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

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

2) Is it possible that there is another maximal ideal $m'$ of $A$ such that $m_i = m'_i, \, \forall i=1,\cdots,n,$ and $m \neq m'$? 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.

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You should quantify $i$ in 1) and 2). –  Georges Elencwajg Oct 11 '12 at 22:33
    
@GeorgesElencwajg: Done. –  Manos Oct 11 '12 at 22:54
    
Dear @Makoto Kato, your deleted answer to 1) was fantastic: please consider resuscitating it, I very much want to upvote it! Also 2) contained a great idea: maybe you can still use it. –  Georges Elencwajg Oct 11 '12 at 23:30
    
@GeorgesElencwajg Thanks. I will rethink about it. –  Makoto Kato Oct 11 '12 at 23:55
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1 Answer

up vote 3 down vote accepted

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

2) Let $Specm(A)$ be the set of maximal ideals of $A$. Let $\bar k$ be an algebracic closure of $k$. Let $Homalg_k(A, \bar k)$ be the set of $k$-homomorphisms. Let $f \in Homalg_k(A, \bar k)$. Then $Ker(f) \in Specm(A)$. Conversely let $m \in Specm(A)$. It is easy to see that there exists $f \in Homalg_k(A, \bar k)$ such that $m = Ker(f)$.

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

Similar results hold for $A_i$.

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

Let $m_i \in Specm(A_i)$ for $i = 1, \dots, n$. There exists a monic polynomial $f_i(x_i) \in A_i$ such that $m_i = (f_i)$ 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 $G = Aut(\bar k/k)$. $G$ acts on $P$ in the obvious way. Let $P/G$ be the set of $G$-orbits. Let $M$ be the set of maximal ideals of $A$ lying over $m_i$ for all $i$. Let $\alpha = (\alpha_1,\dots,\alpha_n) \in P$. There exists a unique $f \in Homalg_k(A, \bar k)$ 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 $m_i$ for all $i$.

Now we construct an example such that $m \neq m'$ and $m_i = m'_i$ for all $i$. Let $\alpha, \beta \in \bar k$ be separable over $k$. Suppose $k(\beta) \supset k(\alpha)$ and $[k(\beta) : k(\alpha)] > 1$ 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 $m = Ker(f), m' = Ker(g)$. Suppose $m = m'$. 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 $m \neq m'$. Clearly $m_1 = m'_1$ and $m_2 = m'_2$.

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Dear Makoto: this is a fantastic answer and I'm sorry I can upvote it only once . I hope many users will upvote it too and that Manos will accept it. Anyway, your insight and your counter-example are really brilliant: congratulations!. –  Georges Elencwajg Oct 12 '12 at 6:54
    
@navigetor23 Different proofs are always welcomed. Please feel free to write one(s). –  Makoto Kato Oct 12 '12 at 10:01
    
@MakotoKato: Dear MakotoKato, i have been studying your proof and have resolved all little details, but i am still not seeing your argument about the canonical bijection between the orbits and the maximal ideals lying over $m_i$. Could you please elaborate? –  Manos Oct 12 '12 at 19:18
    
@Manos I added an explanation about it. Please feel free to ask me if you still have a question. –  Makoto Kato Oct 12 '12 at 20:07
    
@MakotoKato: I see your argument now. Beautiful piece of algebra :) Thanks!! –  Manos Oct 12 '12 at 22:28
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