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I saw this, but I don't understand why we can't use the genus-degree formula for this curve. I think this curve is $V(X^4 + Z^4 - Y^2Z^2)$ in $\mathbb{P}^2_k$, so by the genus degree formula, the genus is $3$.

But the answers of this question say this curve is in actually a cubic curve in a projective space. If so, this curve has the genus $1$.

And one answer says this curve has two points at infinity. But I think this has only $[0:1:0]$ for a point at infinity.

What happen?

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First, let's see why the degree-genus formula fails. Remember that the degree-genus formula tells you the arithmetic genus (i.e., $h^1(X,\mathcal{O}_X)$), but if the curve is singular, this may not be equal to the geometric genus $h^0(X,\omega_{X/k}).$

Considered as a projective plane curve in the usual way (via homogenization), your curve $V(X^4 + Z^4 - Y^2Z^2)$ is singular at infinity. Let's use the Jacobian criterion: we see that \begin{align*} \frac{\partial F}{\partial X} &= 4X^3\\ \frac{\partial F}{\partial Y} &= -2YZ^2\\ \frac{\partial F}{\partial Z} &= 4Z^3 - 2Y^2Z. \end{align*} At $[0:1:0],$ all of these partials vanish, so that the curve is not smooth at $\infty,$ and hence the degree-genus formula will not necessarily compute the geometric genus of $V(X^4 + Z^4 - Y^2Z^2)$ without modification (see in particular Hartshorne exercise IV.1.8).

Now, let's compute the (geometric) genus, considering instead of the model $V(X^4 + Z^4 - Y^2Z^2),$ the corresponding nonsingular projective curve $X.$ There's a double cover $\pi : X\to\Bbb P^1_k$ of $\Bbb P^1_k$ by this curve given by $[x: y: z]\mapsto [x : z],$ so we can use the Riemann-Hurwitz formula. This tells us \begin{align*} 2h^0(X,\omega_X) - 2 &= 2(2h^0(\Bbb P^1_k,\mathcal{O}(-2)) - 2) + \deg R\\ \implies 2h^0(X,\omega_X) - 2 &= -4 + \deg R\\ \implies h^0(X,\omega_X) &= -1 + \frac{1}{2}\deg R, \end{align*} where $R = \sum_{P\in X} (e_P - 1)P$ is the ramification divisor of $X.$ Because the morphism to $\Bbb P^1$ is a double cover, the ramification index $e_P$ of any point is at most $2.$ In particular, if $P$ is ramified, then $e_P = 2,$ and otherwise $e_P = 1.$ The ramified points on $X$ are precisely the points $[x : y : z]$ such that $y = 0$ (these are the points with only one preimage under our map $\pi$). There are four of these (the four roots of $x^4 + 1$), so we have $$ h^0(X,\omega_X) = -1 + \frac{4}{2} = 2 - 1 = 1. $$

In computing the geometric genus, I assumed that $X$ was smooth. When one refers to the curve $y^2 = x^4 + 1$ (or any hyperelliptic curve given by an equation of the form $y^2 = f(x)$), one often implicitly means the nonsingular projective curve corresponding to this affine curve. Again, as we've shown, this is not simply $V(X^4 + Z^4 - Y^2Z^2)$: this plane curve is singular. Using weighted projective space, or by gluing affine models, you can obtain the desired nonsingular curve, see here for example.

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  • $\begingroup$ Your very nice answer gave rise to this question : math.stackexchange.com/questions/2964300/…. You may have a look at it :-) $\endgroup$ – Alphonse Oct 21 '18 at 12:20
  • $\begingroup$ Namely, could you please describe explicitly what the corresponding nonsingular projective curve $X$ is? For instance, what equations describe $X$ as projective variety? $\endgroup$ – Alphonse Oct 21 '18 at 13:56
  • $\begingroup$ According to MAGMA, the code $$ \text{ P2<X,Y,Z> := ProjectiveSpace(Rationals(),2); \\ C := Curve(P2, Y^2*Z^2 - X^4 - Z^4); \\ ArithmeticGenus(C); \\ GeometricGenus(C); }$$ gives arithmetic genus 3 and geometric genus 1 for the singular projective curve $X_0 = V(X^4+Z^4-Y^2Z^2)$. $\endgroup$ – Alphonse Oct 21 '18 at 18:05
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That genus-degree formula you mentioned is for non-singular curves. That one point at infinity, there are two "places" above it ("places" are points in the desingularized model, which can for instance be obtained by the blow-up process).

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  • $\begingroup$ Thank you very much! I've misunderstand whether the curve is smooth. So, does the word "this curve is an elliptic curve" means that the desingularization of the curve becomes an elliptic curve? $\endgroup$ – k.j. Sep 10 '18 at 15:53

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