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< Over there, you'll find my website.
2d

awarded  Nice Answer 
Aug
30 
reviewed  Close Intuition for the tensor algebra? 
Aug
30 
revised 
Lower bounding the multiplicative order of 2 modulo p
added 13 characters in body 
Aug
30 
answered  Lower bounding the multiplicative order of 2 modulo p 
Aug
28 
comment 
Learning the exponents in a sum of two modular roots of unity
Arguing about dimension is a bit of a red herring. But I have to take care of the condition of being $n$th roots of unity, which I forgot. I consider $f(x^m)f(y^m)/(x^my^m)=0, m=(q1)/n$. Now, I have to be careful with the degree,which is $m\max(a,b)$. If that's smaller than $q^{1/4}$, I can use Weil. So $f$ shouldn't be injective if $a,b$ are small. 
Aug
28 
comment 
Learning the exponents in a sum of two modular roots of unity
$f$ should not be injective if $q>n^4$ is prime, as the curve $(f(x)f(y))/(xy) = 0$ will have points in $\mathbb{F}_q$ by Weil. 
Aug
19 
comment 
Hyperelliptic curves with fixed genus and many rational points
ams.org/journals/jams/19971001/S0894034797001951 
Aug
19 
comment 
Isomorphism classes of curves $x^{m}+y^{n}=constant$
If $m < n/2$ then $y$ (which is a function of degree $m$) is unique up to scalar, by the Castelnuovo genus inequality. Maybe that allows you to conclude in this case. In general, I don't know, but what you are aiming for is uniqueness up to scalar of $x,y$. 
Aug
18 
comment 
Paper of BoutotCarayol in `Courbes modulaires et courbes de Shimura'
I was surprised to see that Asterisque is not in Numdam. Can somebody in the know explain why? 
Aug
12 
comment 
On $a+b+c= abc = n$, elliptic curves, and solvable Galois groups
Not in general. Your $f(x)$ is the $x$coordinate of an endomorphism of the elliptic curve and $h$, for instance, has roots giving the $x$coordinates of the kernel of the endomorphism and the Galois group is a subgroup of the automorphism group of said kernel. Sometimes this is solvable but if the endomorphism is multiplication by $m$ for some large $m$, it won't be. 
Aug
9 
comment 
Elementary proof for Hilbert's irreducibility theorem
This is only for specializations from $n>2$ variables to $2$ variables. As the author himself points out (first sentence of the conclusion) the harder case of $2$ to $1$ variables over the integers is not covered by his argument. 
Aug
5 
revised 
When do partial derivatives $p_x$, $p_y$ of a polynomial over $\mathbb{C}$ not have any common factor?
added 2 characters in body 
Aug
5 
answered  When do partial derivatives $p_x$, $p_y$ of a polynomial over $\mathbb{C}$ not have any common factor? 
Aug
5 
comment 
When do partial derivatives $p_x$, $p_y$ of a polynomial over $\mathbb{C}$ not have any common factor?
@Mohan Something is wrong with your example, as $p_x=2xy1, p_y = x^2$ don't have a factor in common. 
Aug
4 
comment 
Meromorphic functions on $U^2 = T^3 + 1$, cokernel of $O_S \to F_\infty/O_\infty$
The cokernel has dimension one and is generated by the image of a element of $F_{\infty}$ with a simple pole. This follows from RiemannRoch or can be proved by hand by simplifying the steps of a proof of RiemannRoch. Not MO, though. 
Aug
4 
comment 
When do partial derivatives $p_x$, $p_y$ of a polynomial over $\mathbb{C}$ not have any common factor?
If $g^2$ divides $p$ then $g$ divides all partial derivatives of $p$. I am not sure if it is the only case, though. 
Jul
20 
comment 
When is the image of an integral polynomial contained in the image of another?
In particular, a paper of Bilu and Tichy (Acta Arith, XCV 2000) is relevant. 
Jul
20 
comment 
When is the image of an integral polynomial contained in the image of another?
The curve $f(x)=g(y)$ will have infinitely many integral points so will be reducible or have genus zero. Google irreducibility of $f(x)g(y)$ for many papers on this. 
Jul
9 
reviewed  Reopen Exotic group topologies on the affine group $ax+b$ 
Jul
6 
comment 
Bound of Chebyshev function and zeros of zeta function
No such $\alpha$ is known unconditionally. 