Morphism of locally free sheaves over integral scheme I'm trying to prove the following: let $X$ be an integral scheme with generic point $x$. Then a morphism of locally free sheaves
$$
\varphi: F \to G
$$
on $X$ is a monomorphism iff the map of stalks $\varphi_x : F_x \to G_x$ is injective. 
Any help on this would be great, I feel like what I should be doing is looking at some open set $U$ and noting that since $x$ is generic then $x$ lies in $U$, so if 
$$
\varphi(U) : F(U) \to G(U)
$$
has non-zero kernel then we get a germ in $F_x$ which maps to zero, but I can't quite see how to do this. Thanks for any help!
 A: Let $K$ be the kernel. We know $K_x=0$, where $x$ is the generic point and we have to show $K_y=0$ for any $y \in X$. Pick any affine open $U=\operatorname{Spec} A$  with $y \in U$. We have $x \in U$, since $x$ is contained in any open set. Thus we have reduced to the affine case and we only have to show the following:
Let $A$ be an integral domain with fraction field $F$ and $M$ a submodule of a free $A$-module. If $M \otimes_A F=0$, then $M=0$. This is a very easy statement in commutative algebra.
A: The statement you ask about is equivalent to: $$Ker(\phi)=0\iff (\operatorname {Ker}\phi)_x=0$$ Since $Ker(\phi)$ is a sheaf, its being zero can be checked on any covering of $X$ by open subsets so that, recalling that any open subset of $X$ contains $x$, we may assume that $X$ is affine of the form $X= \operatorname {Spec}(R)$, where $R$ is a domain.
We may also assume that $F,G$ are free on $X$ so that $\phi$ corresponds to an $R-$linear map $\Phi: R^n\to R^m$.
But then the assertion is clear: the map $\Phi$ is given by a $m\times n$ matrix $(r_{i,j})$ and the same matrix also describes $\Phi_x=\operatorname {Frac}\Phi: K^n\to K^m \; (K=\operatorname {Frac}R)$, so that indeed: $$Ker(\Phi)=0\iff (\operatorname {Ker}\Phi_x)=0$$ 
