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bio website math.u-psud.fr/~chambert
location Université Paris-Sud (Orsay)
age 44
visits member for 5 years, 3 months
seen 2 days ago

Jul
6
comment About Abhyankar's conjecture
The statement you give is Raynaud's contribution. Harbater's theorem is for covering of arbitrary algebraic curves — it says that a group $G$ is the Galois group of a covering of a curve (genus $g$, minus $r$ points) over an algebraically closed field of characteristic $p$ iff its quotient $G/p(G)$ by the subgroup $p(G)$ generated by $p$-subgroups is such a group. By Grothendieck the latter property means that it is a quotient of the fundamental group of a compact Riemann surface of genus $g$ deprived of $r$ points.
Jul
5
comment Volume of arithmetic quotients of symmetric spaces
The deck transformation group preserves the measure. See Paul Garrett's answer.
Jul
5
comment Volume of arithmetic quotients of symmetric spaces
I am a bit puzzled by the amount of notation, but look at the canonical map from $\Gamma(\mathfrak P^{ek})\backslash G_\infty$ to $\Gamma\backslash G_\infty$. It should be a (possibly ramified) covering of degree $[G:G_k]$, and this should imply your claim.
Jul
2
comment N-th root of unity in N-th division field of abelian variety?
@JoeSilverman: Unfortunately, I made a confusion and (see Zarhin's comment above), it is $(A\times A^{\vee })^4$ which is principally polarized.
Jul
2
comment N-th root of unity in N-th division field of abelian variety?
My mistake! Sorry for the confusion and thank you for your answer and your reply to my comment.
Jul
1
comment N-th root of unity in N-th division field of abelian variety?
It seems that one can also combine Joe Silverman's answer with your observation that $A^4$ has a principal polarization.
Jul
1
comment N-th root of unity in N-th division field of abelian variety?
But one knows (“Zarhin's trick”) that $A^4$ has a principal polarization. Since $K(A[n])=K(A^4[n])$, this implies that the result holds in general!
Jun
12
comment When does a modular form satisfy a differential equation with rational coefficients?
@DrorSpeiser: You're right, sorry. I misunderstood your question.
Jun
6
comment discrete valuation ring and ring of witt vectors
Notation: In the context of Witt vectors, the field with $p$ elements should definitely not be denoted by $\mathbb Z_p$!
Jun
2
awarded  Nice Answer
Jun
1
answered Solving algebraic problems with topology
May
14
comment Group schemes, adeles, double cosets, and étale cohomology
In my paper with Yuri Tschinkel, Torseurs arithmétiques et espaces fibrés, we have a similar description (Proposition 1.2.6). NB. As remarked by Philippe Gille, there is a slight mistake there, that we consider torsors which are locally trivial for the Zariski topology, without saying so.
May
11
comment Irreducible/prime/indivisible elements
The product of <= 1 field does not admit prime elements.
Apr
24
comment Constants sheaves on an open subset
Actually, the formulation of the question is incorrect, which led to the two different answers. You write that $\mathbb Z_U$ is a sheaf on $U$, while $F$ is a sheaf on $X$; so $\mathop{\rm hom}(\mathbb Z_U,F)$ does not make sense. You must either restrict $F$ to $U$ or extend $\mathbb Z_U$ to $X$. In the first case, the answer is yes (Zhen Lin's comment), and in the second it depends on the choice of an extension (direct image or extension by zero).
Apr
22
comment Base change of regular schemes
And if you want an equal-characteristic example, take $R=\mathbf C[[t]]$ and $X$ be the closed subscheme of $\mathbf A^2_R$ defined by $xy=t$; it is regular, but its base change to $\mathbf C[[t^{1/2}]]$ is not.
Apr
20
comment Must an algebraic variety with trivial tangent bundle be an abelian variety?
And no if you don't assume some properness condition.
Apr
18
revised Reference for a lemma on étale maps
Typo. tale->étale
Apr
17
comment Reference for a lemma on étale maps
With all due respect to the established litterature, the Stacks project cannot be considered less respectable than anything else published.
Apr
15
comment Interpretation of the monomorphism $H^2(\pi_1(X),\mathbb{Z}) \rightarrow H^2(X,\mathbb{Z})$
What about removing the zero-section and considering its fundamental group?
Apr
15
awarded  Yearling