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Does the Zariski topology on a ring (not commutative in common) form a compact or paracompact space and why?

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For a commutative ring, yes Zariski is compact (although most algebraic geometers would say quasi-compact). For a noncommutative ring, I'm not even sure what the definition is. –  Donu Arapura Dec 12 '10 at 18:49
    
Yeah, that's what I'm wondering in this. What is Spec R for a non-commutative ring? –  Simon Rose Dec 12 '10 at 18:50
    
To elaborate just a smitch on Donu's answer, the point it compactness in its traditional sense is rather less of a remarkable thing in the Zariski topology for if $X\subseteq\mathbb{P}^N$ (this should be taken to be real or complex projective space) is a nonsingular variety then it is a closed subset of a compact set, hence compact in the topologist's sense. The same thing with more details can be found in Shefarevich Basic Algebraic Geometry vol. 1 on p. 105-106. –  Adam Hughes Dec 12 '10 at 19:00
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@Adam: it's somewhat more surprising than that. Compactness holds for arbitrary commutative rings, with no Noetherianness or any other assumptions, e.g. it holds for arbitrary Boolean rings (and this is equivalent to the compactness theorem in propositional logic). –  Qiaochu Yuan Dec 12 '10 at 19:47
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@Martin: Given that a lot of us here are interested in the definition of non-commutative Spec and don't know it, would you be willing to do us the favor of providing a specific link or -- better yet -- summarizing the definition in a CW answer? –  Pete L. Clark Dec 12 '10 at 22:23
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2 Answers 2

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It is compact (or quasi-compact?).

Let $R$ be a ring. The Zariski topology has closed sets given by $V(I) = \{P \mid I \subset P\}$. So let $I_\alpha$ be a family of ideals such that $\bigcap V(I_\alpha) = \emptyset = V(R)$.

This means that $\sum I_\alpha = R$. However, as $1 \in R$, we can write $1 \in \sum I_\alpha$, and so we have that $1 = i_{j_1} + \cdots + i_{j_k}$ for some indices ${j_\ell}$. But then $V(I_{j_1}) \cap \cdots \cap V(I_{j_k}) = V(R) = \emptyset$ and so Spec $R$ satisfies the finite intersection property, and is compact.

I admit that I'm worried that I'm subtly using commutativity in here somewhere.

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Are these two-sided primes? How does the deduction that sum I_a = R work in an arbitrary ring? –  Qiaochu Yuan Dec 12 '10 at 19:50
    
thank u for this answer, it was very helpfull to me and i think that is quite correct put you dont give the complete of question about what if it is paracompact thank you again khaled ahmad –  khaled ahmad Dec 23 '10 at 20:42
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It looks like the accepted answer to this question didn't address some of the questions asked in the comments above, so I thought I'd fill in some details. This is really just a long comment; sorry I had to include it as an answer.

It seems that the question is regarding the Zariski topology on two-sided prime ideals. The word "ideal" below refers to two-sided ideals.

A prime ideal of a noncommutative ring $R$ is a proper ideal such that, for any ideals $I$ and $J$ of $R$, if the ideal product $IJ$ is contained in $P$, then either $I \subseteq P$ or $J \subseteq P$. Let $\mathrm{Spec}(R)$ denote the set of prime ideals of $R$.

The Zariski topology on $\mathrm{Spec}(R)$ can be defined just as in the commutative case, as stated in Simon Rose's answer: the closed subsets of $\mathrm{Spec}(R)$ are precisely those of the form $V(I) := \{P \in \mathrm{Spec}(R) : I \subseteq P\}$. It's easy to see that $\bigcap_j V(I_j) = V(\sum I_j)$ for any set of ideals $\{I_j\}$ of $R$, where $\sum I_j$ is the smallest ideal of $R$ containing the $I_j$ (and is equal to the set of all finite sums of elements from the $I_j$). To see that these are closed under finite unions, one uses the definition of prime ideal to see that $V(I) \cup V(J) = V(IJ)$. That is, for any prime $P$ of $R$ and any ideals $I$ and $J$ of $R$, the statement $IJ \subseteq P$ is equivalent to the statement that either $I \subseteq P$ or $J \subseteq P$.

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