# The degree of a map onto a rational normal curve

Source: Miranda's Exercise J Page 167

If $v^{2}=h(u)$ defines a hyperelliptic curve of genus $g$, then $\phi=[1:u:u^{2}:\dots,u^{g-1}]$ defines a degree $2$ map onto a rational normal curve of degree $g-1$ in $\mathbb{P}^{g-1}$.

I can't understand why the degree must be $2$. Why is this?

I think that the image of $X$ is the smooth projective curve local complete intersection $Y$ with equations $x_i x_j -x_{i-1}x_{j-1}=0$, so it's a rational normal curve of degree $g-1$. Now, I know that the degree of the pull-back of a hyperplane divisor on $Y$ equals the product $\deg (\phi) \deg(Y)$.

• I think that the image of $X$ is the smooth projective curve local complete intersection $Y$ with equations $x_i x_j -x_{i-1}x_{j-1}=0$, so it's a rational normal curve of degree $g-1$. Now, I know that the degree of the pull-back of a hyperplane divisor on $Y$ equals the product $deg (\phi) deg(Y)$. – TheWanderer Oct 24 '13 at 15:49