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Suppose that $f_{4}(x)$ is a polynomial of degree 4 with no multiple roots, $C$ is the curve defined by $y^{2}=f_{4}(x)$, I want to show that there is a polynomial $f_{3}(x)$ of degree 3 with no multiple root such that $C$ is birational equivalent to the curve defined by $y^{2}=f_{3}(x)$, I completely don't know how to do it, can anyone help me

YuerWu
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1 Answers1

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As Alexandre pointed it out, just use the linear-fractional transformation to send one point to infinity. Explicitly, denote the four roots of $f_4$ to be $a,b,c,d$, then there is a unique linear-fractional transformation

$$\sigma:\mathbb P^1\to \mathbb P^1,$$ sending $a,b,c$ to $0,1,\infty$. Denote $\lambda$ to be $\sigma(d)$. Let $C'$ be the curve defined by $y^2=x(x-1)(x-\lambda)$. Then its direct to show that $C$ and $C'$ are birational. Indeed, $C$ with $\sigma^{-1}(\infty)$ removed is isomorphic to $C'$ with $\sigma(\infty)$ removed.

In general, a hyper-elliptic curve of genus $g$ can be represented by the equation $y^2=f(x)$ for some polynomial $f(x)$ with all roots being simple and the degree of $f(x)$ is either $2g+1$ or $2g+2$, depending on whether the infinity point is ramified when considered as a double cover to the $x$-line.

AG learner
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  • Can you do the same with $y^2=f_n(x)$ for $n>4$? You seem to be using $n=4$ here. – markvs Dec 25 '21 at 03:25
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    @markvs With roots $a_1,\dots, a_n$ you would send $a_1$ to $0$, $a_2$ to $1$, $a_3$ to $\infty$, and then $a_4,\dots, a_n$ to $\lambda_1,\dots, \lambda_{n-3}$ or something like that. – Will Sawin Dec 25 '21 at 03:27
  • You can't do it all by one linear fractional transformation. Also what are $\lambda_i$? – markvs Dec 25 '21 at 03:29
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    @markvs The automorphism is only determined by three points, all other parameters are free. $\lambda_i=\sigma(a_{n-3})$ in Will's notation. – AG learner Dec 25 '21 at 03:59
  • But then you won't get a cubic equation? You have to repeat the procedure? I also wonder if it is all described in Milne. – markvs Dec 25 '21 at 04:02
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    @markvs Here is a more precise statement: If we denote $C$ to be the affine curve $y^2=(x-a_1)\cdots(x-a_{2g-2})$, there is $\sigma$ a linear-fractional transformation sending $a_1,a_2,a_3$ to $0,1,\infty$. Let $\lambda_n=\sigma(a_{n-3})$, $4\le n\le 2g-2$. Denote $C'$ to be the curve $y^2=x(x-1)(x-\lambda_1)\cdots(x-a_{2g-5})$. Then $C-{\sigma^{-1}(\infty)}$ is isomorphic to $C'-{\sigma(\infty)}$. I'm not sure if there is a reference for this. – AG learner Dec 25 '21 at 04:20
  • @AGlearner: I am sure I have seen it already. Not in Milne?? – markvs Dec 25 '21 at 04:27
  • @markvs Griffiths has a book Introduction to algebraic curves. On page 137-143, there are some related discussion on hyper-elliptic curves. (There are typos on indexes in the comments above, perhaps you already found them.) – AG learner Dec 25 '21 at 04:47
  • I might have read Griffiths. – markvs Dec 25 '21 at 04:52