Do we have $B\otimes_AB_{\mathfrak q}\cong B_{\mathfrak q}$?

Question

Let $f: A\to B$ be a homomorphism of commutative rings with unity, let $\mathfrak q$ be a prime ideal of $B$, do we have $B\otimes_AB_{\mathfrak q}\cong B_{\mathfrak q}$ under the canonical homomorphism $b_1\otimes \frac{b_2}{t}\mapsto \frac{b_1b_2}{t}$?

Answer 1

No, you do not have such an isomorphism. Let $B=A[x]$ with $A$ an integral domain, and $\mathfrak q=(x+1)$, say. Then $$ x\otimes_A \frac1{x} $$ is not the same element as $1\otimes_A 1$. No amount of tensor property manipulation will give you an expression with only constant terms on the left-hand side of the products.

You do have such an isomorphism when $B$ is, for instance, a localisation of $A$. Then we have $$ B\otimes_AB_{\mathfrak q}\cong B\otimes_BB_{\mathfrak q}\cong B_{\mathfrak q} $$ where the first isomorphism comes from the fact that moving a factor from $A$ to one side corresponds to moving its inverse to the other side.