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



Show that a polynomial of degree 4 is birational equivalent to a polynomial of degree 3
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.


