How to use Nakayama's lemma here?

Question

Let $X$ and $Y$ be algebraic varieties, with $X$ complete, and let $\pi: X\times Y \to Y$ be the projection onto $Y$. Let $L$ be a line bundle over $X\times Y$ and consider the natural map $$ \alpha: \pi^* \pi_* L \to L. $$

Suppose that over every fiber of $\pi$ the map $$ \alpha_y: (\pi^* \pi_* L)_y \to L_y $$ is an isomorphism. Then an application of Nakayama's lemma shows that $\alpha$ is surjective.

Could you please make it explicit (with as much details as you can) how Nakayama's lemma is used in the above argument?

Answer 1

The cokernel $F$ of $\alpha$ is an $\mathcal{O}_{X \times Y}$-module of finite type whose fibers vanish. Now Nakayama's Lemma exactly states that such a module has to vanish.

Answer 2

Thanks to Martin I now understand the matter. I will write here a more complete/naive/low-level answer, hoping this will be helpful to future non-expert readers. In the following $F_y$ will denote the stalk at $y$, while $F_{\mid y} = F_y/m_y F_y$ will be the fiber at $y$.

We have the exact sequence of $\mathcal{O}_{X\times Y}$- modules $$ L' \overset{\alpha}\to L \to F \to 0 $$ and taking fibers at $y\in Y$ we obtain the exact sequence $$ L'_{\mid y} \overset{\alpha_y}\to L_{\mid y} \to F_{\mid y} \to 0 $$ of modules over the local ring $(\;\mathcal{O}_{X\times Y,\, y}\,,\; m\;)$, where $F_{\mid y}$ is actually $0$ since $\alpha_y$ is an isomorphism.

Now, a version of Nakayama's lemma is the following:

Let (A,m) be a local ring and $M$ a finitely generated A-module. If (the image of) $f_1, \dots, f_n$ generates $M/m M$, then $f_1, \dots, f_n$ generate $M$.

In our case the module $F_{\mid y} = F_y/mF_y$ over the local $(\;\mathcal{O}_{X\times Y,\, y}\,,\; m\;)$ is generated by $0$. Therefore the lemma implies that the for every $y\in Y$ the stalks $F_y$ are generated by $0$, so they are trivial.

The conclusion follows from the fact that a sheaf is the zero sheaf if and only if all of its stalks are trivial.