Prime ideals in tensor products of algebras and their pullbacks Suppose $\mathfrak{p}$ is a prime ideal in $B\otimes_CA$, and $\mathfrak{p}_A,\mathfrak{p}_B,\mathfrak{p}_C$ are its pullbacks in $A,B,C$.

Does it hold: $(B\otimes_CA)_{\mathfrak{p}}\cong ({B}_{\mathfrak{p}_B} \otimes_{C_{\mathfrak{p}_C}}{} A_{\mathfrak{p}_A})$?

To construct a map from left to right, I have to show each element not in $\mathfrak{p}$ is invertible in the right, how to see this? 
 A: No, it doesn't! Take $C=k$, $A=k[x]$ and $B=k[y]$.
You have $A\otimes_C B=k[x,y]$, now take $p=(x+y)$: all the pullbacks are $(0)$, and you have $A_{p_A}\otimes_{C_{p_C}}B_{p_B}=k(x)\otimes_k k(y)$. Here, $(x+y)$ and $(x-y)$ are both maximal ideals (because the quotient is a field), so this is not a local ring and can't be isomorphic to $(B\otimes_CA)_p=k[x,y]_{(x+y)}$.
Anyway, the universal property of localization by $p$ ensures that you have maps $A_{p_A},B_{p_B},C_{p_C}\to (A\otimes_CB)_p$. These maps are $C_{p_C}$-linear, hence they give you $\phi: A_{p_A}\otimes_{C_{p_C}}B_{p_B}\to (A\otimes_CB)_p$ thanks to the universal property of tensor product.
A: What is true is that for any $A$-modules $M$ and $N$, $(M\otimes_AN)_\mathfrak{p}\simeq M_\mathfrak{p}\otimes_{A_\mathfrak{p}}N_\mathfrak{p}$ as $A_\mathfrak{p}$-modules. More precisely we have canonical isomorphisms
$$(M\otimes_AN)_\mathfrak{p}
=(M\otimes_AN)\otimes_AA_\mathfrak{p}\simeq
M\otimes_AN_\mathfrak{p}\simeq M_\mathfrak{p}\otimes_{A_\mathfrak{p}}N_\mathfrak{p}\text{.}$$
The first isomorphism is transitivity of tensor product, and the second is a standard isomorphism, given explicitly by sending $m\otimes y$ to $(m\otimes 1)\otimes y$. 
