21,971 reputation
13265
bio website facebook.com/paterz1
location Montreal, Canada
age 23
visits member for 3 years, 8 months
seen yesterday

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.


Jan
27
comment If a continuous function on $\mathbb{R}$ $f$ receives an extremum at a single point, it must be the global extremum.
@GEdgar : You missed an important part of the question, which is to show that $x_0$ is the global maximum, which is different than showing that $x_0$ is a global maximum.
Jan
27
answered If a continuous function on $\mathbb{R}$ $f$ receives an extremum at a single point, it must be the global extremum.
Jan
27
revised If $m^4+4^n$ is prime, then $m=n=1$ or $m$ is odd and $n$ even
deleted 1 character in body
Jan
23
awarded  Popular Question
Jan
22
awarded  Enlightened
Jan
22
awarded  Nice Answer
Jan
20
comment Showing affinity of a function - proof help
@elbarto : Would have been probably very problematic to deal with this concept since $V(A)$ is not finite-dimensional!
Jan
19
answered Showing affinity of a function - proof help
Jan
18
answered Why Are There No Solutions To $2^x \equiv 3\pmod{9}$?
Jan
17
comment True or False: $2^{2^{2011}} \text{ divides } 2^{2^{2012} }$
Try proving that $(a^n,b^m) = (a,b)^{\min \{n,m \}}$, where $(a,b)$ denotes the g.c.d. of $a$ and $b$.
Jan
16
answered $a^2=1$ in a field with positve characteristic
Jan
16
answered True or False: $2^{2^{2011}} \text{ divides } 2^{2^{2012} }$
Jan
11
comment Fundamental group of Topologists sine curve
@qwert4321 : I don't see how you've shown this, because it's not what you want. The homeomorphism is simply given by $(x, \sin(1/x)) \mapsto x$, so that the topologist sine curve is homeomorphic to $]0,\infty[$, which is homeomorphic to the real line (but if I told you it was homeomorphic to $]0,\infty[$, it would have been kind of a bigger hint!). But the contractible part and the rest of your argument is fine!
Jan
11
answered Fundamental group of Topologists sine curve
Jan
9
revised The number of $n\times n$ matrix over integer modulo $p$ field with determinant equal $1$
edited body
Jan
7
comment cohomology of classifying space of cyclic group
@anomaly : No, but in algebraic geometry yes (where we also compute cohomology often) where $\mathbb Z_{(p)}$ is the stalk of the scheme $\mathrm{Spec}(\mathbb Z)$ at $(p)$, and its importance is not to be denied.
Jan
6
comment Tangent vectors: arrows vs. derivatives
@LittleBrownOne I'm talking about your imagination, I know you guys do serious things. You don't picture the abstract manifold, you picture some $n$-dimensional potato-shaped thing. But I tend to think about a manifold as a bunch of $n$-disks separated from each other at first and then glued together with a bunch of charts in some kind of network. The embedding is some kind of representation of this network.
Jan
6
comment Tangent vectors: arrows vs. derivatives
I mean in physics everything is embedded right? Because in physics you work with the real life so to say, a curve on a manifold for you is not a function $\gamma : [0,1] \to M$ which satisfies whatever but a particle moving on $M$ whatever $M$ is, and it so happens that the mathematical abstract thing models well what you see in physics. But in mathematics we do not work in your context, in fact we work with no context because we want to be as general as possible for our results to hold everywhere. Therefore this abstract definition is necessary!
Jan
6
comment Which entire functions satisfy $\,\lvert\,f(z)\rvert \leq \lvert z\rvert^k$?
@user160738 : The point is that a non-zero polynomial never has bounded values as $N \to \infty$, whatever its coefficients are.
Jan
6
answered Tangent vectors: arrows vs. derivatives