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Falting's theorem states that a non-singular algebraic curve with genus $g>1$ only has finite many rational points. Apparently, the degree-formula (see https://en.wikipedia.org/wiki/Genus%E2%80%93degree_formula) allows to determine the genus if the singularities are known.

  • If the formula gives $g>1$, can we apply Falting's theorem also in the case of a singular algebraic curve ?

  • How can Falting's theorem be used to determine whether a diophantine equation in two variables has finite or infinite many rational points ? In particular, for which integer polynomials $p(x)$ can we guarantee that $y^2=p(x)$ has only finite many rational solutions ?

Community
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Peter
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2 Answers2

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Faltings' theorem applies when the genus of $C$ is $>1$. If $p(x)$ is a square-free polynomial of degree $\ge5$ then $g\ge2$ for the curve $y^2=p(x)$ and this has just finitely many rational points.

If $p(x)$ is squarefree of degree $3$ or $4$ then $y^2=p(x)$ is an elliptic curve, which may or may not have infinitely many integer points.

Angina Seng
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  • In other words, for every squarefree polynomial $p(x)$ with degree at least $5$, we only have finite many rational solutions of $y^2=p(x)$. Can these be found efficiently (I ask because in the case of elliptic curves no efficient bounds are known) ? – Peter Mar 15 '18 at 09:48
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One may also apply Siegel's Theorem, see here.

Theorem 5 (Siegel’s theorem on integer points): Let $P \in {\bf Q}[x,y]$ be an irreducible polynomial of two variables, such that the affine plane curve $C := \{ (x,y): P(x,y)=0\}$ either has genus at least one, or has at least three points on the line at infinity, or both. Then $C$ has only finitely many integer points $(x,y) \in {\bf Z}^2$.

"Siegel’s theorem can be compared with the more difficult theorem of Faltings, which establishes finiteness of rational points (not just integer points), but now needs the stricter requirement that the curve $C$ has genus at least two (to avoid the additional counterexample of elliptic curves of positive rank, which have infinitely many rational points)."

Dietrich Burde
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