Maximal ideal of $\mathbb{Q}[x_1,...,x_n]$ is contained in only finitely maximal ideals of $\mathbb{C}[x_1,...,x_n]$ 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.
 A: We can actually describe the maximal ideals containing $M$. Let me outline a more "geometric" approach (which might be useful if you are interested in algebraic geometry):


*

*The quotient  $\mathbf Q[x_1, \dots, x_n]/M$ is a field, hence identified with a finite extension $L$ of $\mathbf Q$ (by Zariski's lemma). Let $\alpha_i$ be the image of $x_i$ under this identification.

*Similarly, for a maximal ideal $N = (x_1 - \beta_1,\dots,x_n - \beta_n)$ of $\mathbf C[x_1, \dots, x_n]$, the quotient $\mathbf C[x_1, \dots, x_n]/N$ is identified with $\mathbf C$ via the isomorphism $$b_N: \mathbf C[x_1, \dots, x_n]/N \rightarrow \mathbf C$$ given by  $b_N(x_i) = \beta_i$. 

*An inclusion $M \subset N$ induces an embedding $$\varphi: \mathbf Q[x_1, \dots, x_n]/M \rightarrow  \mathbf C[x_1, \dots, x_n]/N$$ of $L$ into  $ \mathbf C.$ 

*Conversely, for an embedding  $$\sigma: L \rightarrow  \mathbf C,$$ the maximal ideal $$(x_1 - \sigma(\alpha_1), \dots, x_n - \sigma(\alpha_n) )$$ of $ \mathbf C[x_1, \dots, x_n]$ contains $M$.


Therefore, there exist exactly $[L : K]$ maximal ideals in $\mathbf C[x_1, \dots, x_n]$ containing $M$ and they are the ones described in the last bullet above.

Geometrically, in the language of schemes, this is what's going on. The base change morphism  $$\mathbf A^n_\mathbf C \rightarrow \mathbf A^n_\mathbf Q $$ sends a closed point $p$ of $\mathbf A^n_\mathbf C$ to its Galois orbit $G_\mathbf Q p$ regarded as a closed point of $\mathbf A^n_\mathbf Q$. The fiber of $G_\mathbf Q p$ consists the points $\sigma(p)$, for $\sigma \in G_\mathbf Q =  \operatorname{Gal} (\overline{\mathbf Q}/\mathbf Q)$.
A: If I haven't made a mistake, the result is true if $\mathbb{Q} \subseteq \mathbb{C}$ is replaced by any field extension, so you don't need the Nullstellensatz.  
Let $A = \mathbb{Q}[X_1, ... , X_n], B = \mathbb{C}[X_1, ... , X_n]$.  Then $B$ is a flat extension of $A$, and so the going-down property holds.
Since $M$ is a maximal ideal, and $A$ is an $n$-dimensional ring, we know by the dimension formula for finitely generated algebras over a field that there is a chain of $n$ inclusions of prime ideals $0 \subset \mathfrak p_1 \subset \cdots \subset \mathfrak p_{n-1} \subset M$.  
Let $P$ be a prime ideal of $B$ which contains $MB$.  Then $P \cap A \supseteq MB \cap A \supseteq M$, which implies $P \cap A = M$, because $M$ is a maximal ideal of $A$.  By the going down property, there exist prime ideals $P_{n-1} \supset \cdots \supset P_1$, contained in $P$, such that $P_i \cap A = \mathfrak p_i$.
Now we have a chain of $n$ inclusions of prime ideals $0 \subset P_1 \subset \cdots \subset P_{n-1} \subset P$.  Since $B$ is also an $n$-dimensional ring, $P$ must be a maximal ideal.
We have shown that every prime ideal of the quotient ring $B/MB$ is maximal.  Hence $B/MB$ is a zero dimensional Noetherian ring.  But this is the same as saying that $B/MB$ is artinian, and an artinian ring has only finitely many maximal ideals.  So $MB$, and therefore $M$, is contained in only finitely many maximal ideals of $B$.
A: 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.
