5,911 reputation
2040
bio website math.u-psud.fr/~chambert
location Université Paris-Sud (Orsay)
age 43
visits member for 4 years, 8 months
seen 9 hours ago

2d
answered Link: Serre's intersection formula <-> Bloch-Quillen Thm / When only intersecting divisors, is there 'shorter' approach of proof known?
2d
comment Collecting proofs that finite multiplicative subgroups of fields are cyclic.
I love this proof which furnishes an explicit generator. A slight variant of it consists in observing that if $x$ is chosen so that $x^{n/p}\neq 1$ (roughly one element over $p$ will do the job), then $x^{n/q}$ is of exact order $q$. The rest is as in Paul's answer.
Dec
23
reviewed Approve Where is the Erdős–Rado theorem stated in Erdős and Rado's Bull AMS paper?
Dec
21
comment Pathological behavior of Lie algebra under a map of abelian schemes
Raynaud's theorem also requires that $F'$ be semi-abelian. As Bosch-Lütkebohmert-Raynaud say at the end of example 7/5.9, their example shows that this hypothesis cannot be relaxed.
Dec
21
awarded  Enlightened
Dec
21
awarded  Nice Answer
Dec
17
reviewed Reject nontrivial theorems with trivial proofs
Dec
17
comment Coherent cohomology of an abelian scheme and base change
Berthelot, Breen and Messing, Théorie de Dieudonné cristalline, II (LNM 930), Prop. 2.5.2 compute the De Rham cohomology of an abelian scheme - the proof uses a bit of spectral sequences.
Dec
17
comment (Affine) Schemes and the point of view of morphisms with values in a field
It depends on the set-theoretic framework you place yourself. Within ZFC, you have no other option than fixing a set of fields; within a theory with classes, the morphisms from $A$ to a field will constitute a class; within Bourbaki's theory, there is a notion of an equivalence relation without mentioning an underlying set, but you will need to prove (this is easy) that this relation is "collectivisante" so that its equivalence classes form a set.
Dec
16
answered p-adic Stein spaces
Dec
16
comment Least supersingular prime
@MichaelStoll: You're right (twice)...
Dec
16
answered Least supersingular prime
Dec
11
comment How to prove that two univariate polynomials are always algebraically dependent?
Actually, the way Ma and Marinescu prove this theorem of Siegel, consists in considering the map $K[T_1,...,T_k]\to \mathscr M(X)$ induced by the $f_i$, and to show that it cannot be injective if $k>\dim(X)$ by inspecting its jets at a non-degenerate point $x$, making use of the Schwarz lemma. It is thus very close to Michael Stoll's approach which can be thought of as a kind of baby-case.
Dec
10
answered How to prove embedded copies of a curve using different base points in its Jacobian are algebraically equivalent
Dec
10
comment Tensor product-definition-balanced versus bilinear maps
Take a module $P$ which has two distinct structures of an $R$-module. ($R=P=\mathbf C$ is an acceptable choice.) Then it seems that a bilinear map for the first structure is only balanced for the second one.
Dec
8
answered Is the Weil–Deligne representations coming from $\ell$-adic cohomology independent of $\ell$?
Dec
3
comment Is there a scheme parametrizing the closed subgroups of an algebraic group?
Take $G=\mathbf G_m^2$ (or, to keep your notation, restrict to subgroups of a maximal torus). Then connected subgroups of dimension $1$ correspond to rational lines in $\mathbf Q^2$ — the only relevant schematic structure seems to be the discrete one.
Nov
27
comment How to see that this pairing of line bundles is multiplicative?
The Deligne pairing has been studied in the mid 80s — in relation with Arakelov geometry. You could have a look at the papers of Deligne (Le déterminant de la cohomologie), Moret-Bailly (in Seminaires sur les pinceaux arithmétiques, Astérisque), or, for a higher dimensional generalization, Elkik (Fibrés d'intersection et intégrales de classes de Chern). All of this is in French, though.
Nov
26
awarded  Pundit
Nov
25
comment Fermat's last theorem over larger fields
@ACL: I forgot one hypothesis in the quoted theorem, namely that the curve has $\mathbf Q_p$ points for all primes $p$.