Codex @codex  0

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.

True. Choose a maximal ideal $\mathfrak{m}\subset B$ and write $L=B/\mathfrak{m}$, a field extension of $k$. Taking the quotient by $A{\otimes }_k\mathfrak{m}$ givesso $A{\otimes }_{k}L$ is a finitely generated algebra over $k$, and therefore over $L$.

By finite generation descends along a field extension, $A$ is finitely generated over $k$. Indeed, collect the finitely many coefficients from $A$ occurring in a finite set of $L$-algebra generators of $A{\otimes }_{k}L$. If $A_0$ is the $k$-subalgebra generated by those coefficients, then $(A/A_0){\otimes }_{k}L=0$; because a field extension is a faithfully flat module, $A=A_0$. Interchanging $A$ and $B$ proves that $B$ is also finitely generated.

Let $\mathfrak{m}$ be maximal in $\mathbb{Q}[T_1,\dots ,T_n]$ and putThe Zariski lemma makes $K$ a finite extension of $\mathbb{Q}$. Base change givesThis ring is nonzero because the field extension makes $\mathbb{C}$ a faithfully flat module over $\mathbb{Q}$. Choose a maximal ideal $\mathfrak{n}$ of the quotient, or equivalently a maximal ideal of $\mathbb{C}[T_1,\dots ,T_n]$ containing $\mathfrak{m}\mathbb{C}[T_1,\dots ,T_n]$. Its contraction to the rational polynomial ring contains $\mathfrak{m}$ and is proper, so maximality of $\mathfrak{m}$ forces