Throughout this post, we fix a field $k$ (of characteristic zero if makes our lives easier :)) and $X$ an arbitrary variety (scheme) over $k$. I understand what's the pathology/discrepancy, behind the theory of covering spaces for a fixed $X$ in algebraic geometry, and why we come up with the particular "étale" setup to mimic algebraic topology (into a certain extend of course). We cannot even define the notion of fundamental group $\pi_1(X,x)$, in a proper pointed (topological) category; so, we construct the so-called finite étale morphisms and exploit the isomorphism $\pi_1(X,x) \cong \text{Aut}(\tilde{X})$, to obtain the algebraic (étale) fundamental group.

Now my question has to do with the intuition behind the idea of geometric points. While in the category of topological spaces, we can easily think of a singleton $x \in X$ and form the pointed space $(X,x)$ (hence $\pi_1(X,x)$ makes sense) the same isn't true in the category of schemes, where additional (algebraic) data is considered. So if I understand correctly, the idea of choosing a point here, becomes equivalent with the choice of a morphism $\overline{x} \to X$ (which in turn tries somehow to mimic as well the idea of a loop lifting in the pure topological setup??). The latter though, we don't want to be arbitrary, but due to J.S. Milne - "Lectures on Étale Cohmology" the chosen geometric point, should be coming from the choice of a morphism of schemes $\text{Spec}(L) \to X$, where $L$ ($L/k$ is a finite extension), some separably algebraically closed field. So why do we need these assumptions for $L$? Why there is no other "canonical" way to choose a point in $X$?

P.S. Any reference which deals with this question is welcomed. Thanks!


Minhyong Kim wrote something very nice about this which I can't find at the moment. His point, as I understand it, is that in topology you can use any map from a simply connected space into $X$ as a "basepoint" for the fundamental group, up to and including a universal cover. The analogous statement in algebraic geometry is that you can use any map from a scheme with no nontrivial etale covers into $X$ as a "basepoint" for the etale fundamental group, and separably closed fields are the easiest examples of such things.

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    $\begingroup$ mathoverflow.net/questions/546/… $\endgroup$ – Eoin Jan 9 '18 at 4:19
  • $\begingroup$ @Eoin: yep, that's the one, thanks! $\endgroup$ – Qiaochu Yuan Jan 9 '18 at 5:57
  • $\begingroup$ Thanks for the answer Qiaochu. Indeed for my understanding the answer you quoted is exactly what I wanted. As a matter of fact I don't think that the answer per se, answers the original question, but rather gives a reason/motivation why the chosen is the right one. In topology under reasonable assumptions (something that contains many times the word connected inside usually works :), locally path-connected, semi-locally simply connected, connected), a space X, admits a nice covering structure hence the whole theory can be unwrapped. $\endgroup$ – user321268 Jan 17 '18 at 14:05
  • $\begingroup$ ...cont. In algebraic geometry these ideas are not applicable for many reasons (over the complex numbers say, we cannot even define the fundamental group properly), therefore someone wants to choose points on schemes $x \in X$, and "neighborhood" $(U,u)$, $U \to X, u \mapsto x$, which has the same property as in the topological setup, i.e., trivial finite coverings, which in case of algebraic geometry transfornms to finite etale coverings. So for a field $k$, this can be happened only if we choose $L\k$, separably algebraically closed. This is what I understood. Thanks again. $\endgroup$ – user321268 Jan 17 '18 at 14:14

Not an answer, but too long for a comment. Lots of hand-weaving, very few of actual facts. I just hope it will help a bit.

Look at $\mathbb{A}^1_{\mathbb{F}_p}$. But instead of a line, picture it as a line times a circle "$L\times S$", endowed with an awful topology such that if $U$ is an open subset and $(x,y)\in U$, then the whole circle $x\times S$ is also in $U$. So the topology won't ever be able to distinguished two points in the same circle. Still, assume that if $S'\rightarrow S$ is a covering, you can consider the covering $L\times S'\rightarrow L\times S$.

With this topology, the smallest closed subset (i.e. the points) are the circles $x\times S$, and they have non-trivial coverings. Well, this is not a good thing to call them points... Instead, of the smallest closed subset, you could look at their universal cover. This looks much more like points from a $\pi_1$-perspective.

So fix $p=x\times S$ "a point" and $\overline{p}=x\times\tilde{S}$ its universal cover "a geometric point lying over $x\times S$". Recall from topology that $\overline{x}$ is unique up to isomorphism, but not up to unique isomorphism. In fact, $\overline{p}$ will have many automorphisms. And this is by definition $\pi_1(p,\overline{p})$. (We need a chosen geometric point before speaking of $\pi_1$, peaking another one will of course give the same group, but not canonically).

(Note that, $\operatorname{Spec}(\mathbb{F}_p)$ is indeed a kind of a circle with a fundamental group spanned by the Frobenius, but the picture above has lots of limits to properly describe $\mathbb{A}^1_{\mathbb{F}_p}$, in particular wild ramification does not enter the picture)

Then enter some additional difficulties : instead of a universal cover, we actually needs the pro-system of finite etale maps...

  • $\begingroup$ Roland thanks a lot. Interesting example I think, quite instructive and reveals the conceptual complexity of algebraic geometry when topology shows up. I do understand what's your point, though one question if I may: where did you find that example? Did you think it on your own, or there is a text-book which deals with this sort of question? $\endgroup$ – user321268 Jan 17 '18 at 14:48
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    $\begingroup$ I made that up. I mean, because $\pi_1(\operatorname{Spec}\mathbb{F}_p)=\hat{\mathbb{Z}}$, one can indeed visualize this scheme as a circle instead of a point (that I did not invent it). $\endgroup$ – Roland Jan 17 '18 at 15:07
  • $\begingroup$ Very good example! Thank you $\endgroup$ – GRE Sep 6 at 14:37

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