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<-------- Over there, you'll find my website.


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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 $f(x^m)-f(y^m)/(x^m-y^m)=0, m=(q-1)/n$. 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 $(f(x)-f(y))/(x-y) = 0$ 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/1997-10-01/S0894-0347-97-00195-1
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 Boutot-Carayol 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?
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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 $p_x=2xy-1, p_y = x^2$ 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 Riemann-Roch or can be proved by hand by simplifying the steps of a proof of Riemann-Roch. 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.