Tensor product and localisation

Question

Let $k$ be an algebraically closed field and $K$ an extension field of $k$. Suppose $A$ is a finitely generated $k$-algebra which is a domain. Then we have a natural map $A \rightarrow A \otimes _ k K $. Now if $\mathfrak p$ is a prime ideal in $A \otimes _ k K $ that contracts to a prime ideal $\mathfrak q $ in $A$, then is it possible to prove that $ (A \otimes _ k K)_ \mathfrak p$ is isomorphic to $ A_ \mathfrak q \otimes _ k K$? If so can anyone let me know how to prove this?

Answer 1

A special case is $\mathfrak{p}=0$, hence $\mathfrak{q}=0$. Then the question is if $Q(A \otimes_k K) \cong Q(A) \otimes_k K$ holds. But this already fails for $A=k[x]$ and $K \neq k$ (this implies that $K/k$ is trancendental since $k$ is assumed to be algebraically closed), because here $Q(A) \otimes_k K$ is not a field:

Fact. If $k$ is any field and $K/k$ is a field extension, then the integral domain $k(x) \otimes_k K$ is a field iff $K/k$ is algebraic.

Proof. If $K/k$ is algebraic, then $k(x) \otimes_k K$ is an integral ring extension of $k(x)$, which is a field, and hence also a field. Conversely, assume that $k(x) \otimes_k K = (k[x] \setminus \{0\})^{-1} K[x]$ is a field and take some element $t \in K$. Then $\frac{1}{x-t} = \frac{p}{q}$ for some $p \in K[x]$ and $q \in k[x] \setminus \{0\}$, i.e. $q=p \cdot (x-t)$. It follows that $q(t)=0$ and hence $t$ is algebraic. $\square$