Let $A \to B$ be a ring extension.

What is the definition of $B/A$ étale ?

When $A$ is a field, do we get a nice characterization ?

  • $\begingroup$ This question should rather be: "please list some standard textbooks defining etale morphisms". $\endgroup$ – user2035 Dec 10 '09 at 14:04
  • $\begingroup$ False, an etale ring map is different. The only place I could find that explicitly describes it is Stacks-GIT $\endgroup$ – Harry Gindi Dec 10 '09 at 14:14
  • $\begingroup$ Judging from the second question, nicojo does not care about this particular variant. $\endgroup$ – user2035 Dec 10 '09 at 15:35
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    $\begingroup$ Etale morphisms are morphisms of schemes, while etale maps are maps of rings that induce etale morphisms under the functor spec. However, obtaining an explicit characterization of such maps is not in general easy, and I believe it uses Zariski's main theorem. $\endgroup$ – Harry Gindi Dec 10 '09 at 15:54
  • $\begingroup$ You need Zariski's main theorem to prove that all of these characterizations are equivalent, that is. $\endgroup$ – Harry Gindi Dec 10 '09 at 16:31

You say that a ring homomorphism $\phi: A \to B$ is étale (resp. smooth, unramified), or that $B$ is étale (resp. smooth, unramified) over $A$ is the following two conditions are satisfied:

  • $A \to B$ is formally étale (resp. formally smooth, formally unramified): for every square-zero extension of $A$-algebras $R' \to R$ (meaning that the kernel $I$ satisfies $I^2 = 0$) the natural map $$\mathrm{Hom}_A(B, R') \to \mathrm{Hom}_{A}(B, R)$$ is bijective (resp. surjective, injective).
  • $B$ is essentially of finite presentation over $A$: $A \to B$ factors as $A \to C \to B$, where $A \to C$ is of finite presentation and $C \to B$ is $C$-isomorphic to a localization morphism $C \to S^{-1}C$ for some multiplicatively closed subset $S \subset C$.

The second condition is just a finiteness condition; the meat of the concept is in the first one. Formal smoothness is often referred to as the infinitesimal lifting property. Geometrically speaking, it says that if the affine scheme $\mathrm{Spec} \ B$ is smooth over $\mathrm{Spec} \ A$, then any map from $\mathrm{Spec} \ R$ to $\mathrm{Spec} \ B$ lifts to any square-zero (and hence any infinitesimal) deformation $\mathrm{Spec} \ R'$. Moreover, if $\mathrm{Spec} \ B$ is étale over $\mathrm{Spec} \ A$ this lifting is unique.

Differential-geometrically, unramifiedness, smoothness and étaleness correspond to the tangent map of $\mathrm{Spec} \ \phi$ being injective, surjective and bijective, respectively. In particular, étale is the generalization to the algebraic case of the concept of local isomorphism.

There are two references you might want to consult. The first one, in which you can read all about the formal properties of these morphisms, is Iversen's "Generic Local Structure in Commutative Algebra". The second one, Hartshorne's "Deformation Theory", will give you a lot of information about the geometry; section 4 of chapter 1 (available online) talks about the infinitesimal lifting property.

EDIT: The EGA definition of étale morphism of rings is slightly different from the above, in the sense that it requires finite presentation, not just locally of finite presentation: see the comments below.

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    $\begingroup$ It's always been my understanding that etale morphisms are necessarily flat. Is this true, and if so how can one see it follows from this definition? $\endgroup$ – Charles Rezk Dec 10 '09 at 16:33
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    $\begingroup$ @Charles: yes, it is true. Smooth (and hence étale) morphisms are flat. The proof uses the Jacobian criterion: see EGA IV.4, 17.5.1 $\endgroup$ – Alberto García-Raboso Dec 10 '09 at 17:18
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    $\begingroup$ This is not the same as the standard EGA definition. There you require that the map be finitely presented. For instance, the map Z->Q is not etale according to the usual defininition, but it is according to the one in this comment. $\endgroup$ – JBorger Dec 10 '09 at 21:50
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    $\begingroup$ @James: Indeed the definition in EGA (IV.4 17.3.2(ii)) requires finite presentation. However, the definition I presented (ha ha) is also used often: apart from Iversen's book, I have also found it in Sernesi's "Deformation of Algebraic Schemes" and in several papers online. Hochster comments (www.math.lsa.umich.edu/~hochster/711F04/L09_20.ps) that "this creates only small differences in the theory and is convenient in some ways, while adding a few complications in other ways". I guess your example shows such a difference. $\endgroup$ – Alberto García-Raboso Dec 11 '09 at 0:15
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    $\begingroup$ This is shocking! I had no idea there was more than one definition of etale floating around. $\endgroup$ – Charles Rezk Dec 14 '09 at 16:28

Apparently $B$ should be finitely generated as a ring over $A$ and be a flat $A$-module, and the module of Kaehler differentials of $B$ over $A$ should vanish. When $A$ is a field, the characterization is that $B$ should be a finite direct sum of finite separable field extensions of $A$.

  • $\begingroup$ In your second sentence: that is the case when both A and B are fields. $\endgroup$ – Mariano Suárez-Álvarez Dec 10 '09 at 13:49
  • $\begingroup$ Yes, thanks for this correction. And in general, $B$ is a direct sum of such fields. I will edit my answer now. $\endgroup$ – Leonid Positselski Dec 10 '09 at 13:53
  • $\begingroup$ This is not the same as the standard EGA definition. There you require the map to be finitely presented. Exercise: find an example that shows the two definitions are not equivalent. $\endgroup$ – JBorger Dec 10 '09 at 21:55
  • $\begingroup$ According to Alberto's definition it only needs to be essentially of finite presentation over A, but I don't know if this changes the above comment. $\endgroup$ – Harry Gindi Dec 10 '09 at 22:31
  • $\begingroup$ @James: OK. But when A is Noetherian, for B to be finitely generated over A or finitely presented over A means the same thing. $\endgroup$ – Leonid Positselski Dec 10 '09 at 23:31

That should be read "B is etale over A". This happens when the map from A->B is an etale ring map, which means that its dual map is an etale morphism of affine schems from SpecB->SpecA, which is defined:


As with most things in ring theory, this condition is somewhat more trivial when A is a field. We get flatness since the only stalk of specA is A (spec A has one point), which is a field, so all of its modules are free, and hence flat. Unramifiedness will not always hold, but it's also lot easier because k is a field. If k is of characteristic zero, the extension is automatically separable, so then we only need to restrict to it being finite.

There's a more direct definition which says that the morphism A->B is a smooth ring map with relative dimension zero.

If you'd like to read a section on them in more generality, you can check out Stacks-Git Chapter 7 section 85 (7.85) on page 366 .


I'm sure it's also in Hartshorne.


If f is a map of local rings $$f:A\rightarrow B$$ is étale iff it is flat and unramified (check out Bhargav Bhatt's notes at the stacks project link text). If A is a field and B is finite over A, then f is étale iff B is isomorphic to a finite product of separable field extensions of A (see proposition I.3.1 of Milne's book "Étale cohomology"). More generally, for f any ring homomorphism, check out definition II.1.1 of SGA 4.5 (B is a finitely presented A-algebra and satisfies a Jacobian criterion is a possible definition. Or B is a finitely presented A-algebra and B is flat and the relative differentials are trivial). The definition comes down to "smooth of relative dimension 0".


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