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This is probably a silly question, but I still can't figure it out:

Acording to Faltings Theorem, a non-singular algebraic curve in $\mathbb{Q}[x,y]$ with genus $g>1$ has a finite number of rational points.

But the curve $y=x^4-1$, for example, is non-singular and, from the genus-degree formula, $g=\frac{(4-1)(4-2)}{2}=3>1$, but of course all $(r, r^4-1)$ with $r\in\mathbb{Q}$ are rational points.

What am I getting wrong?

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    $\begingroup$ The question is not at all silly! I had a similar question about how to determine the genus in general and did not receive a satisfying answer. But I remember that we have to homogenize the equation first (but I don't remember the exact details) $\endgroup$
    – Peter
    Commented Sep 6, 2017 at 21:43

1 Answer 1

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Your curve $C$ has genus $0$, not $3$. One can see it isomorphic to $\mathbb{A}^1$ via the projection \begin{align*} C &\to \mathbb{A}^1\\ (x,y) &\mapsto x \end{align*} which has inverse \begin{align*} \mathbb{A}^1 &\to C\\ x &\mapsto (x, x^4 - 1) \, . \end{align*} Similarly, the projective closure $\overline{C}$ of $C$ is birational to $\mathbb{P}^1$ via \begin{align*} \overline{C} &\mapsto \mathbb{P}^1\\ [X:Y:Z] &\mapsto [X:Z] \end{align*} and \begin{align*} \mathbb{P}^1 &\to \overline{C}\\ [S:T] &\mapsto [ST^3 : S^4 - T^4 : T^4] \, . \end{align*}

The problem with your computation is that the genus-degree formula applies to nonsingular projective plane curves, and $\overline{C}$ has a singular point at infinity, which we can see as follows. Homogenizing, we have that the projective curve $\overline{C}$ is given by the vanishing of the polynomial $F := YZ^3 - X^4 + Z^4$, and the partials $F_X = -4X^3$, $F_Y = Z^3$, $F_Z = 3YZ^2 + 4Z^3$ all vanish at the point $[0:1:0]$, so this is a singular point of the curve.

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    $\begingroup$ Just to nitpick : I wouldn't say that $\overline{C}$ is isomorphic to $\mathbb{P}^1$ since the former has a singular point. Only the normalization of $\overline{C}$ is. Your morphism $\overline{C}\rightarrow\mathbb{P}^1$ is not defined at $[0:1:0]$, which is the singular point. $\endgroup$
    – Roland
    Commented Sep 7, 2017 at 8:27
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    $\begingroup$ Ah, that's a good point. I guess I should say birational. Thanks! $\endgroup$ Commented Sep 7, 2017 at 9:55

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