bio | website | math.u-psud.fr/~chambert |
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location | Université Paris-Sud (Orsay) | |
age | 43 | |
visits | member for | 4 years, 9 months |
seen | 22 hours ago | |
stats | profile views | 3,126 |
Dec 30 |
comment |
The etale fundamental group of a field
@ChandanSinghDalawat: The two volumes have been republished as a single one (with a new chapter): cassini.fr/#algebre_et_theories_galoisiennes |
Dec 30 |
reviewed | Approve The etale fundamental group of a field |
Dec 30 |
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Completion of a local ring of a curve
On the other hand, the hypothesis that $k$ be algebraically closed is not necessary. For example, assuming that the residual extension $k\to A/\mathfrak m$ is finite and separable, one can use Hensel's lemma to lift the residue field $A/\mathfrak m$ into a subfield of $\widehat{A_{\mathfrak m}}$. |
Dec 30 |
revised |
Completion of a local ring of a curve
typo. thenor -> then for |
Dec 30 |
reviewed | Edit Companion to theoretical physics for working mathematicians |
Dec 30 |
revised |
Companion to theoretical physics for working mathematicians
Adding HB Callen wonderful book on Thermodynamics (the only one I know with deals with the subject from a mathematical point of view) |
Dec 25 |
answered | Link: Serre's intersection formula <-> Bloch-Quillen Thm / When only intersecting divisors, is there 'shorter' approach of proof known? |
Dec 25 |
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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 |
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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 |
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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 |
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(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 |
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Least supersingular prime
@MichaelStoll: You're right (twice)... |
Dec 16 |
answered | Least supersingular prime |
Dec 11 |
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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 |