5,593 reputation
1738
bio website math.u-psud.fr/~chambert
location Université Paris-Sud (Orsay)
age 43
visits member for 4 years, 6 months
seen 31 mins ago

20h
answered A proof from Lang's undergraduate analysis
Oct
17
answered Equidistribution of rational points on an algebraic variety
Oct
17
answered Proofs of the uncountability of the reals.
Oct
16
revised About the hypothesis of Zorn's lemma
Adding a summary of comments.
Oct
16
awarded  Revival
Oct
16
answered Categorifications of Zorn's lemma
Oct
16
asked About the hypothesis of Zorn's lemma
Oct
6
comment Why do roots of polynomials tend to have absolute value close to 1?
I don't agree, for two reasons. 1) Partial sums furnish nice sequence of polynomials. 2) There is a generalization of Jentzsch-Szegö where the limit behaviour of the zeroes can be more general (see the book of Andrievskii-Blatt, or a paper of mine, dx.doi.org/10.1142/S1793042111004691, where I discuss a generalization to Riemann surfaces of arbitrary genus).
Oct
5
comment Why do roots of polynomials tend to have absolute value close to 1?
A theorem of Jentzsch-Szegö gives a result in a similar spirit.
Sep
30
comment Vanishing of the module of differentials of a extension of perfect fields
My impression is that the indicated maps from the tensor products do not exist. Am I wrong?
Sep
30
awarded  Explainer
Sep
28
comment Equivalence of definitions of the Milnor $K$-groups
Math is like litterature or philosophy. One should never forget to study the classics!
Sep
28
comment Galois groups and prescribed ramification
You're right. Knowing that it splits allows to understand precisely those extensions which are regular, is contain no nontrivial extension of the finite ground field.
Sep
26
answered Equivalence of definitions of the Milnor $K$-groups
Sep
26
comment Galois groups and prescribed ramification
(followed) : It thus seems that the Galois group of any extension of $k(X)$ which is unramified above $X$ is generated by $2g+r$ elements.
Sep
26
comment Galois groups and prescribed ramification
I'm not really sure of the following argument, but let's try something. If $X$ is an affine curve over a field $k$, and $\overline X$ is the base change to an algebraic closure $\overline k$, there is an exact sequence of fundamental groups, which is split if $X$ has a $k$-rational point: $1\to \pi_1(\overline X)\to \pi_1(X) \to \mathop{\rm Gal}(\overline k/k)\to 1$. It follows that $\pi_1(X)$ is the direct product of $\widehat{\mathbf Z}$ with a profinite free group on $2g+r-1$ generators.
Sep
25
answered Galois groups and prescribed ramification
Sep
23
comment Minimum of two plurisubharmonic functions
In the real case, take $D=\mathopen]-1;1\mathclose[$, $u(x)=x$ and $v(x)=-x$. One has $\Phi(x)=-|x|$, $\Phi|_{\partial D}=-1$ hence $\Phi(x)>\sup_{\partial D}\Phi$ for every $x\in D$.
Sep
22
comment Minimum of two plurisubharmonic functions
It suffices that min {u,v} = u everywhere. More seriously (it is easier to visualize the real analogue of convex functions), if $u$ and $v$ are distinct linear functions on the real line and $x\in\mathbf R$ is such that $u(x)=v(x)$, then $\min\{u,v\}$ is concave and is not convex.
Aug
4
comment Connections between Standard, Hodge and Tate conjectures on algebraic cycles?
The Tate conjecture furnishes $\mathbf Q_\ell$-linear combinations of cycles. Is it enough?