8,191 reputation
12840
bio website mathematik.uni-wuerzburg.de/…
location Würzburg
age 49
visits member for 3 years, 9 months
seen 2 hours ago
I'm a professor of mathematics at the University of Würzburg (Germany). For further infos see my web page.

20h
comment Weierstrass form of genus one $y^{10} z^{30} - 8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10} - 64=0$
@joro: What do you mean? In your question, you ask for a Weierstrass form! So what does your comment mean that you have a Weierstrass form which is birationally equivalent to the given curve? By the way, I still believe that I correctly answered your question.
21h
revised Weierstrass form of genus one $y^{10} z^{30} - 8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10} - 64=0$
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22h
comment Weierstrass form of genus one $y^{10} z^{30} - 8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10} - 64=0$
A non-singular curve given in Weierstrass form is absolutely irreducible, hence so is every curve which is birationally equivalent (no matter over which field) to this curve. But your curve factors over $\mathbb Q(\sqrt[5]{2})$.
22h
comment Weierstrass form of genus one $y^{10} z^{30} - 8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10} - 64=0$
@joro: I just noticed that your curve isn't absolutely irreducible, while it is reducible over the rationals! So that tremendously simplifies things, because only singularities can be rational points! (But non of them, except for two points at infinity, are rational.) I guess this also explains the error messages in Maple when trying to find the Weierstrassform, because the program certainly expects an absolutely irreducible curve of genus $1$.
23h
comment Weierstrass form of genus one $y^{10} z^{30} - 8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10} - 64=0$
@joro: You are right, $C$ indeed has only finitely many rational points. I amended my answer.
23h
revised Weierstrass form of genus one $y^{10} z^{30} - 8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10} - 64=0$
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1d
comment Genus of a plane curve of the form $\prod_{i=1}^n (a_iX+b_iY+Z) = Z^n$
@Jason Starr: In which sense can, for $n\ge3$, the $n$ pairs $(\bar a_i,\bar b_i)\in\overline{\mathbb F_q}^2$ be linearly independent?
1d
answered Genus of a plane curve of the form $\prod_{i=1}^n (a_iX+b_iY+Z) = Z^n$
1d
revised Weierstrass form of genus one $y^{10} z^{30} - 8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10} - 64=0$
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2d
revised Weierstrass form of genus one $y^{10} z^{30} - 8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10} - 64=0$
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2d
answered Weierstrass form of genus one $y^{10} z^{30} - 8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10} - 64=0$
Jul
22
awarded  Enlightened
Jul
22
comment When is the image of an integral polynomial contained in the image of another?
@David: I tried to clarify my argument. The point is that for some fixed $i$, $A_i(h,Y)$ is irreducible and at the same time has a rational root for infinitely many integers $h$. And this of course implies $Y$-degree $1$.
Jul
22
revised When is the image of an integral polynomial contained in the image of another?
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Jul
22
awarded  Nice Answer
Jul
22
answered When is the image of an integral polynomial contained in the image of another?
Jul
15
comment Upper density of the set of $n$'s such that $p(n)$ is prime, where $p$ is polynomial
Does anyone of those who downvoted the question or opted to close it care to tell the reason? To me it seems a reasonable question, and despite some of the comments it seems that an accurate answer isn't that easy.
Jun
30
comment Need explicit formula for certain “$q$-numbers” involving gcd's
@OP: What is $\langle e_0,e_0\rangle$? It seems to me that the problem is not well formulated, as $gcd(0,0)=\infty$. In particular, I don't know how to interpret your assertion $0=\langle o_0,o_1\rangle=\langle e_0,e_1-qe_0\rangle=q-q\cdot q^{gcd(0,0)}$.
Jun
21
comment How to prove this polynomial always has integer values at all integers?
Comparing the leading coefficient of $P_m(x)$ with that of $\binom{x}{2m}$, one gets that $\frac{3}{(2m+1)(m-1)}\binom{2m}{m}\sum_{i=0}^m\binom{m}{i}^2\frac{1}{2i-1}$ is an integer provided that $P_m(\mathbb Z)\subseteq\mathbb Z$. Is this integrality known? I don't know if there is a closed expression for the sum.
Jun
8
comment Existence of polynomials of degree $\geq 2$ which represent infinitely many prime numbers
@Gerhard Paseman: $x^2+y^2$ even assumes all the infinitely many primes in $4\mathbb N+1$ ...