Fraction field commutes with extension of scalars? Let $K$ be a field, $A$ a $K$-algebra, and $L$ an extension of $K$. Denote $B:= A \otimes_K L$, the extension of scalars of $A$, and assume that $B$ is an integral domain (and therefore $A$ too). Under what conditions do we have
$$B_{(0)}\cong A_{(0)} \otimes_K L,$$
where the index ${(0)}$ denotes taking the fraction field? Or, more in general, what is the relation between $B_{(0)}$ and $A_{(0)} \otimes_K L$? It seems to be quite a specific situation that relies on $L$ being a field and on $B$ being an integral domain, but at the same time it looks very natural to consider this question.
 A: Assuming as KReiser points out in the comments that $A, K, L$ have been chosen so that $B$ remains integral, which is not automatic, this is still very rarely true. A simple example is $A = K[x], B = L[x]$, which is about as nice as it gets, with fraction fields $K(x), L(x)$. There's a natural injection
$$K(x) \otimes_K L \to L(x)$$
and it is never a surjection unless $L = K$. We can see this as follows: if $L \neq K$ then there is some $\alpha \in L$ not in $K$, and $\frac{1}{x - \alpha}$ is not in the image of $K(x) \otimes_K L$ because it cannot be written as an element of $L$ times a quotient $\frac{p(x)}{q(x)}, p, q \in K[x]$. (It is perhaps easier to split into the cases that $\alpha$ is algebraic and $\alpha$ is transcendental: in the transcendental case $x - \alpha$ does not divide any polynomial in $K[x]$, and in the algebraic case $\alpha$ has some minimal polynomial with other roots besides $\alpha$, so the partial fraction decomposition of $\frac{p(x)}{q(x)}$ over $\overline{K}$ necessarily contains $\frac{1}{x - \alpha'}$ where $\alpha'$ is one of these other roots and hence cannot contain $\frac{1}{x - \alpha}$ alone.) 
Edit, 10/6/22: This argument is incorrect as Eoin points out in the comments. The image of $K(x) \otimes_K L$ can be described exactly as $\frac{L[x]}{K[x]}$, by which I mean the set of rational functions that can be expressed as the quotient of an element of $L[x]$ by an element of $K[x]$. If $L$ is algebraic over $K$ then every polynomial in $L[x]$ divides a polynomial in $K[x]$ so this is all of $L(x)$ and the above map is an isomorphism.
To get a counterexample we need to assume that $L$ is transcendental over $K$ and then if $\alpha \in L$ is transcendental over $K$ then $\frac{1}{x - \alpha}$ is not in the image of the above map since $x - \alpha$ does not divide any polynomial in $K[x]$ by hypothesis.
