 Finite generation descends along a field extension

Let $L/k$ be a field extension and $A$ a $k$-algebra. If $A{\otimes }_{k}L$ is a finitely generated algebra over $L$, then $A$ is finitely generated over $k$.

Choose finitely many generators of $A{\otimes }_{k}L$. Each is a finite sum of pure tensors, so only finitely many coefficients from $A$ occur. Let $A_0\subseteq A$ be the $k$-subalgebra generated by those coefficients. The image of $A_0{\otimes }_{k}L$ contains every chosen generator and all scalars, hence equals $A{\otimes }_{k}L$. ThereforeHere $A/A_0$ is a quotient vector space, not a quotient algebra: $A_0$ need not be an ideal. The extension $L$ is a faithfully flat module over $k$, so this vanishing forces $A/A_0=0$. Thus $A=A_0$, proving finite generation.