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visits member for 2 years, 9 months
seen Feb 7 '13 at 18:18

Jan
25
comment Doesn't Abhyankar's lemma contradict faithful flat descent (no, but I'm confused)
I think I understand what I'm doing wrong now. Starting with $Y\to X$ as in the question, the base change gives $Y\times_X X^\prime \to X^\prime$. This is not necessarily etale, but once you normalize you obtain $Y^\prime\to X^\prime$ and this morphism is etale by Abhyankar. Now, if by some chance $Y^\prime = Y\times_X X^\prime$ (i.e., the base change is normal) then faithfully flat descent implies that the morphism we started out with was already etale. Thank you very much QiL!!
Jan
25
comment Doesn't Abhyankar's lemma contradict faithful flat descent (no, but I'm confused)
By "some maximal ideal" do you mean "any maximal ideal", or does Abhyankar really mean that there is maybe just "one" maximal ideal?
Jan
24
asked Is there a construction known for associating a K3 surface to a curve or cover of curves
Jan
24
asked Doesn't Abhyankar's lemma contradict faithful flat descent (no, but I'm confused)
Jan
24
accepted Is a group with only finitely many subgroups of index n (for all n) finitely generated
Jan
24
comment Is a group with only finitely many subgroups of index n (for all n) finitely generated
Great! I guess one can also use this to obtain a non-abelian example by taking the product of $\mathbf Q/\mathbf Z$ with a finite non-abelian group $G$. Then $\mathbf Q/\mathbf Z \times G$ is not finitely generated, but it also has only finitely many subgroups of index $n$, right?
Jan
24
asked Is a group with only finitely many subgroups of index n (for all n) finitely generated
Jan
24
accepted Is a finite field extension of a imperfect field imperfect
Jan
23
asked Is a finite field extension of a imperfect field imperfect
Jan
20
comment Is $M_g$ a subvariety of $M_{h}$ for some $h>g$
No, but maybe something like Knudsen clutching can be used?
Jan
19
asked Is $M_g$ a subvariety of $M_{h}$ for some $h>g$
Jan
19
asked Are there generalizations of Prym varieties to higher dimensions
Jan
19
comment Is the relative rank function with respect to an ample line bundle non-decreasing
Ok. I should have thought of that...I can still save the situation. In my case, $\mathcal L = \Omega^1_{X/K}$. What can we do in this case? Of course, if you take a rational variety this is not going to work, but let's assume $X$ is such that $h^0(\Omega_{X/K}) >0$.
Jan
19
comment Is the relative rank function with respect to an ample line bundle non-decreasing
I fixed some typos and inaccuracies. The answer is still incomplete though. I need to figure out why $\mathcal L \cong \mathcal O_X(D)$ with $D$ effective. Do you know if this is true?
Jan
19
revised Is the relative rank function with respect to an ample line bundle non-decreasing
deleted 46 characters in body
Jan
18
answered Is the relative rank function with respect to an ample line bundle non-decreasing
Jan
18
revised Is the relative rank function with respect to an ample line bundle non-decreasing
added 516 characters in body
Jan
18
asked Is the relative rank function with respect to an ample line bundle non-decreasing
Jan
13
comment Are there moduli spaces of higher-dimensional varieties
When I say moduli stack I mean a Deligne-Mumford stack solving the moduli problem. Sorry for not saying that explicitly.
Jan
13
asked Are there moduli spaces of higher-dimensional varieties