Currently working on algebraic surfaces over the complex numbers. I did a course on schemes but at the moment just work in the language of varieties.

Now i encounter the term "étale morphism" every now and then (in the book by Beauville). I know Hartshorne's definition as a smooth morphism of relative dimension zero, and wikipedia states a bunch of equivalent ones. I can work with this, so no problem there. However, some more intuition about the concept would also be nice.

So basically, if you have worked with étale morphisms, could you explain what's your personal intuition for such things, in the case of varieties? If in your answer you could also mention smooth and flat morphisms, that would be really appreciated.

Thanks in advance! Joachim

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    $\begingroup$ I think of étale morphisms as being the algebraic version of topological covering maps. It seems to work. $\endgroup$ Aug 2, 2012 at 18:04
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    $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ Aug 2, 2012 at 18:27
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    $\begingroup$ In differential geometry, “étale” is a (chic but somewhat rare) synonym of local diffeomorphism, which are definitely more general than covering maps (noteworthy exception: a local diffeomorphism between compact manifolds is a covering map). In the algebraic setting, isn't the inclusion of an open subscheme étale? $\endgroup$
    – PseudoNeo
    Aug 2, 2012 at 18:27
  • $\begingroup$ Dear @Qiaochu: how did you arrange that when one clicks on your link, one obtains a Wikipedia page which displays the exact section you refer to (rather than the beginning of that page) ? $\endgroup$ Aug 2, 2012 at 19:11
  • $\begingroup$ @Georges: the section is encoded in the link. Just click on the corresponding section in the table table of contents and you'll get the appropriate link. $\endgroup$ Aug 2, 2012 at 19:22

3 Answers 3


Instead of answering your question with general results easily obtainable from the literature, online or traditional, I'll give you a few morphisms.
Deciding whether they are are étale may contribute to developing your intuition.
(Of course I'll gladly help you or anybody else if you had any problem with these morphisms)

a) $\mathbb A^1_\mathbb C\to \operatorname {Spec}(\mathbb C[X,Y]/(Y^2-X^3)): t\mapsto (t^2,t^3)$
b) $\mathbb A^1_\mathbb C\to \operatorname {Spec}(\mathbb C[X,Y]/(Y^2-X^2-X^3)): t\mapsto (t^2-1,t^3-t)$
c) $\mathbb A^2_\mathbb C\to \mathbb A^2_\mathbb C: (x,y)\mapsto (x,xy)$
d) $\operatorname {Spec}\mathbb C[T]\to \operatorname {Spec}\mathbb C[T^2,T^3]$
e) $\operatorname {Spec}\mathbb Q[T]/(T^2-4)\to \operatorname {Spec}\mathbb Q$
f) $\operatorname {Spec}\mathbb Q[T]/(T^2+4)\to \operatorname {Spec}\mathbb Q$
g) $\operatorname {Spec}\mathbb Q[T]/(T^2)\to \operatorname {Spec}\mathbb Q$
h) $\operatorname {Spec}\mathbb F_9\to \operatorname {Spec}\mathbb F_3$

Edit (one day later) : Two useful theorems and how they settle the question of étaleness of the above morphisms

Theorem 1 Given a field $k$ and a $k$-algebra $A$, the morphism $\operatorname {Spec}(A)\to \operatorname {Spec}(k) $ is étale iff $A$ is isomorphic as a $k$-algebra to a finite product $A\cong K_1\times...\times K_n$ of finite separable field extensions $K_i/k$.
Remark In the étale case, $A$ must be reduced ( i.e. $\operatorname {Nil}(A)=0$ )
Example Every finite Galois extension $K/k$ gives rise to an étale morphism $\operatorname {Spec}(K)\to \operatorname {Spec}(k) $. This is the kernel of Grothendieck's famous geometrization of Galois theory
Illustration The morphisms e), f), h) are étale but g) is not because $\operatorname {Spec}\mathbb Q[T]/(T^2)$ is not reduced.

Theorem 2 A morphism of schemes $f:X\to Y$ is étale iff it is flat and unramified.
Illustration The morphisms a) , b), c) and d) are not étale because they are not flat
For the sake of completeness let me mention that a), c) and d) are ramified but that b) is unramified.
Let me also mention that a) and d) are two different presentations of the same morphism.

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    $\begingroup$ Dear @PseudoNeo , I'll let users answer/comment and I promise that I'll explain all the examples that haven't been satisfactorily addressed within a reasonable time. $\endgroup$ Aug 2, 2012 at 20:08
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    $\begingroup$ I'll make one contribution to the war effort: I suspect a) and d) to be the same ;-) $\endgroup$
    – PseudoNeo
    Aug 2, 2012 at 20:11
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    $\begingroup$ Dear @PseudoNeo, yes you are absolutely right: you have already foiled my most diabolical trap! Congratulations (and +1)! $\endgroup$ Aug 2, 2012 at 20:14
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    $\begingroup$ Dear @PseudoNeo, if $f$ is étale then $X$ is smooth iff $Y$ is smooth (I only assume here that $X$ and $Y$ are complex algebraic varieties, not necessarily affine). So if $X$ is smooth and $Y$ is not smooth the characterization you request is very easy: $f:X\to Y$ is never étale! $\endgroup$ Aug 2, 2012 at 21:03
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    $\begingroup$ As promised to @PseudoNeo, I have now explained (after leaving 24 hours for discussion) why the mentioned morphisms are or are not étale. $\endgroup$ Aug 3, 2012 at 21:28

For four of Georges examples I can give a general hint. If $X\to Spec \ k$ is étale where $k$ is a field, then $X$ must just be a disjoint union $\coprod Spec \ L_i$ where each $L_i$ is a finite separable field extension of $k$.

Proving this fact is a good exercise and it goes back to the analogy people have been making. A "covering space" of a point should just topologically be a discrete set of points, but this is algebraic geometry so there should be algebraic information as well. The algebraic information is that the field extensions are all separable. This has to do with the fact that $X$ being smooth implies it is "geometrically reduced" and hence you see that we can't pick up nilpotents when base changing.

Now it should be pretty easy to do (e)-(h). Unfortunately, all the base fields there are perfect, so we don't have weird unnecessary complication. I'll throw in

i) $\displaystyle Spec \left(\frac{\mathbb{F}_p(t)[x]}{(x^p-t)}\right)\to Spec \ \mathbb{F}_p(t)$

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    $\begingroup$ +1 for the characterization of étale morphisms $X\to \text{Spec} \:(k)$. As to your morphism $\displaystyle Spec \left(\frac{k[x]}{(x^p-t)}\right)\to Spec \ k$, it is never étale if $k$ is a field of characteristic $p\gt 0$ (in your example $k=\mathbb F_p(t)$ ), independently of the fact that $k$ is perfect or not and independently of the choice of $t\in k$ (you may choose $t=0$ or $1$ if you like). $\endgroup$ Aug 2, 2012 at 21:27
  • $\begingroup$ @Matt: the word “characterization” in Georges's comment seems to indicate that your “if” is an “iff”. Is that so? $\endgroup$
    – PseudoNeo
    Aug 2, 2012 at 22:13
  • $\begingroup$ @PseudoNeo Yes. It is an if and only if condition. $\endgroup$
    – Matt
    Aug 2, 2012 at 22:18
  • $\begingroup$ So both e) and f) are étale, aren't they? $\endgroup$
    – PseudoNeo
    Aug 3, 2012 at 0:57
  • $\begingroup$ Yes, you are right @PseudoNeo: e) and f) are indeed étale. $\endgroup$ Aug 3, 2012 at 9:13

To elaborate on Gunnar's comment: covering spaces provide good intuition, at least in characteristic zero. In positive characteristic you have to keep the Frobenius in mind at all times. Flatness expresses the fact that all the fibers of a connected covering have the same cardinality. Smoothness means that etale morphisms are surjections, hence isomorphisms (this is the relative dimension zero part), on tangent spaces, so in the complex topology they are local homeomorphisms by the inverse function theorem.

  • $\begingroup$ Maybe also useful: en.wikipedia.org/wiki/… $\endgroup$
    – Andrew
    Aug 2, 2012 at 18:25
  • $\begingroup$ Dear @Andrew: how did you arrange that when one clicks on your link, one obtains a Wikipedia page which displays the exact section you refer to (rather than the beginning of that page) ? $\endgroup$ Aug 2, 2012 at 19:10
  • $\begingroup$ Dear @GeorgesElencwajg: I made sure to click the relevant hyperlink in the table of contents at the top of the page before copying the url. Pretty low-tech really ;-) $\endgroup$
    – Andrew
    Aug 2, 2012 at 19:23
  • $\begingroup$ Thanks a lot, @Andrew. Low-tech: I don't know. Helpful and simple: definitely! $\endgroup$ Aug 2, 2012 at 20:00

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