Is there any reference for resolution of equations in a polynomial ring, such as $x^2+y^2=z^2$ in $\mathbb{C}[t]$? Thanks!
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$\begingroup$ I think, but am not sure, that it is an open question whether solving polynomial equations over $\mathbb{C}[t]$ is decidable or not. See math.mit.edu/~poonen/papers/aws2003.pdf, in particular Table 1, for some (possibly outdated) related results. $\endgroup$– Qiaochu YuanCommented Aug 25, 2014 at 17:13
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2 Answers
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In general this is a difficult problem, but in special cases like $x^n+y^n=z^n$ in polynomials we can use Mason's theorem, which is about an analogue of the $abc$-conjecture for polynomials in $\mathbb{C}[t]$. It implies the following result:
Theorem: Let $n ≥ 2$ be an integer, and suppose $a, b, c \in \mathbb{C}[t]$ are pairwise relatively prime polynomials, at least one of which is not a constant, satisfying $a^n + b^n = c^n$. Then $n = 2$.
For the case $n=2$ we have the basic solutions $(a(t),b(t),c(t))=(m(t)^2-n(t)^2,2m(t)n(t),m(t)^2+n(t)^2)$ with polynomials $m(t),n(t)$.
answered Aug 25, 2014 at 18:24
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2$\begingroup$ Is Mason's theorem is needed for this? In fact, there's a more general result: If $a,b,c\in\mathbb{C}(t)$ are not all identically zero and satisfy $a^n+b^n=c^n$, then $[a,b,c]\in\mathbb{CP}^2$ consists of a single point (i.e., $a$, $b$, and $c$ are constant multiples of a single rational function). This follows because, when $n>2$, the nonsingular curve plane $a^n+b^n-c^n=0$ has genus $g=\tfrac12(n{-}1)(n{-}2)>0$, so there is no non-constant holomorphic map from $\mathbb{CP}^1$ to this curve. The same conclusion would follow for any plane curve of positive genus, not just the Fermat curves. $\endgroup$ Commented Aug 25, 2014 at 21:01
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$\begingroup$ No, it is not needed, you are right. There are many other ways to prove Fermat's theorem for polynomials. But usually the proof refers to Mason's theorem. $\endgroup$ Commented Aug 25, 2014 at 21:14
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The question is vague about what equations are to be considered. As Qiaochu points out in his comment, in the most general possible interpretation, the problem is undecidable. As Robert points out in his comment, if the equation defines a curve over $\mathbb{C}$ then it can be decided a simple genus calculation. For an arbitrary curve over $\mathbb{C}(t)$ there are effective versions of the Mordell conjecture. This was proved originally by Manin and Grauert and effectively by Szpiro, with further improvements by several others (e.g. Vojta, Compositio 78 (1991), 29-36).
answered Sep 3, 2014 at 20:35