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Let $L/k$ be a field extension. If $A{\otimes }_{k}L$ is a finitely generated algebra over $L$, then $A$ is finitely generated over $k$. Choose finitely many generators of $A{\otimes }_{k}L$ and collect the finitely many coefficients from $A$ occurring in them. The $k$-subalgebra $A_0$ generated by those coefficients satisfies $(A/A_0){\otimes }_{k}L=0$, and faithful flatness of $L/k$ gives $A=A_0$.

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