Let \(X\) be an integral scheme. Prove that every nonzero morphism of invertible sheaves on \(X\) is injective, and that every generically injective morphism of locally free sheaves on \(X\) is injective.
Hint
First prove that a locally free sheaf on \(X\) has no nonzero torsion subsheaf, where a torsion sheaf is one whose support has positive codimension.
Solution
Let \(\eta\) be the generic point of the integral scheme \(X\) and put \[ K=K(X)=\OO_{X,\eta}. \]
1. If \(\mce\) is locally free and \(\mct\subseteq\mce\) satisfies\[ \mct_\eta=0, \]then \(\mct=0\).
It suffices to work on an affine open \[ U=\Spec A \] on which \[ \ro{\mce}{U}\cong\widetilde{A^r}. \] Because \(X\) is integral, \(A\) is a domain and its fraction field is \(K\).
Let \[ s\in\mct(U)\subseteq A^r. \] The condition \(\mct_\eta=0\) says that the image of \(s\) in \[ A^r\otimes_AK\cong K^r \] is zero. By the definition of localization, there is a nonzero \[ a\in A \] such that \[ as=0 \] in \(A^r\). Since \(A\) is a domain, the free module \(A^r\) is torsion-free, so \(s=0\).
Thus \(\mct(U)=0\) on every such trivializing affine open. These opens cover \(X\), hence \(\mct=0\).
2. Every generically injective morphism of locally free sheaves\[ \varphi:\mce\longrightarrow\mcf \]is injective.
Let \[ \mck=\ker\varphi. \] Generic injectivity means precisely that \[ \varphi_\eta:\mce_\eta\longrightarrow\mcf_\eta \] is injective, so \[ \mck_\eta=0. \] Since \(\mck\subseteq\mce\) and \(\mce\) is locally free, step 1 gives \[ \mck=0. \] Therefore \(\varphi\) is injective.
3. A nonzero morphism of invertible sheaves is nonzero at the generic point.
Let \[ \varphi:\mcl\longrightarrow\mcm \] be a morphism of invertible sheaves. Such a morphism is equivalently a global section \[ s_\varphi\in H^0\bigl(X,\mcm\tensor\dualof{\mcl}\bigr), \] where \(\mcm\tensor\dualof{\mcl}\) is again invertible.
Suppose \(\varphi_\eta=0\). Then the corresponding section has zero germ at \(\eta\). The image subsheaf generated by \(s_\varphi\) is a subsheaf of the locally free sheaf \(\mcm\tensor\dualof{\mcl}\) with zero generic stalk, so step 1 implies that this image subsheaf is zero. Hence \[ s_\varphi=0 \] and therefore \(\varphi=0\).
Thus a nonzero \(\varphi\) must satisfy \[ \varphi_\eta\ne0. \]
4. Every nonzero morphism of invertible sheaves on \(X\) is injective.
At the generic point, \(\mcl_\eta\) and \(\mcm_\eta\) are one-dimensional \(K\)-vector spaces. By step 3, a nonzero morphism \[ \varphi:\mcl\longrightarrow\mcm \] induces a nonzero linear map \[ \varphi_\eta:\mcl_\eta\longrightarrow\mcm_\eta. \] A nonzero map between one-dimensional vector spaces is injective. Therefore \(\varphi\) is generically injective, and step 2 gives \[ \boxed{\varphi\text{ is injective}.} \]
5. Q.E.D.
Step 1 shows that a subsheaf of a locally free sheaf with zero generic stalk is zero. Step 2 proves the locally free assertion, and steps <1>3–<1>4 prove the invertible-sheaf assertion.