A map is injective if it is nonzero at the generic point?

Question

Proposition IV 2.1 in Hartshorne's states that if $f: X\to Y$ is a finite separable morphism of curves. Then $f^{*}\Omega_Y\to \Omega_X$ is injective.

And he proves this by saying that it will be sufficient to show the map is nonzero at the generic point, since both $f^{*}\Omega_Y$ and $\Omega_X$ are invertible sheaves on $X$.

I don't know why this proof can lead to the conclusion of injective. Could anyone explain? Thanks!

Answer 1

$\newcommand{\ms}[1]{\mathscr{#1}}$$\newcommand{\h}{\mathcal{O}}$Since you can check injectivity at stalks $x$ we may as well work over some affine open $\mathrm{Spec}(A)\subseteq Y$ with preimage $x\in \mathrm{Spec}(B)\subseteq X$. Note then that we are trying to show that

$$ (f^\ast \Omega^1_{Y/k})_x\to (\Omega^1_{X/k})_x$$

is injective. But, concretely if $x=\mathfrak{p}\in\mathrm{Spec}(B)$ and $f(x)=\mathfrak{q}\in \mathrm{Spec}(A)$ this is just trying to show that

$$M:=\Omega^1_{A_\mathfrak{q}/k}\otimes_{A_\mathfrak{q}}B_{\mathfrak{p}}\to \Omega^1_{B_\mathfrak{p}/k}=:N$$

a map of $B_\mathfrak{q}$-modules is injective. But, since $\Omega^1_{X/k}$ and $f^\ast \Omega^1_{Y/k}$ are line bundles, we know that $M$ and $N$ are just free $B_\mathfrak{p}$-modules of rank $1$. Note then that this implies that $M$ and $N$ are flat, so the maps $M\to M\otimes_{B_\mathfrak{p}} K$ and $N\to N\otimes_{B_\mathfrak{p}}K$ are injective. So, it suffices to show that $M\otimes_{B_\mathfrak{p}} K\to N\otimes_{B_\mathfrak{p}}K$ is injective which, since these are maps of one-dimensional vector spaces, is true iff the map is non-zero.

Writing it more geometrically, suppose that $f:\ms{E}\to \ms{E}'$ is a map of vector bundles on an integral scheme $X$ (assumed Noetherian for simplicity) with generic point $\eta$. For any point $x\in X$ we have by definition that

$$\ms{E}_x:=\varinjlim_{x\in U}\ms{E}(U)$$

and

$$\ms{E}_\eta:=\varinjlim_{U}\ms{E}(U)$$

Since $\{x\in U\}$ is a subsystem of just $\{U\}$ we get a natural map $\ms{E}_x\to \ms{E}_\eta$. We claim that this map is injective. But, this is clear since on any affine open $\mathrm{Spec}(A)$ of $x$ if $\ms{E}\mid_{\mathrm{Spec}(A)}\cong \widetilde{M}$ this is the map $M\otimes_A A_x\to M\otimes_A \mathrm{Frac}(A)$ which is injective since $A_x\to \mathrm{Frac}(A)$ is injective and $M$ is a flat $A$-module (since it's projective).

So we get the diagram

$$\begin{matrix} 0 & & 0\\ \downarrow & & \downarrow\\ \ms{E}_x & \xrightarrow{f_x} & \ms{E}'_x\\ \downarrow & & \downarrow\\ \ms{E}_\eta & \xrightarrow{f_\eta} & \ms{E}'_\eta\end{matrix}$$

Since $\ms{E}_x$ is faithully flat (in fact finite free) over $\mathcal{O}_{x,x}$ we see that $f_x$ is injective if and only if $f_\eta$ is. Since $x$ was arbitrary, we see that $f$ is injective if and only if $f_\eta$ is.

If, in addition, $\ms{E}$ and $\ms{E}'$ are line bundles so that $\ms{E}_\eta$ and $\ms{E}'_\eta$ are one-dimensional $k(\eta)$-spaces we know that $f_\eta$ is injective if and only if it's non-zero. Thus, $f$ is injective if and only if $f_\eta$ is non-zero.

NB: Just a word of warning, the obvious analogue of the above statement for surjectivity is patently false. For example, the map $\mathbb{Z}\to \mathbb{Z}$ given by $x\mapsto 2x$ is certainly generically surjective, but not surjective. Intuitively, the point is that the kernel of a map of vector bundles is torsion free, so can't be killed by passage to the generic stalk (i.e. can't be killed by tensoring with the fraction field) but the cokernel needn't be torsion free.