Filippo Alberto Edoardo

less info
2,286 reputation
923
bio website perso.univ-st-etienne.fr/…
location Saint-Etienne, France
age 32
visits member for 2 years, 6 months
seen 1 hour ago

I am currently a Maître de Conférences in Saint-Étienne, in France. I am particularly interested in Number Theory and Arithmetic Geometry.


Apr
8
reviewed Approve suggested edit on Resources for learning domain theory?
Apr
7
reviewed Approve suggested edit on Can anyone give an example of Ricci flat Riemannian or Lorentzian Manifold that is not flat?
Mar
31
reviewed Approve suggested edit on On the Positive Definiteness of a Linear Combination of Matrices
Mar
29
reviewed Approve suggested edit on Is this set a Riesz Basis of $L^2(0,\pi)$
Mar
28
reviewed Approve suggested edit on If y forms Pythagorean triples with two different x, can some other y also form Pythagorean triples with those two x?
Mar
26
awarded  Custodian
Mar
26
reviewed Approve suggested edit on Are there models for homotopy colimits and limits of simplicial sets that generalize Kan's suspension and loop functors?
Mar
22
awarded  Informed
Mar
20
comment Dimension of the nilpotent centralizer of a nilpotent matrix
Welcome to MO, Michael!
Mar
1
revised An algebraic number is not a root of unity?
Improved formatting and corrected two grammar mistakes.
Mar
1
awarded  Enlightened
Mar
1
awarded  Nice Answer
Feb
28
comment Open subgroups of the etale fundamental group of $P^1_\mathbb Q\setminus\{0,\infty\}$
@Donu: I don't understand what do you mean by "The kernel of $H\rightarrow \mathrm{Gal}_K$ is $\hat{\mathbb{Z}}$" indeed, one line above you wrote this kernel as $N\hat{\mathbb{Z}}$. I agree this is isomorphic to $\hat{\mathbb{Z}}$, but are you claiming $N=1$? In fact, I do not understand if the OP would like that every open subgroup contains the whole $\hat{\mathbb{Z}}(1)$ or something isomorphic to it.
Feb
28
answered An algebraic number is not a root of unity?
Feb
28
comment An algebraic number is not a root of unity?
In your examples, you take $n=7$ or $n=3^2$: since it makes things easier, I am wondering if it is enough for you that $n$ be a prime-power or whether you might want to take things like $n=60.
Feb
16
comment Misunderstanding in the hypotheses of Schlessinger's criterion
I see: well, you are right in saying that if $u'$ is surjective then we are done (so, condition H2 in Schlessinger's paper is vacuous for us): it is false for very general functors like the ones in Schlessinger's papers but true for deformation ones and your argument is right. For injectivity, this has really to do with the fact that in deformation business you only allow conjugation by matrices which are $1\pmod{m_A}$. If you check the references in my answer you find the full argument (and understand where injectivity can fail, Tilouine's proof is quite transparent).
Feb
16
revised Misunderstanding in the hypotheses of Schlessinger's criterion
Added small remark
Feb
16
comment Misunderstanding in the hypotheses of Schlessinger's criterion
My be my answer was not very clear. What I want to stress is that your functor is different from $A\mapsto\mathrm{GL}_n(A)$ and that this causes troubles when performing the fiber product. Now I do not understand exactly your question (or, at least, it seems that you found the answer yourself...)
Feb
16
answered Misunderstanding in the hypotheses of Schlessinger's criterion
Feb
7
revised About principal ideal theorem in number fields
Added reference