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Let $k$ be a field and $D = k[X_1, . . . , X_n]$ the polynomial ring in $n$ variables over $k$.

Show that:

a) Every maximal ideal of $D$ is generated by $n$ elements.

b) If $R$ is ring and $\mathfrak m\subset D=R[X_1,\dots,X_n]$ is maximal ideal such that $\mathfrak m \cap R$ is maximal and generated by $s$ elements, then $\mathfrak m$ is generated by $s + n$ elements.

The days that I am trying to solve. Help me.

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marked as duplicate by YACP, Julian Kuelshammer, O.L., Amzoti, TMM Jun 22 '13 at 12:31

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Is $R$ related in some way to $k$ and/or $D$? –  Gerry Myerson Jun 22 '13 at 4:40
    
Just some quick hints: I think the important idea is how maximal ideals behave in finitely generated algebras over a field. Say I have a maximal ideal $\mathfrak{m}$ of $k[X_1, X_2]$; think about the ideal $\mathfrak{m} \cap k[X_1]$ of the subring $k[X_1]$. Is it maximal? Can you relate $\mathfrak{m}$ to an ideal of $(k[X_1]/(\mathfrak{m} \cap k[X_1]))[X_2]$? Why does all this help? –  TTS Jun 22 '13 at 5:03
    
I think the question b) becomes stupid because of a) ; if $\mathfrak m \subseteq D$ is maximal, a) says it is generated by $n$ elements. And then we have to show that $\mathfrak m$ is generated by $s+n$ elements? We've already done that! –  Patrick Da Silva Jun 22 '13 at 5:32
    
@TTS : "Is it maximal?" You sound like you think it is. You know it's not, right? –  Patrick Da Silva Jun 22 '13 at 5:33
    
@PatrickDaSilva I'm somewhat confident that it is [I have a theorem in mind and I'm almost certain it's true when $k = \bar{k}$ because I can visualize that case.], but now you have me worried — do you have a counterexample? –  TTS Jun 22 '13 at 5:38

2 Answers 2

The answer to question a) can be found as Corollary 12.17 in these notes. The proof is left as an exercise, but the proof of it is just collecting together the previous results in the section.

(As Patrick DaSilva has mentioned, as written your question b) follows trivially from part a). I'm guessing it's not what you meant to ask.)

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For b) I guess the OP considers $D=R[X_1,\dots,X_n]$. Of course, b) follows easily from a) by factoring out the ideal $(\mathfrak m\cap R)[X_1,\dots,X_n]$. –  user26857 Jun 22 '13 at 18:09

We induct on $n$, the result $n=1$ being clear since $k[X_1]$ is a PID. Now consider the inclusion

$$k[X_1,\ldots,X_{n-1}] \subseteq K[X_1,\ldots,X_{n-1}][X_n].$$

Choose a maximal ideal $\mathfrak{m}$ in the ring on the right. Then by Zariski's lemma we have that $k[X_1,\ldots,X_n]/\mathfrak{m}$ is a finite algebraic extension of $k$. Since we have the inclusion

$$k \hookrightarrow k[X_1,\ldots,X_{n-1}]/\mathfrak{m}^c \hookrightarrow k[X_1,\ldots,X_n]/\mathfrak{m}$$

it follows that $k[X_1,\ldots,X_{n-1}]/\mathfrak{m}^c$ is an integral domain that is also a finite dimensional vector space, thus a field. It will now follow that $\mathfrak{m}^c$ is maximal and the induction hypothesis implies that it is generated by $n-1$ elements. We now need a Lemma (as suggested by Jeff Tolliver below):

Lemma: $$\left(k[X_1,\ldots,X_{n-1}]/\mathfrak{m}^c\right) [X_n] \cong k[X_1,\ldots,X_n]/\mathfrak{m}^ck[X_1,\ldots,X_n]$$

Proof: $$\begin{eqnarray*} \left(k[X_1,\ldots,X_{n-1}]/\mathfrak{m}^c\right) [X_n] &\cong& k[X_1,\ldots,X_{n-1}][X_n] \otimes_{k[X_1,\ldots,X_{n-1}]} k[X_1,\ldots,X_{n-1}]/\mathfrak{m}^c \\ &\cong& k[X_1,\ldots,X_n]/\mathfrak{m}^c k[X_1,\ldots,X_n] \hspace{5mm} \end{eqnarray*}$$ where the first line is using exercise 3.6 of Atiyah - Macdonald, the second line exercise 3.2 of the same book. Explicitly if we trace through the isomorphisms, this sends $$\overline{a_0} + \overline{a_1}X_n + \ldots + \overline{a_k}X_n^k \mapsto \overline{a_0 + a_1X_n\ldots + a_kX_n^k}$$ where the bar on the left is the residue class mod $\mathfrak{m}^c$ and on the right mod $\mathfrak{m}^ck[X_1,\ldots,X_n]$. Thus the lemma is proven.

Back to the problem. It is clear that $\mathfrak{m}$ corresponds to a maximal ideal $\overline{\mathfrak{m}}$ in $k[X_1,\ldots,X_n]/\mathfrak{m}^ck[X_1,\ldots,X_n]$ and by the lemma to some maximal ideal in $\left(k[X_1,\ldots,X_{n-1}]/\mathfrak{m}^c\right)[X_n]$. Since this is the polynomial ring over a field, this maximal ideal is generated by some $P = \overline{a_0} + \ldots + \overline{a_k}X_n^k$. Then in $k[X_1,\ldots,X_n]/\mathfrak{m}^ck[X_1,\ldots,X_n]$, the element corresponding to $P$ is $\overline{a_0 + \ldots + a_kX_n^k}$ and this we know generates $\overline{\mathfrak{m}}$. If we write $$f = a_0 + \ldots + a_kX_n^k$$ it will now follow by the induction hypothesis that $\mathfrak{m}$ is generated by $f$ and the $n-1$ elements of $\mathfrak{m}^c$. This completes the proof of the problem.

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I think this is along the lines of what I sketched out in the comments ["Zariski's lemma"; I'll have to remember that] but it seems like we conclude that $X_n$ is always in $\mathfrak{m}$, which worries me. –  TTS Jun 22 '13 at 7:52
    
@TTS Yes my proof as it stands (the last part is wrong, I am in the process of trying to rectify that. –  user38268 Jun 22 '13 at 7:59
    
$\mathfrak{m}$ corresponds to some maximal ideal $P$ of $(k[X_1,\ldots,X_{n-1}]/\mathfrak{m}^c)[X_n]$. But this last ring is a PID, so $P$ is generated by the class of some polynomial $f(X_1,\ldots,X_n)$. You can show $\mathfrak{m}=(a_1,\ldots,a_{n-1},f)$. –  Jeff Tolliver Jun 22 '13 at 8:06
    
@JeffTolliver Thanks Jeff. However I don't really understand the first part of your comment, what do you mean by $\mathfrak{m}$ corresponds to some maximal ideal in that polynomial ring, in what way? Regards, –  user38268 Jun 22 '13 at 8:16
    
$(k[X_1,\ldots,X_{n-1}]/\mathfrak{m}^c)[X_n]$ is canonically isomorphic to $k[X_1,\ldots,X_{n}]/\mathfrak{m}^ck[X_1,\ldots,X_{n}]$. Since $\mathfrak{m}$ contains the set $\mathfrak{m}^c$, it also contains the ideal it generates, namely $\mathfrak{m}^ck[X_1,\ldots,X_{n}]$. There is a correspondence between maximal ideals of $k[X_1,\ldots,X_n]$ containing $\mathfrak{m}^ck[X_1,\ldots,X_{n}]$ and maximal ideals of $k[X_1,\ldots,X_{n}]/\mathfrak{m}^ck[X_1,\ldots,X_{n}]=(k[X_1,\ldots,X_{n-1}]/ \mathfrak{m}^c)[X_n]$, so $\mathfrak{m}$ corresponds to such an ideal. –  Jeff Tolliver Jun 22 '13 at 8:24

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