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.

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    $\begingroup$ $A=\mathbb Q[x_1,\dots,x_n]$ and $B=\mathbb C[x_1,\dots,x_n]$. We have: $\dim A=\dim B=n$, the extension $A\subset B$ is (faithfully)flat, and from dimension formula (see Bruns and Herzog, Theorem A.11) we get $\dim B/MB=0$, so $B/MB$ is artinian hence it has only finitely many maximal ideals. $\endgroup$ – user26857 Jan 4 '17 at 15:24
  • $\begingroup$ Can you explain why $B$ is flat and faithfully flat over $A$ ? $\endgroup$ – user371231 Jul 4 at 13:40

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 $\sigma \in G_\mathbf Q = \operatorname{Gal} (\overline{\mathbf Q}/\mathbf Q)$.

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    $\begingroup$ Note that the very first step where you conclude that $L$ is a finite extension of $\mathbb{Q}$ is the hardest step of this proof (and is basically a form of the Nullstellensatz). $\endgroup$ – Eric Wofsey Jan 5 '17 at 4:55
  • $\begingroup$ @EricWofsey yes :) $\endgroup$ – Alex Macedo Jan 5 '17 at 4:56

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 $B$.


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.


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