TITLE: Maximal ideal of $\mathbb{Q}[x_1,...,x_n]$ is contained in only finitely maximal ideals of $\mathbb{C}[x_1,...,x_n]$
QUESTION [6 upvotes]: Suppose $M\subset \mathbb{Q}[x_1,...,x_n]$ is maximal. I'd like to show that as a subset of $R= \mathbb{C}[x_1,...,x_n]$, $M$ is contained in only finitely many maximal ideals. By Hilbert-Nullstellensatz, I know max ideals of $R$ are of the form $(x_1-a_1,...,x_n-a_n)$, so perhaps if $M$ were contained infinitely many of them, one could obtain a contradiction? I'm not quite sure where to go with this.

REPLY [3 votes]: Here's another approach.  By Zariski's lemma, the quotient $L=\mathbb{Q}[x_1,\dots,x_n]/M$ is a finite extension of $\mathbb{Q}$.  Now tensoring the exact sequence $$M\to\mathbb{Q}[x_1,\dots,x_n]\to L\to 0$$ with $\mathbb{C}$ over $\mathbb{Q}$ gives an exact sequence $$M\otimes\mathbb{C}\to \mathbb{C}[x_1,\dots,x_n]\to L\otimes\mathbb{C}\to 0.$$  The image of the first map is just the ideal in $\mathbb{C}[x_1,\dots,x_n]$ generated by $M$, so this says that $L\otimes\mathbb{C}$ is the quotient of $\mathbb{C}[x_1,\dots,x_n]$ by the ideal generated by $M$.
Now $L$ is finite-dimensional as a $\mathbb{Q}$-vector space, and hence $L\otimes\mathbb{C}$ is finite-dimensional (of the same dimension) as a $\mathbb{C}$-vector space.  In particular, this means $L\otimes\mathbb{C}$ is an artinian ring, and so it has only finitely many maximal ideals.  But maximal ideals of $L\otimes\mathbb{C}$ are in bijection with maximal ideals of $\mathbb{C}[x_1,\dots,x_n]$ containing $M$, so we're done.