# $A \rightarrow B$ injective implies $f^{\#}$ an injective morphism of sheaves?

$\DeclareMathOperator{\Spec}{Spec}$ Had something I was thinking about and wanted to check if I was mistaken anywhere. Let $Y = \Spec A, X = \Spec B$ be affine schemes, with $\phi: A \rightarrow B$ a ring homomorphism with corresponding morphism $(f,f^{\#}): X \rightarrow Y$.

Suppose that $\phi$ is injective. Then for every prime $\mathfrak p$ of $A$, we get an injective ring homomorphism $A_{\mathfrak p} = A_{\mathfrak p} \otimes_A A \rightarrow A_{\mathfrak p} \otimes_A B = S^{-1}B$, where $S = A - \mathfrak p$.

Then $f^{\#}: \mathcal O_Y \rightarrow f_{\ast}\mathcal O_X$ is an injective morphism of sheaves, since if we check at each stalk, we have

$$(f_{\ast} \mathcal O_X)_\mathfrak p = \varinjlim_{V \ni \mathfrak p} \mathcal O_X(f^{-1}\mathfrak p) = \varinjlim_{x \in A - \mathfrak p} \mathcal O_X(D_B(x)) = \varinjlim_{x \in S} B_x = S^{-1}B$$

and $f^{\#}_{\mathfrak p}$ is just this ring homomorphism $A_{\mathfrak p} \rightarrow S^{-1}B$, which is injective.

So the injectivity of $\phi$ is equivalent to the injectivity of the morphism of sheaves $f^{\#}$?

• Yes ; see Exercise II.2.18 b) in Hartshorne. Commented Dec 12, 2017 at 19:56
• Oh..I did that exercise before and didn't remember it
– D_S
Commented Dec 12, 2017 at 19:58

Basic commutative algebra: if $M$ and $N$ are $A$-modules and $f:M\to N$ is an $A$-module homomorphism, then $f$ is injective if and only if $f_{\mathfrak p}:M_{\mathfrak p}\to N_{\mathfrak p}$ is injective for all primes $\mathfrak p\subset A$.
Going to our case, the ring morphism $f:A\to B$ is also a homomorphism of $A$-modules when we consider $B$ as an $A$-module via $f$. Since the "localization" at $\mathfrak p\subseteq A$, i.e. $A_{\mathfrak p}\to B_{\mathfrak p}$, is the same as the stalk at $\mathfrak p$ of the sheaf morphsim $\mathcal O_Y\to f_*\mathcal O_X$, and a morphism of sheaves is injective if and only if its injective at all stalks, the result follows immediately from our above cited fact.