18,438 reputation
12655
bio website facebook.com/paterz1
location Montreal, Canada
age 22
visits member for 2 years, 11 months
seen 1 hour ago

I finished my undergraduate studies in mathematics at University of Montreal. I am doing my Ph.D at the Berlin Mathematical School in algebraic geometry. My current interests, besides algebraic geometry, are number theory, analysis, topology, measure theory, representation theory and commutative algebra.


Apr
10
revised Convergence/divergence of a couple of infinite series
added 100 characters in body
Mar
27
comment an isomorphism of extension functors
ahhhh, so it's a *local ring, not a local ring. Sorry, I didn't notice that the * had a meaning! So it's like a ''locally homogeneous'' ring, if we would want to give it a fancy name. (Although I don't like it.)
Mar
26
comment an isomorphism of extension functors
Is there any other sense to $m$ being maximal? You could've just written "Let $(R,m)$ be a Noetherian local ring". I thought that $R$ and $R_m$ would be isomorphic since localizing at $m$ only adds units. Similarly for $M$ and $M_m$.
Mar
26
comment Relation between divisibility of polynomials in different rings, $h | f$ in $\mathbb{Z}[x], \mathbb{Z}/p^k\mathbb{Z}[x]$ and $\mathbb{F}_p[x]$
You will definitely need to understand Hensel's lemma to understand these algorithms, that's for sure. I don't know if it'll be of use for this particular question, but it'll help you read the paper.
Mar
26
comment Relation between divisibility of polynomials in different rings, $h | f$ in $\mathbb{Z}[x], \mathbb{Z}/p^k\mathbb{Z}[x]$ and $\mathbb{F}_p[x]$
@BoZenKhaa : The point is that the factor $\pmod p$ may not be a factor in $\mathbb Z$. We want $h_0$ to divide $f$ in $\mathbb Z$ and $h$ to divide $h_0$ mod $p$. The second part we have already acheived, but not the first. Think about a quadratic residue for instance, i.e. an irreducible factor of $x^2+1 \pmod p$, or in other words, a root. I'm quite worried about this question right now.
Mar
26
comment Relation between divisibility of polynomials in different rings, $h | f$ in $\mathbb{Z}[x], \mathbb{Z}/p^k\mathbb{Z}[x]$ and $\mathbb{F}_p[x]$
If $h$ divides $f$ modulo $p^k$, then $f \equiv gh \pmod {p^k}$, hence $f \equiv gh \pmod p$. If $h$ is irreducible $\pmod p$, this means $h$ divides some irreducible factor of $f$ modulo $p$. Call this factor $h_0$. Then you most probably use Hensel's lemma to lift your factor $\pmod p$ to a factor in the $p$-adics, and then argue that your factor has integer coefficients. I can't really do this argument right now, but that's my first guess.
Mar
26
comment Relation between divisibility of polynomials in different rings, $h | f$ in $\mathbb{Z}[x], \mathbb{Z}/p^k\mathbb{Z}[x]$ and $\mathbb{F}_p[x]$
You can just write $h \, | \, f \pmod p$ for short. Your notation is very heavy to read.
Mar
26
comment Is there a canonical isomorphism between these two vector spaces?
@user99680 : My point was that in the finite dimensional case, injective and bijective are the same. So if he wants to work with a general vector space of non-necessarily finite dimension, then the correct assumption (as the counter examples have shown) is not injective, but rather bijective. In this case both spaces are zero though, so the isomorphism is quite canonical. I admit my first comment didn't realize the only example gave zero... -_-
Mar
25
comment Is there a canonical isomorphism between these two vector spaces?
You need to suppose $T$ bijective to make your finite dimensional case work out with this context.
Mar
25
comment How to prove the uniqueness of a continuous extension of a densely defined function?
@fgp : $x \in f^{-1}(U) \cap g^{-1}(V)$.
Mar
25
revised L'Hospital's rule problem
added 6 characters in body
Mar
24
comment Find the nontrivial proper subgroups of $\Bbb Z_{2}\times \Bbb Z_{2}$.
@DonAntonio : What if $a^2 = b$? I don't see any inconsistency with the cyclic group of order $4$.
Mar
24
answered Determining the degree of a field extension.
Mar
24
comment Find the nontrivial proper subgroups of $\Bbb Z_{2}\times \Bbb Z_{2}$.
You need to assume $a^2 = b^2 = 1$. Otherwise you get the free abelian group on two generators, i.e. $\mathbb Z^2$. Just saying. =)
Mar
22
comment Fractional Calculus: Motivation and Foundations.
I don't understand how you replace $n$ with fractional $q$ by leaving $n$'s there still. Can you please clear this up? Otherwise the rest is amazing.
Mar
21
answered Solving functional equation 1
Mar
20
answered Prove that if $m$, $n$ $\in \mathbb {N}$ with $g=\gcd(m, n)$, then $\phi(mn) = \frac{g\phi(m)\phi(n)}{\phi(g)}$
Mar
17
comment Rank of sum of projections
Sorry about the delete/undelete of my answers. Everytime I wrote the answer I supposed something you removed from the assumptions... haha.
Mar
11
comment automorphisms of the free group on two generators
What is $F_2$? It's not clear.
Mar
11
comment Notation Question regarding Ring-mod-Number and Ring-mod-Some Kernel
I'm glad it did.