Faithfully flat ring homomorphism properties This is from Liu's Algebraic Geometry and Arithmetic Curves exercise 1.2.19 a.
Let $f:A\to B$ be a faithfully flat ring homomorphism. How can I show that $f$ is injective and that $I\to I\otimes_AB$ is injective for every ideal $I$ of $A$.
 A: Consider the exact sequence $0\to \ker f\to A \to A/\ker f\to 0$ and tensor it with $B$ (over $A$): we get the commutative diagram with exact rows
$$\require{AMScd}
\begin{CD}
0 @>>> \ker f\otimes B @>>> A\otimes B @>>> A/\ker f\otimes B @>>> 0 \\
@. @VVV @VVV @VVV @. \\
0 @>>> (\ker f)B @>>> B @>>> B/(\ker f)B @>>> 0
\end{CD}
$$
of $A$-modules.
The rightmost vertical arrow is an isomorphism, and also $A\otimes B\to B$ is an isomorphism. Since $(\ker f)B=0$ by definition, we obtain $\ker f\otimes B=0$ and, since $B$ is faithfully flat, $\ker f=0$.
If $I$ is an ideal of $A$, we have, using the obvious morphism of $A$-modules $M\to M\otimes B$ given by $x\mapsto x\otimes 1$, the commutative diagram with exact rows
$$
\begin{CD}
0 @>>> I @>>> A @>>> A/I @>>> 0\\
@. @VVV @VVV @VVV @. \\
0 @>>> I\otimes B @>>> A\otimes B @>>> (A/I)\otimes B @>>> 0
\end{CD}
$$
The central vertical arrow is injective (why?) so also the leftmost vertical arrow is injective.
A: fact: let $f: R\to S$ be a ring homomorphism and $S$ be a faithfully flat $R$-algebra. If $M$ is an $R$-module, then the map $\bar{f}:M\to M‎\otimes S$ defined by $\bar{f} (x)= x ‎\otimes 1$ is a monomorphism. In particular $f$ is a monomorphism.  
proof: Suppose $0\neq x\in M $ then $0\neq Rx‎\otimes S \subset M‎\otimes S$ since $S$ is 
faithfully flat. So $Rx‎\otimes S =(x ‎\otimes 1)S\neq 0$ and thus $x ‎\otimes 1\neq 0$
(all tensors are on R)
