TITLE: Are tensor products cancellative?
QUESTION [5 upvotes]: Let $k$ be a field, $F \subseteq k$ a subfield, and $A_0$ a finitely generated $F$-algebra.  Then $k \otimes_F A_0$ is a finitely generated $k$-algebra.  If $B_0$ is another finitely generated $F$-algebra, and there exists some $k$-algebra isomorphism $\phi:k \otimes_F A_0 \rightarrow k \otimes_F B_0$, do $A_0$ and $B_0$ have to be isomorphic as $F$-algebras?  Or at least $F$-vector spaces?  
I know that if $\phi$ were induced by some existing homomorphism $A_0 \rightarrow B_0$, then the claim would follow from faithful flatness.  If $\phi$ is any old isomorphism, I'm not so sure.  What do you think?

REPLY [2 votes]: No. For example, let $k = \mathbb{R}, F = \mathbb{C}$. The $\mathbb{R}$-algebras $\mathbb{H}$ and $M_2(\mathbb{R})$ are nonisomorphic, but both of them complexify to $M_2(\mathbb{C})$. For a commutative example, $\mathbb{C}$ and $\mathbb{R} \times \mathbb{R}$ both complexify to $\mathbb{C} \times \mathbb{C}$. 
If $k \to F$ is a Galois extension, then questions of this form fall under the heading of what is called Galois descent. If $A$ is an $F$-algebra, then $k$-algebras $A_0$ such that $A_0 \otimes_k F \cong A$ are called $k$-forms of $A$, and in general these need not either exist or be unique.