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I'm trying to show that the curve $f(z,w)=z^4-w^2+1$ has genus 1. The curve is clearly non-singular, so I tried using the degree formula \begin{equation*} g=\frac{(d-1)(d-2)}{2}=\frac{3\cdot2}{2}=3, \end{equation*} but it should be equal to one. I'm not sure where my error is.

Thanks!

mj_indefinite
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  • Why do you say it's "clearly nonsingular"? First of all, you don't have an equation yet. Is the curve $f(z,w)=0$? Next, to apply the degree formula, you need the entire curve in the projective plane, not just the part in the affine plane. This curve will in fact have a singular point at infinity. – Ted Shifrin Apr 01 '17 at 22:04
  • so the projective curve is $F(Z,W,X) = Z^4-W^2 X^2+X^4= 0$ and the singular point is $(0,1,0)$ ? @TedShifrin – reuns Apr 01 '17 at 22:09
  • Yes, the curve is $f(z,w)=0$. Since the only way for $f_z=0$ and $f_w=0$ is for $z=w=0$ and this doesn't hold for $f=0$, the equation is non-singular, right? How do I account for the point at infinity? – mj_indefinite Apr 01 '17 at 22:10
  • @user1952009: Yes. And, indeed, we see that, setting $W=1$, we have the curve $z^4-x^2+x^4=0$, and this curve has a tacnode at $x=z=0$. – Ted Shifrin Apr 01 '17 at 22:13
  • I see. How can I show that the curve has genus 1 without using the degree formula? – mj_indefinite Apr 01 '17 at 22:21
  • We have absolutely no idea of what you know and what your context is, @mj_indefinite. I suspect you need to ask the professor who assigned you the exercise. – Ted Shifrin Apr 01 '17 at 22:38

1 Answers1

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There are a couple of ways of doing this. First, review what the genus of a curve is. This and this question+answers might help. In particular, in the second link, look at the formula of how to adjust the formula $g=(d-1)(d-2)/2$ when there are singularities.

In this particular case, however, I would suggest you find a birational map from $C: w^2=z^4+1$ to a curve of (clearly) genus $1$. For this, look at Theorem 6, page 17, in this paper of mine.

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Álvaro Lozano-Robledo
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