How to prove $(B \backslash P)^{-1} (A \otimes_B B)$ is isomorphic to $A \otimes_B B_P$

Question

I am trying to prove $(B \backslash P)^{-1} (A \otimes_B B)$ is isomorphic to $A \otimes_B B_P$ for a commutative ring with unity $B$, $A$ a $B$-algebra and a prime ideal $P$ of $B$. I have been trying to prove this by the universal property of the localisation map and the tensor products.. maybe there is a simpler way? Thank you.

Edit: I think one way of proving this is to define a map $$ (B \backslash P)^{-1} (A \otimes_B B) \to A \otimes_B B_P $$ by sending $\frac{a \otimes b}{x}$ to $a \otimes \frac{b}{x}$, and extending it by linearity. We can also define a map
$$ A \otimes_B B_P \to (B \backslash P)^{-1} (A \otimes_B B) $$ by sending $a \otimes \frac{b}{x}$ to $\frac{a \otimes b}{x}$, and extending it by linearity. We can see that these maps are inverses to each other so they are isomorphic. However, how can I show these maps are well defined? Thank you!

Answer 1

Let \begin{align} &\lambda:B\to B_{\mathfrak P}& &b\mapsto b/1 \end{align} denote the canonical ring homomorphism. Then we get a $B$-algebra homomorphism $\varphi=1_A\otimes_B\lambda:A\otimes_BB\to A\otimes_BB_{\mathfrak P}$. Since for every $s\in B\setminus\mathfrak P$, the element $\varphi(1\otimes s)=1\otimes(s/1)$ is invertible in $A\otimes_BB_{\mathfrak P}$, with inverse $1\otimes (1/s)$, by universal property of ring localization, $\varphi$ extends uniquely to a $B_{\mathfrak P}$-algebra homomorphism \begin{align} &\bar\varphi:(B\setminus\mathfrak P)^{-1}(A\otimes_BB)\to A\otimes_BB_{\mathfrak P}& &(a\otimes b)/s\mapsto a\otimes(b/s) \end{align}

Conversely, we have $B$-algebra homomorphisms \begin{align} &\alpha:A\to(B\setminus\mathfrak P)^{-1}(A\otimes_BB)& &a\mapsto(a\otimes 1)/1\\ &\beta:B\to(B\setminus\mathfrak P)^{-1}(A\otimes_BB)& &b\mapsto(1\otimes b)/1 \end{align} Since for every $s\in B\setminus\mathfrak P$ the element $\beta(s)=(1\otimes s)/1$ is invertible in $A\otimes_BB_{\mathfrak P}$ with inverse $(1\otimes 1)/s$, by universal property of ring localization, $\beta$ extends uniquely to a $B$-algebra homomorphism $\bar\beta:B_{\mathfrak P}\to(B\setminus\mathfrak P)^{-1}(A\otimes_BB)$. By universal property of tensor product, we get a $B$-algebra homomorphism \begin{align} &A\otimes_BB_{\mathfrak P}\to(B\setminus\mathfrak P)^{-1}(A\otimes_BB)& &a\otimes(b/s)\mapsto(a\otimes b)/s \end{align}