Function field of variety Let $k$ be a field, let $X/k$ be a geometrically integral variety and let $k(X)$ be its function field.
If $L/k$ is an algebraic extension, is it true that the function field of $X \times_k L$ is $L \otimes_k k(X)$?
 A: Geometric integrality tells you that $k(X)\otimes_k\bar{k}$ is a domain. If $U=\mathrm{Spec}(A)$ is a non-empty affine open of $X$, then $k(X)$ is the quotient field of $A$. The ring $k(X)\otimes_kL$ is a domain because it is a subring of $k(X)\otimes_k\bar{k}$ ($k(X)\otimes_kL\rightarrow k(X)\otimes_k\bar{k}$ comes by tensoring the injection $L\rightarrow\bar{k}$ with the $k$-flat $k(X)$). Moreover, since $k\rightarrow L$ is integral (i.e. algebraic), $k(X)\rightarrow k(X)\otimes_kL$ is integral. A domain integral over a field is a field, so $k(X)\otimes_kL$ is a field. We have the injection $A\otimes_kL\rightarrow k(X)\otimes_kL$, and this is a localization. This implies that $k(X)\otimes_kL$ is the quotient field of $A\otimes_kL$. Since $\mathrm{Spec}(A\otimes_kL)$ is a non-empty affine open of $X_L$, it follows that $k(X)\otimes_kL$ is indeed the function field of $X_L$.
A: That is true. In fact, there is not much to prove here because it is, more or less, the definition of $X\times_kL$. W.l.o.g., we can assume that $X=\mathrm{Spec}(A)$ is affine (because $k(X)=\mathrm{Quot}(A)$ for any such affine subset of $X$), then the definition of the fibre product is $X\times_kL:=\mathrm{Spec}(A\otimes_k L)$ and $\mathrm{Quot}(A\otimes_k L)=\mathrm{Quot}(A) \otimes_k L = k(X)\otimes_k L$. 
