If $A\rightarrow B$ be faithfully flat ring homomorphism, then $IB\cap A=I$ Let $A,B$, be rings, $I\subset A$ an ideal, and $A\rightarrow B$ be faithfully flat ring homomorphism. Show that $IB\cap A=I$.
By our assumptions, I know that for any $A$-module $M$, we have an injection $M\rightarrow M\otimes_A B$, given by $m\mapsto m\otimes 1$. I've read we could apply $M=A/I$ to this, so I guess we get an injection $A/I \rightarrow A/I\otimes_A B\cong  B/IB$ but I don't really see how this will do any good?
 A: First of all, as there was confusion in the comments, let us quickly show the claim that $A \rightarrow B$ is universally injective if $A \rightarrow B$ is faithfully flat (i.e. for any $A$-module $M$, the induced morphism $M \rightarrow B \otimes_A M$ is injective).
Take $M$ to be any $A$-module and consider $M \rightarrow M \otimes_A B$. As $B$ is faithfully flat over $A$, we can show injectivity after tensoring with $B$. That is, we have to show $M \otimes_A B \rightarrow M \otimes_A B \otimes_A B$, $m \otimes b \mapsto m \otimes 1 \otimes b$ is injective. But this morphism has a section $M \otimes_A B \otimes_A B \rightarrow M \otimes_A B$, $m \otimes b \otimes b' \mapsto m \otimes bb'$, and we're done with the claim.
Now the statement of the exercise. It is clear that $I \subseteq IB \cap A$. Further, let us write $\varphi \colon A \rightarrow B$ for the given faithfully flat morphism. We have by definition $IB \cap A = \varphi^{-1}(IB)$. Tensoring $\varphi$ with $A/IA$ we obtain an injection $\varphi \colon A/IA \rightarrow B/IB$, which translates to $\mathrm{ker}(A \rightarrow B \rightarrow B/IB) \subseteq I$, which in turn translates to $\varphi^{-1}(IB) \subseteq I$.
