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S Jul 30, 2020 at 17:06 history suggested RobPratt
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S Jul 30, 2020 at 17:06
Jul 30, 2020 at 4:41 comment added Will Chen A $\pi_1(X)$-set is a set (finite in our case) equipped with an action by the group $\pi_1(X)$. By the Galois correspondence, if you fix a geometric point $x\in X$, taking fibers gives an equivalence of categories between finite etale covers of $X$ and finite $\pi_1(X,x)$-sets. So any question internal to the category of finite etale covers of $X$ (e.g., existence of automorphisms), can be translated into a combinatorial problem involving sets with group-action.
Jul 30, 2020 at 4:38 comment added user190964 Could explain what is the corresponding $\pi_1(X)$-set? How can say that the automorphism is trivial in your example?
Jul 30, 2020 at 4:37 comment added user190964 I just checked your additional comments and try to make the picture clear. $K$ means the base field, and my comments comes from the correspondences between 1. smooth algebraic curves and its function field $X\leftrightarrow K(X)$, 2. finite etale morphism $Y\to X$ and finite separable field extension $K(Y)/K(X)$, and 3. the automorphism groups $\mathrm{Aut}(Y/X)\leftrightarrow \mathrm{Aut}(K(Y)/K(X))$.
Jul 30, 2020 at 4:30 comment added Will Chen What is $F(X)$?
Jul 30, 2020 at 4:24 comment added user190964 I mean finite étale maps. I am confused that there may exist a nontrivial finite separable extension $F$ of $K(X)$ with trivial automorphism group $\mathrm{Aut}(F/K(X))$ and $F$ corresponds to a curve $Y$ by $Y=F(X)$.
Jul 30, 2020 at 4:17 history edited user190964 CC BY-SA 4.0
edited title
Jul 30, 2020 at 4:17 comment added Will Chen By etale covering do you mean finite etale map? If so, this is just Galois theory. The map corresponds to a finite index index $d$ subgroup $H\le\pi_1(X)$, and the automorphism group is just the automorphism group of the corresponding $\pi_1(X)$-set. IIRC this should just be the quotient of the normalizer of $H$ by $H$. Sometimes it'll be trivial. For example you can fix a surjection $f : \pi_1(X)\rightarrow S_3$ (symmetric group on 3 things), and let $H$ be the preimage of an order 2 subgroup of $S_3$. The resulting cover will have trivial aut group for the same reason that $Q(\sqrt{3})$ does
Jul 30, 2020 at 4:11 history asked user190964 CC BY-SA 4.0