Let $X, Y$ be smooth curves over a field of characterstic other than 2. Also, let $f: X \to Y$ be a finite etale morphism of of degree 2. It is claimed in Hartshorne exercise IV.2.7(a) that the cokernel of the natural map $\mathcal{O}_Y \to f_* \mathcal{O}_X$ is an invertible sheaf.
I am having trouble seeing why this is true. On one hand, $f_*\mathcal{O}_X$ is a locally free coherent sheaf of rank $2$, and so the natural map $f^+: \mathcal{O}_Y \to f_*\mathcal{O}_X$ is locally a smooth map $A \to A \oplus A$ where $A$ is a 1-dimensional integral domain. Perhaps there's some way to use flatness or smoothness to conclude that the quotient is a rank 1 projective module?
I am very stuck on this point. The only idea was that this map picks out some $(a, a') \in A \oplus A$ so we can look at the quotient concretely, but it seems that unless these are units, quotients of this form aren't usually projective. Is there some way to see that these are units in this special case? Is $f^+$ perhaps not the natural map Hartshorne was intending?
Thanks!
Best Answer
Since $f$ is finite flat,$\def\cO{\mathcal{O}} \def\cL{\mathcal{L}}$ $f_*\cO_X$ is locally free of rank equal to $\deg f=2$, so $(f_*\cO_X)_y$ is a free $\cO_{Y,y}$-module of rank two for any $y\in Y$. Letting $\cL$ be the cokernel of $\cO_Y\to f_*\cO_X$, we may reduce the exact sequence $0\to \cO_{Y,y} \to (f_*\cO_X)_y\to \cL_y\to 0$ modulo $\mathfrak{m}_y$ to get $k\to k^2 \to \cL_y/\mathfrak{m}_y\cL_y\to 0$. Since the map $k\to k^2$ may be represented as the pullback of the constant functions on $y$ to the constant functions on $f^{-1}(y)=\{p,q\}$ for two distinct points (this is where we use that $f$ is unramified), the map is injective. So $\cL$ is of rank one at every closed point and therefore is a line bundle by exercise II.5.8(c).