Intersecting maximal ideals of $k[x_1,\dots,x_n]$ with $k[x_i]$

Question

Let $\mathfrak m$ be a maximal ideal of $A=k[x_1,\dots,x_n]$, where $k$ is a field, not necessarily algebraically closed. Define $\mathfrak m_i = \mathfrak m \cap k[x_i], \, \forall i=1,\dots,n$.

1) How can i see that $\mathfrak m_i$ is a maximal ideal of $k[x_i], \, \forall i=1,\dots,n$?

2) Is it possible that there is another maximal ideal $\mathfrak m'$ of $A$ such that $\mathfrak m_i = \mathfrak m'_i, \, \forall i=1,\dots,n,$ and $\mathfrak m \neq \mathfrak m'$? Under what conditions this is not the case?

Any relative references in Atiyah-MacDonald? (I have the book but can not find anything relevant).

Thanks.

Answer 1

1) Let $A_i = k[x_i]$. There exists a canonical injection $A_i/\mathfrak m_i \rightarrow A/\mathfrak m$. Since $A/\mathfrak m$ is finite algebraic over $k$, so is $A_i/\mathfrak m_i$. Hence $\mathfrak m_i$ is maximal.

2) Let $\operatorname{Specm}(A)$ be the set of maximal ideals of $A$. Let $\bar k$ be an algebraic closure of $k$. Let $\operatorname{Homalg}_k(A, \bar k)$ be the set of $k$-homomorphisms. Let $f\in\operatorname{Homalg}_k(A,\bar k)$. Then $\ker(f) \in \operatorname{Specm}(A)$. Conversely let $\mathfrak m \in \operatorname{Specm}(A)$. It is easy to see that there exists $f \in \operatorname{Homalg}_k(A, \bar k)$ such that $\mathfrak m = \ker(f)$.

Let $f, g \in \operatorname{Homalg}_k(A, \bar k)$. We write $f \equiv g$ if there exists $\sigma \in \operatorname{Aut}(\bar k/k)$ such that $\sigma f = g$. This is an equivalence relation on $\operatorname{Homalg}_k(A, \bar k)$. Let $f, g \in \operatorname{Homalg}_k(A, \bar k)$. Let $\mathfrak m = \ker(f), \mathfrak m' = \ker(g)$. Then $\mathfrak m = \mathfrak m'$ if and only if $f \equiv g$. Hence $\operatorname{Specm}(A)$ is identified with the set of equivalence classes on $\operatorname{Homalg}_k(A, \bar k)$.

Similar results hold for $A_i$.

Paraphrasing above results we get the following criteria:
Let $f, g \in \operatorname{Homalg}_k(A, \bar k)$. Let $\mathfrak m = \ker(f), \mathfrak m' = \ker(g)$. Let $\alpha_i = f(x_i), \beta_i = g(x_i)$ for all $i$. Then $\mathfrak m = \mathfrak m'$ if and only if there exists $\sigma \in \operatorname{Aut}(\bar k/k)$ such that $\sigma(\alpha_i) =\beta_i$ for all $i$. For each $i$, $\mathfrak m_i = \mathfrak m'_i$ if and only if there exists $\tau_i \in \operatorname{Aut}(\bar k/k)$ such that $\tau_i(\alpha_i) =\beta_i$.

Let $\mathfrak m_i \in \operatorname{Specm}(A_i)$ for $i = 1, \dots, n$. There exists a monic polynomial $f_i(x_i) \in A_i$ such that $\mathfrak m_i = (f_i)$ for each $i$. Let $P_i$ be the set of roots of $f_i(x_i)$ in $\bar k$. Let $P = \prod_i P_i$. Let $G = \operatorname{Aut}(\bar k/k)$. $G$ acts on $P$ in an obvious way. Let $P/G$ be the set of $G$-orbits. Let $M$ be the set of maximal ideals of $A$ lying over $\mathfrak m_i$ for all $i$. Let $\alpha = (\alpha_1,\dots,\alpha_n) \in P$. There exists a unique $f \in \operatorname{Homalg}_k(A, \bar k)$ such that $f(x_i) = \alpha_i$ for all $i$. We denote by $\psi(\alpha) = \ker(f)$. Clearly $\ker(f) \in M$. Hence we get a map $\psi\colon P \rightarrow M$. By the above result, $\psi$ induces a bijection $\bar \psi\colon P/G \rightarrow M$. Hence $|P/G| = 1$ if and only if there exists only one maximal ideal of $A$ lying over $\mathfrak m_i$ for all $i$.

Now we construct an example such that $\mathfrak m \neq \mathfrak m'$ and $\mathfrak m_i = \mathfrak m'_i$ for all $i$. Let $\alpha, \beta \in \bar k$ be separable over $k$. Suppose $k(\beta) \supset k(\alpha)$, $[k(\beta) : k(\alpha)] > 1$ and $[k(\alpha) : k] > 1$. Let $p(x)$ be the minimal polynomial of $\beta$ over $k$. Let $q(x)$ be the minimal polynomial of $\beta$ over $k(\alpha)$. Clearly $\deg p(x) > \deg q(x)$. Since $p(x)$ is divisible by $q(x)$, there exists a root $\beta'$ of $p(x)$ such that $\beta'$ is not a root of $q(x)$.

Let $A = k[x_1, x_2]$. Let $f\colon A \rightarrow \bar k$ be the map defined by $f(x_1) = \alpha$, $f(x_2) = \beta$. Let $g\colon A \rightarrow \bar k$ be the map defined by $g(x_1) = \alpha$, $g(x_2) = \beta'$. Let $\mathfrak m = \ker(f), \mathfrak m' = \ker(g)$. Suppose $\mathfrak m = \mathfrak m'$. There exists an automorphism $\sigma$ of $\bar k/k$ such that $\sigma(\alpha) = \alpha$, $\sigma(\beta) = \beta'$. Hence $\beta'$ is a root of $q(x)$. This is a contradiction. Hence $\mathfrak m \neq \mathfrak m'$. Clearly $\mathfrak m_1 = \mathfrak m'_1$ and $\mathfrak m_2 = \mathfrak m'_2$.

Answer 2

To give a reference regarding the first point: one says a commutative ring $R$ is a Hilbert ring if for every maximal ideal $\mathfrak n\subseteq R[X]$, it follows that $\mathfrak m:=R\cap \mathfrak n$ is maximal in $R$. It is a theorem that

$(1)$ If $R$ is Hilbert and $f:R\to S$ is an $R$-algebra of finite type, then $S$ is Hilbert.

$(2)$ If $\mathfrak n$ is maximal in $S$, then $\mathfrak m=\mathfrak n^c:= f^{-1}(\mathfrak n)$ is maximal in $R$, and

$(3)$ $[R/\mathfrak m:S/\mathfrak n]$ is finite.

It is trivial to see a field $k$ is Hilbert, so that any polynomial ring $k[X_1,\ldots,X_n]$ is also Hilbert, so your first point follows.

One also has the following theorem:

A ring $R$ is Hilbert if and only if whenever $K$ is a field that is of finite type as an $R$-algebra, then $K$ is of finite type as an $R$-module. This says essentially that Hilbert rings are those which satisfy Zariski's lemma. (Note this gives another proof of Zariski's lemma, since fields are trivially Hilbert rings!)