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Solving conics over functions fields. (English) Zbl 1129.11053

Let \(F\) be a field of characteristic other than 2, and \(a,b,c\in F[t] \backslash \{0\}\) with \(abc\) square free. Let \(u\) be a new variable and set \(f_a=bu^2+c\), \(f_b=cu^2+a\), and \(f_c=au^2+b\). For an irreducible polynomial \(p\in F[t]\) we denote \(L_p=F[t]/(p)\). Further, if \(f\in F[t][u]\), then we denote by \(f\) mod \(p\) the image of \(f\) in \(L_p [u]\). Finally, we denote by supp\((a)\) the set of all monic irreducible \(p\in F[t]\) that divide \(a\). A solubility certificate for the equation \[ aX^2+ bY^2+cZ^2=0\tag{1} \] is a list containing the following:
\(\bullet\) For every \(p\in\text{supp}(a)\), a root of \(f_a\bmod p\) in \(L_p\);
\(\bullet\) For every \(p\in\text{supp}(b)\), a root of \(f_b\) mod \(p\) in \(L_p\);
\(\bullet\) For every \(p\in\text{supp}(c)\), a root of \(f_c\) mod \(p\) in \(L_p\);
\(\bullet\) If \(\deg(a)\equiv\deg(b)=\equiv\deg(c) \pmod 2\) and \(abc\) has no root in \(F\), then either a solution or a solubility certificate for the equation \[ l_aX^2+l_bY^2+l_xZ^2=0, \] over \(F(t)\), where \(l_a,l_b,l_c\) are the leading coefficients of \(a,b\), and \(c\), respectively. In the paper under review, the authors prove that equation (1) has a solution in the projective plane \(P^2 (F(t))\) if and only if a solubility certificate exists; in this case they give a simple algorithm which determines a such solution. Moreover, they describe an algorithm that reduces an equation \(aX^2+bY^2+cZ^2=0\), where \(a,b,c\in F(t)^*\), to an equation of the same form with \(a,b,c\in F[t]\setminus\{0\}\) and \(abc\) square free.

MSC:

11R58 Arithmetic theory of algebraic function fields
11Y50 Computer solution of Diophantine equations
68W30 Symbolic computation and algebraic computation

Software:

Magma; Maple
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References:

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