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For an ideal $I \subset \mathbb{C} [x_1, ... , x_n]$ show that dim$_{\mathbb{C}}R/I$ is finite iff $I$ is contained in only finitely many maximal ideals.

Thoughts so far: I'm not sure how to get started, so a hint to get things going would be appreciated!

user26857
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user19817
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    If $I$ is a (fixed: radical) ideal contained in only a finitely many maximal ideals $\mathfrak{m}1, \ldots$ then show that $R/I$ ($R = \mathbb{C}[x_1,\ldots,x_n]$) injects into $R/\mathfrak{m}_1 \oplus \cdots$. Conversely, if $\text{dim}{\mathbb{C}}(R/I) < \infty$ then $R/I$ is an Artinian ring. –  Aug 31 '15 at 20:18
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    @JinShin: Your first sentence is only true if $I$ is radical. – Eric Wofsey Aug 31 '15 at 20:19
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    @EricWofsey: You're right. It's fixed. For m.deslauriers: For a more general ideal $I$, note that if $I$ is in only a finite number of maximal ideals then $(\mathfrak{m}_1)^{e_1} \cdots (\mathfrak{m}_r)^{e_r} \subseteq I$ for some maximal ideals $\mathfrak{m}_1, \ldots, \mathfrak{m}_r$ and nonnegative integers $e_1, \ldots, e_r$. Then make an obvious filtration. –  Aug 31 '15 at 20:37
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    @JinShin So the fact that $R/I$ is Artinian means that it can't be contained in infinitely many maximal ideals, for otherwise (say there are infinitely many max. ideals $I_1, I_2, ... $ you could create an infinite non-terminating descending chain: $I_1 \supset I_1 \cap I_2, ... $ Does this make sense? – user19817 Aug 31 '15 at 23:10
  • @m.deslauriers: That works and is a very sensible way to nail it. –  Aug 31 '15 at 23:15
  • @JinShin Thanks! I'm still unsure about the other direction.. Could you help me a bit further? – user19817 Aug 31 '15 at 23:40
  • @m.deslauriers: For completeness sake I just wrote it in an answer. –  Aug 31 '15 at 23:52

1 Answers1

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Set $R = \mathbb{C}[x_1, \ldots, x_n]$.

If $I \subset R$ is an ideal such that $\dim_{\mathbb{C}}(R/I) < \infty$ then $R/I$ is Artinian, and consider a descending sequence of ideals $\mathfrak{m}_1 \supset \mathfrak{m}_1 \cap \mathfrak{m}_2 \supset \cdots$ where $\mathfrak{m_i} \subset R$ is maximal.

If $I \subset R$ is an ideal that is contained in only a finite number of maximal ideals, then $\sqrt{I} = \mathfrak{m}_1 \cap \cdots \mathfrak{m}_r$ for some max ideals $\mathfrak{m}_i, \ldots$ of $R$, and since $\mathfrak{m}_i$'s are coprime to each other $\sqrt{I} = \mathfrak{m}_1 \cdots \mathfrak{m}_r$, so $\mathfrak{m}_1^{e} \cdots \mathfrak{m}_r^{e} \subseteq I$ for some positive integer $e \geq 1$. But each $R/\mathfrak{m}_i^{e_i}$ is finite dimensional over $\mathbb{C}$ because each factor $\mathfrak{m_i}^v / \mathfrak{m_i}^{v+1}$ is finite dimensional over $\mathbb{C}$ ($\mathfrak{m}_i^v / \mathfrak{m}_i^{v+1}$ is a f.g. module over $R/\mathfrak{m}_i = \mathbb{C}$).

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    "$\sqrt{I} = \mathfrak{m}_1 \cap \cdots \mathfrak{m}_r$ for some max ideals $\mathfrak{m}_i, \ldots$ of $R$" You should explain where that comes from. – Georges Elencwajg Sep 01 '15 at 06:55
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    Hilbert Nullstellensatz. It's a consequence that radical of an ideal in $\mathbb{C}[x_1, \ldots, x_n]$ (or over any field, really) is equal to its Jacobson radical. –  Sep 01 '15 at 07:01
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    Yes, your comment is a perfect reason for your assertion and indeed the result is valid over any field (or even for any Jacobson ring, for users who know the concept). By the way, I had already upvoted your answer just before posting my comment. – Georges Elencwajg Sep 01 '15 at 07:08
  • @user83310 I just came back to this question, and I realize that I didn't understand completely the first time around. Is the claim that the $m_i$'s are comprime? (and so we can therefore apply the Chinese remainder theorem) If so, how do we know that they're coprime? and Then I'm still confused about the fact that $R/m_i^{e_i}$ is finite dimensional over $\mathbb{C}$. Could anyone spell this out explicitly for me? Thank you. – user19817 Sep 09 '15 at 01:38
  • I meant to ask if the $m_i^{e^i}$'s are pairwise coprime.. I know the $m_i$'s are pairwise coprime since they are maximal. – user19817 Sep 09 '15 at 02:39
  • One more question: How do we know that $m_i^v/m_i^{v+1}$ is a f.g. module over $R/m_i$? Thank you! – user19817 Sep 10 '15 at 23:53
  • I just figured out with the $m_i^{e^i}$'s are coprime, so kindly disregard that part of my question. – user19817 Sep 11 '15 at 00:39