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T. Combot's user avatar
T. Combot's user avatar
T. Combot's user avatar
T. Combot
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Parametrizing polynomials with given Galois group
Yes, you are right, probably the question would be better in projective space. The question can indeed be rewritten in terms of building a rational parametrization of the manifold given by the invariants conditions. But even for $A_n$, this is delicate: is $\Delta=r^2$ rationally parametrizable?
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Linear relation between polynomial roots
I tested x^10+x^8+6*x^7-5*x^6+3*x^5+9*x^4-8*x^3+2*x+3 (a random polynomial) in magma online, it failed (out of memory). Factorizing it in $\mathbb{Q}(\alpha)$ with $\alpha$ a root of x^10+x^8+6*x^7-5*x^6+3*x^5+9*x^4-8*x^3+2*x+3 proves the Galois group is 2-transitive, and thus there are no relations except the trace.
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Linear relation between polynomial roots
I read the paper, sadly remark p278 blows it all as complexity is O(n!). Theoretically it should not be better than direct computing the ideal relation I (up to log tems) as it is equivalent to the factorization of a n! degree polynomial.
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Elliptic factors in the Jacobian and zeta function
Thank you, I will read about this. My question was inspired by the question "Algebraic curve mapping to elliptic curve - how to check whether this is possible". Existence of such morphism is related to elliptic factors in the Jacobian, and it was suggested in the comments that the zeta function of its reduction could solve this problem.
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Hyperelliptic equation on a function field
Thank you very much for your answer. Yes if you have a reference for the hyperelliptic result, such a bound seems very nice. By the way, for g=1, is there some hope for a similar bound on a Mordell Weil generators set?
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Solving elliptic equation in rational functions
Do you mean that any non torsion point should be of degree at least C.deg A where C is some universal constant? I seems to me that, assuming $A$ square free, for any solution $X,Y$, the max of their numerators/denominators degree is $\geq 1/4 \deg A$: Either the number of roots of $AY^2$ is at $\geq 3/4 \deg A$ or $Y$ has more that $1/4\deg A$ poles (and then is of degree $\geq \deg A/4$). As the left hanside is cubic in $X$, I get $X$ of degree $\geq \deg A/4$.
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Solving elliptic equation in rational functions
Indeed, I found in Silvermann p230 Th 6.1 the Mordell Weil Theorem and it applies when $A$ is not a square, thank you. For the degree of solutions, the only related result I found is in "THUE'S EQUATION OVER FUNCTION FIELDS" of Schmidt Thm 1, where a degree bound for Thue equation is given. I don't know if it can be used to obtain a bound for my equation, or maybe a hyperelliptic version.
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