if f is surjective.

Let f\colon X\to Y be a dominant morphism, with Y normal and X connected. If f is unramified, then it is also étale, and X is then normal and thus irreducible (since it is connected).

Proof. Let U be the set of points where f is étale. Since U is open, it suffices to show that it is also closed and non-empty. Since U contains the inverse image of the generic point of Y (recall that, for an algebra over a field, unramified = étale), it is non-empty (since X dominates Y). If x belongs to the closure of U, then it belongs to the closure of an irreducible component U_i of U, and thus to an irreducible component X_i=\overline{U_i} of X which intersects U and which thus dominates Y (since every component of U, being flat over Y, dominates Y). Then, if y is the projection of x over Y, {\mathcal{O}}_y\to{\mathcal{O}}_x is injective (taking into account the fact that {\mathcal{O}}_y is integral). Since {\mathcal{O}}_y is normal and {\mathcal{O}}_y\to{\mathcal{O}}_x is unramified, we conclude with the help of 9.5 (ii).

Let f\colon X\to Y be a dominant morphism of finite type, with Y normal and X irreducible. Then the set of points where f is étale is identical to the complement of the support of \Omega_{X/Y}^1, i.e. to the complement of the sub-prescheme of X defined by the different ideal {\mathfrak{d}}_{X/Y}.

(This is the “less trivial” statement which was alluded to in the remark in §4.)

We do not claim that a connected étale cover of an irreducible scheme is itself irreducible if we do not assume the base to be normal. This question will be studied in §11.

## 10. Étale covers of a normal scheme

Let Y be normal and connected of field K, and let X be a separated étale prescheme over Y. Then the connected components X_i of X are integral, their fields K_i are finite separable extensions of K, and X_i can be identified with a non-empty open subset of the normalisation of X in K_i (and thus X with a dense open subset of the normalisation of Y in R(X)=L=\prod K_i, where R(X) is the ring of rational functions on X).

Proof. By 9.10, X is normal, and a fortiori its local rings are integral, and so the connected components of X are irreducible. Since X_i is normal, and also finite and dominant over Y, it follows from a particular (almost trivial, actually) case of the Main Theorem that X_i is an open subset of the normalisation of X in the field K_i of X_i.

Under the conditions of 10.1, X is finite over Y (i.e. an étale cover of Y) if and only if X is isomorphic to the normalisation Y' of Y in L=R(X) (the ring of rational functions on X).

Proof. We know that this normalisation is finite over Y (since Y is normal, and R/K separable); conversely, if X is finite over Y, then it is also finite over Y', and so its image in Y' is closed (and it is also dense).

An algebra L of finite rank over K is said to be unramified over X (or simply unramified over K if X is evident) if L is a separable algebra over K (i.e. a direct sum of separable extensions K_i) and the normalisation Y' of Y in L (i.e. the disjoint sum of the normalisations of Y in the K_i) is unramified (i.e. étale, by 9.11) over Y. Thus:

For every X which is finite over Y and such that every irreducible component of X dominates Y, let R(X) be the ring of rational functions on X (given by the product of the local rings of the generic points of the irreducible components of X), so that X\mapsto R(X) is a functor, with values in algebras of finite rank over K=R(Y). Then this functor establishes an equivalence between the category of connected étale covers of Y and the category of extensions L of K that are unramified over Y.

Proof. The inverse functor is the normalisation functor.

Suppose that Y is affine, and thus defined by a normal ring A with field of fractions K. Let L be a finite extension of K given by a direct sum of fields. Then, by definition, the normalisation Y' of Y in L is isomorphic to \operatorname{Spec}(A'), where A' is the normalisation of A in L. To say that L is unramified over Y implies that A' is unramified (or even étale) over A. If A is local, then it is equivalent to say that the local rings A'_{\mathfrak{n}} (where {\mathfrak{n}} runs over the finite set of maximal ideals of A', i.e. the prime ideals of A' that induce the maximal ideal {\mathfrak{m}} of A) are unramified (i.e. étale) over the local ring A. Finally, note that the discriminant criterion of 4.10 can also be applied to this situation (more generally, a variant of the aforementioned criterion can be stated thusly, without any preliminary flatness condition when X dominates Y, but with Y still assumed to be locally integral: A\to B and B\to B\otimes_A K are injective — then \operatorname{tr}_{L/K} is defined — and \operatorname{tr}_{L/K}(xy) induces a fundamental bilinear form B\times B\to A, i.e. there exists x_i\in B (for 1\leqslant i\leqslant n=\operatorname{rank}_K L) such that \operatorname{tr}(x_ix_j)\in A for all i,j, and \det(\operatorname{tr}(x_i x_j))_{1\leqslant i,j\leqslant n} is invertible in A).

The syllogism 4.6 immediately implies the syllogism of being unramified in the classical case:

Let Y be a normal integral prescheme, of field K. Then

1. K is unramified over Y.

2. If L is an extension of K that is unramified over Y, and if Y' is a normal prescheme, of field L, that dominates Y (e.g. the normalisation of Y in L), and M an extension of L that is unramified over Y', then M/K is unramified over X (this is the transitivity property).

3. Let Y' be a normal integral prescheme that dominates Y, of field K'/K; if L is an extension of K that is unramified over Y, then L\otimes_K K' is an extension of K' that is unramified over Y' (this is the translation property)

Furthermore:

Under the conditions of (iii), if Y=\operatorname{Spec}(A) and Y'=\operatorname{Spec}(A'), then the normalisation \overline{A'} of A' in L'=L\otimes_K K' can be identified with \overline{A}\otimes_A A', where \overline{A} is the normalisation of A in L.

Usually, people (those who are disgusted by the consideration of non-integral rings, even if they are direct sums of fields) state the translation property in the following (weaker) form:

Under the conditions of (iii), let L_1 be a sum extension of L/K (unramified over Y) and of K'/K. Then L_1/K' is unramified over Y'. In the case where Y=\operatorname{Spec}(A) and Y'=\operatorname{Spec}(A'), we further have that \overline{A'} = A[\overline{A},A'] i.e. the normalisation ring \overline{A'} of A' in L_1 is the A-algebra generated by A' and by the normalisation \overline{A} of A in L.

This latter fact is actually false without the unramified hypothesis, even in the case of extensions given by direct sums of number fields…

To finish this section, we are going to give the intuitive interpretation of the notion of étale covers: there should be the “maximal number” of points over the point y\in Y in question, and, in particular, there should not be “multiple points combined” over y. To prove results in this sense, in all desirable generality, we will assume here Proposition 10.7 found below (whose proof will be given in the multiplodoque, Chapter IV, Section 15, and uses Chevalley’s technique of constructible sets, and a little bit of the theory of descent…).

A morphism of finite type f\colon X\to Y is said to be universally open if, for every base extension Y'\to Y (with Y' locally Noetherian), the morphism f'\colon X'=X\times_Y Y'\to Y' is open, i.e. sends open subsets to open subsets. We can actually restrict to the case where Y' is of finite type over y (and even to the case where Y' is of the form Y[t_1,\ldots,t_r], where the t_i are indeterminates). A universally open morphism is a fortiori open (but the converse is false); on the other hand, if f is open, and if X and Y are irreducible, then all of the components of all of the fibres of f are of the same dimension (i.e. the dimension of the generic fibre f^{-1}(z), where z is the generic point of Y). Finally, if Y is normal, then this latter condition already implies that f is universally open (Chevalley’s theorem). It thus follows, for example, that, if f\colon X\to Y is a quasi-finite morphism, with Y normal and irreducible, then f is universally open (or even open) if and only if every irreducible component of X dominates Y. Recall also that a flat morphism (of finite type) is open, and thus also universally open. With these preliminaries, “recall” the following:

Let f\colon X\to Y be a quasi-finite, separated, universally open morphism. For all y\in Y, let n(y) be the “geometric number of points in the fibre f^{-1}(y)”, equal to the sum of the separable degrees of the residue extensions k(x)/k(y) as x runs over the points of f^{-1}(y). Then the function y\mapsto n(y) on Y is upper semi-continuous. For it to be constant on a neighbourhood of the point y (i.e. for it to be the case that n(y)=n(z_i), where the z_i are the generic points of the irreducible components of Y that contain y), it is necessary and sufficient for there to exist a neighbourhood U of y such that X|U is finite over U.17

If y\mapsto n(y) is constant, and if Y is geometrically unibranch18, then the irreducible components of X are disjoint.

Let f\colon X\to Y be a separated étale morphism. With the notation of 10.7, the function n\mapsto n(y) is upper semi-continuous. For it to be constant on a neighbourhood of the point y (i.e. for it to be the case that n(y)=n(z_i), where the z_i are the generic points of the irreducible components of Y that contain y), it is necessary and sufficient for there to exist a neighbourhood U of y such that X|U is finite over U, i.e. such that X|U is an étale cover of U.

For a separated étale morphism f\colon X\to Y (with Y connected) to be finite (i.e. for f to make X an étale cover of Y), it is necessary and sufficient for all of the fibres of f to have the same geometric number of points.

In 10.7 and its corollary, there was no normality hypothesis on Y; if we make such a hypothesis, then we find the following stronger statement (which is usually taken as the definition of unramified for a cover):

Let f\colon X\to Y be a separated quasi-finite morphism. Suppose that Y is irreducible, that every component of X dominates Y, and that X is reduced (i.e. that {\mathcal{O}}_X has no nilpotent elements). Let n be the degree of X over Y (i.e. the sum of the degrees, over the field K of Y, of the fields K_i of the irreducible components X_i of X). Let y be a normal point of Y. Then the geometric number n(y) of points of X over y is \leqslant n, with equality if and only if there exists an open neighbourhood U of y such that X|U is an étale cover of U.

Proof. The “only if” is trivial; we will prove the “if”. Let z be the generic point of Y. Then n(z), which is equal to the sum of the separable degrees of the K_i/K, is \leqslant n, and, by 10.7, we have that n(y)\leqslant n(z); thus n(y)\leqslant n, with equality implying that X|U if finite over U, for some suitable neighbourhood U of y. We can thus suppose that X is finite over Y, and that the function n(y') on Y is constant. Then, by 10.8, X is the disjoint union of its irreducible components, and so, to prove that it is unramified at y, we can restrict to the case where X is irreducible, thus integral. Finally, we can assume that Y=\operatorname{Spec}({\mathcal{O}}_y). The theorem thus reduces to the following classical statement:

Let A be a normal local ring (Noetherian, as always), of field K; let L be a finite extension of K of degree n, and of separable degree n_s; let B be a sub-ring of L that is finite over A, with field of fraction L; let {\mathfrak{m}} be the maximal ideal of A, and n' the separable degree of B/{\mathfrak{m}}B over A/{\mathfrak{m}}A=k (which is equal to the sum of the separable deprees of the residue extensions of this ring). Then n'\leqslant n_s, and a fortiori n'\leqslant n. This latter inequality is an equality if and only if B is unramified (i.e. étale) over A.

Proof. It remains only to show that, if n'=n, then B is étale over A. Recall the proof in the case where k is infinite: we need only show that R=B/{\mathfrak{m}}B is separable over k; if this were not the case, then it would follow (by a known lemma) that there exists an element a of R whose minimal polynomial over k is of degree >n'. This element would come from an element x of B, whose minimal polynomial over K (as an element of L) is of degree \leqslant n; but this minimal polynomial has coefficients in A, since A is normal, and thus gives, by restriction modulo {\mathfrak{m}}, a monic polynomial F\in k[t] of degree \leqslant n=n', such that F(a)=0. But this is a contradiction.

In the general case (where k can be finite), we can again use geometric language: we consider Y'=\operatorname{Spec}(A[t]), which is faithfully flat over Y, and the generic point y' of the fibre \operatorname{Spec}(k[t]) of Y' over y. Then X is unramified over Y at y if and only if X'=X\times_Y Y'=\operatorname{Spec}(B[t]) is unramified over Y' at y', as we immediately see. On the other hand, by the choice of y', its residue field is k(t), and thus infinite. Since y' is a normal point of Y', we are now in the previous case.

## 11. Various addenda

We have already said that a connected étale cover of an integral scheme is not necessarily integral. Here are two examples of this fact.

1. Let C be an algebraic curve with an ordinary double point x, and let C' be its normalisation, with a and b the two points of C' over x. Let C'_1 and C'_2 be copies of C', with a_i (resp. b_i) the point of C'_i corresponding to a (resp. b). In the curve C'_1\coprod C'_2, we identify a_1 with b_2, and a_2 with b_1 (we leave making this process of identification precise to the reader; it will be explained in Chapter VI of the multiplodoque, but, in the case of curves over an algebraically closed field, is already covered in Serre’s book on algebraic curves). We obtain a curve C'' which is connected and reducible, and which is a degree-2 étale cover of C. The reader can verify that, generally, the “Galois” connected étale covers C'' of C whose inverse images C''\times_C C' are trivial covers of C' (i.e. isomorphic to the sum of a certain number of copies of C') are “cyclic” of degree n, and, conversely, for every integer n>0, we can construct a cyclic connected étale cover of degree n. In the language of the fundamental group (which will be developed later), this implies that the quotient of \pi_1(C) by the closed invariant subgroup generated by the image of \pi_1(C')\to\pi_1(C) (the homomorphism induced by the projection) is isomorphic to the compactification of \mathbb{Z}. More precisely, we should show that the fundamental group of C is isomorphic to the (topological) free product of the fundamental group of C with the compactification of \mathbb{Z}. We note that is was questions of this sort that gave birth to the “theory of descent” for schemes.

2. Let A be a complete integral local ring; we know that its normalisation A' is finite over A (by Nagata), and is thus a complete semi-local ring, and thus local, since it is integral. Suppose that the residue extension L/k that it defines is non-radicial (in the contrary case, we say that A is geometrically unibranch; cf. below). This will be the case, for example, for the ring \mathbb{R}[[s,t]]/(s^2+t^2)\mathbb{R}[[s,t]], where \mathbb{R} is the field of real numbers. Then let k' be a finite Galois extension of k such that L\otimes_k k' decomposes; let B be a finite and étale algebra over A corresponding to the residue extension k' (recall that B is essentially unique). Then the residue algebra of B'=A'\otimes_A B over B is L\otimes_k k', which is not local, and so B' is not a local ring, and thus B has zero divisors (since it is complete). Now B' is contained in the total ring of fractions of B (since it is free over A', thus torsion free over A', thus torsion free over A, thus contained in B'\otimes_A K=B'_{(K)}=A'_{(K)}\otimes_K B_{(K)}=B_{(K)}, since A'_{(k)}=K), and so B is not integral. In the case of the ring \mathbb{R}[s,t]/(s^2+t^2)\mathbb{R}[s,t], taking k'/k=\mathbb{C}/\mathbb{R}, we see that B is the local ring of two secant lines at their point of intersection.

We note also that, if there exists a connected étale cover X of Y that is integral but not irreducible, then every irreducible component of X gives an example of an unramified cover X' of Y that dominates Y but is not étale over Y. In the case of example (a), we thus see that C' is unramified over C, without being étale at the two points a and b (note that, directly, by inspection of the completions of the local rings at x and a, from the “formal” point of view, C' at the point a can be identified with a closed subscheme of C at the point x, i.e. one of the two “branches” of C passing through x).

In both (a) and (b), we see that the fact that the conclusions of 9.5 (i) and (ii) fail to hold is directly linked with the fact that a point of Y “blows up” at distinct points of the normalisation (in (b), the fact that the residue extension is non-radicial should be interpreted geometrically in this way). More precisely, we say that an integral local ring A is geometrically unibranch if its normalisation has only a single maximal ideal, with the corresponding residue extension being radicial; a point y of an integral prescheme is said to be geometrically unibranch if its local ring is geometrically unibranch. Examples: a normal point, an ordinary cusp point of a curve, etc. It seems that, if Y admits a point which is not unibranch, then there always exists a non-irreducible connected étale cover of Y; at least, this is what we have shown in case (b), when Y is the spectrum of a complete local ring. We can show, however, that if all the points of Y are geometrically unibranch, then every unramified connected Y-prescheme that dominates Y is étale and irreducible. The proof follows that of 9.5, using the following generalisation of Theorem 8.3, which will be proved later by means of the technique of descent:19

1. cf. EGA IV 16.3.↩︎

2. In EGA II 6.2.3 we further suppose that f is of finite type.↩︎

3. cf. regrets in III 1.2.↩︎

4. cf. IV.↩︎

5. cf. regrets in III 1.2.↩︎

6. cf. IV.5.9↩︎

7. cf. EGA IV 5.8.6.↩︎

8. cf. EGA IV 15.5.1↩︎

9. For the definition, cf. §11↩︎

10. cf. IX 4.10. For a more direct demonstration, cf. EGA IV 18.10.3, using a variant of 9.5 for geometrically unibranch local rings.↩︎