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In the paper "Theta characteristics of an algebraic curve", Mumford has considered the following example: Take an unramified double cover $q:X\rightarrow Y$ and let $L\in Prym$ then $q_*L$ has a "canonical non-degenerate" quadratic form given by the Norm map.

My question is quite easy: Is it necessary to assume that the cover is unramified?

I believe that this still true in the ramified case, unless i have missed something!

Thanks

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For any double covering $q:X \to Y$ and any line bundle $L$ on $X$ there is an embedding $i:X \to \mathbb{P}_Y(E)$, where $$ E = (q_*L)^\vee, $$ such that $i^*O(1) \cong L$ (it is defined by the canonical surjection $q^*E^\vee = q^*q_*L \to L$). The class of $X$ in $\mathbb{P}_Y(E)$ has degree 2 over $Y$, hence provides a global section of the line bundle $p^*M \otimes O(2)$ on $\mathbb{P}_Y(E)$, where $p:\mathbb{P}_Y(E) \to Y$ is the projection and $M$ is a line bundle on $Y$. Since $$ p_*(p^*M \otimes O(2)) \cong M \otimes S^2E^\vee, $$ the equation of $X$ gives a quadratic form on $E$ with values in $M$.

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    Thanks for the nice answer. Just one remark, I think this is true for just one line bundle $M$ ? which should be the Norm of $L$ right? – 2017-01-25
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    The line bundle $M$ is determined by $X$ and $L$. In fact, $M \cong \omega_{X/Y} \otimes \det(E)$. – 2017-01-25
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    I forget to add the condition to be non degenerate, Is this form is non-degenerate? Maybe this is why one should add the unramified condition – 2017-01-25
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    Its degeneration divisor is equal to the branch divisor of the morphism. – 2017-01-25