# Finite morphism of varieties - morphism of sheaves

Let $f:X\rightarrow Y$ be a finite morphism of non-singular projective varieties of degree $d$.

• Consider the map of sheaves $O_Y\rightarrow f_*O_X$. Is this morphism injective? Why? $f_*O_X$ is a rank $d$ vector bundle.
• Suppose the above morphism of sheaves is injective. Consider the exact sequence where $L$ is the cokernel: $$0\rightarrow O_Y\rightarrow f_*O_X\rightarrow L\rightarrow 0.$$ Does this exact sequence split and give us $f_*O_X=O_Y\oplus L$?
• Suppose $f$ is of degree 2 and $X$ and $Y$ are non-singular surfaces. Let $C\subset Y$ be a non-singular curve along which $f$ is branched. Then is there is some relation between $L$ (as above ) and $O_Y(C)$?

Yes, the map is injective assuming the map is dominanat. Since $$1\in\mathcal{O}_Y$$ goes to $$1\in f_*\mathcal{O}_X$$, you should be able to check this.

The exact sequence above may not split in positive characteristic, but it does in zero characteristic (or characteristic not dividing $$d$$), using the trace map.

For the last part (char $$\neq 2$$), one has $$f_*\mathcal{O}_X=\mathcal{O}_Y\oplus L$$ and checking the algebra structure on $$f_*\mathcal{O}_X$$, one can show that $$L^2=\mathcal{O}_Y(-E)$$ where $$E$$ is the branch locus.

• for the last part, it means that the line bundle $O_Y(-E)$ admits a square root? If we start with a line bundle $L'$ the Picard group which does not admit a root , can we not construct a double cover branched along some $D$ where $O_Y(D)=L'$? Commented Jan 22, 2017 at 4:24
• also if $E\subset Y$ is the branch locus of the double cover, then $f^*E$ is a non-reduced divisor of the form $2E'$. Then $E'$ is isomorphic to $E$. Then it looks like $\pi^*L^{-1}$ is $O_Y(-E')$? Is this true? Commented Jan 22, 2017 at 9:04
• The map is not obviously injective. In fact it's obviously not injective, considering any nontrivial closed immersion gives a counterexample...
– Eoin
Commented Jan 25, 2020 at 8:45

For the first part of your question, a morphism of schemes $f : X \to Y$ such that $\mathcal{O}_Y \to f_*\mathcal{O}_X$ is injective in said to be schematically dominant. If $f$ is dominant and $Y$ is reduced then $f$ is schematically dominant (EGA IV3 prop 11.10.4).