How to show the $B_p \to A \otimes B_p$ is injective

Question

Let $A$ and $B$ be commutative rings with unity and suppose we have $\phi: B \to A$. Take a prime ideal $p$ of $B$ and localize $B$ at $p$, $B_p$. Then we can consider $A \otimes B_{p}$. I would like to show that $B_p \to A \otimes B_p$, defined by $b/x \to 1 \otimes b/x$, is injective... Any explanation is appreciated. Thank you.

Edit. The tensor product is over $B$.

Answer 1

It seems that the user is assuming $A\otimes B_p$ to be taken over $B$.

In general, given $B$-algebras $A,C$, then $C$ injects into $A \otimes_B C$ if

1) the morphism $B\longrightarrow A$ is injective.

2) $C$ is flat over $B$.

Now, localisation $B_p$ is flat over $B$: localisation is an exact functor so the functor $B_p\otimes_B-$ is exact as well.

The other assumption is necessary: consider $A = \mathbb Z/p, B = \mathbb Z$, then $B_p\otimes A = \mathbb Z_{(p)} \otimes (\mathbb Z /p) \simeq \mathbb{Z}_{(p)}/p\mathbb{Z}_{(p)}\neq 0$ but the map $Z_{(p)}\longrightarrow \mathbb{Z}_{(p)} \otimes (\mathbb{Z}/p)$ cannot be injective (it sends elements of the maximal ideal to zero, for example).

Answer 2

If the tensor product is over $\mathbb{ Z}$, then the tensor product $A \otimes_{B} B_p$ of $A$ and $B_p$ is considered as $B$-module.

Then $b/x \to 1 \otimes_{B} b/x$ is a group homomorphism whose Kernel is the torsion of $B_p$.

So it is injective iff $B_p$ is torsion free.