Let $R$ be a commutative ring. If $0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow 0$ is an exact sequence of modules, we have that $\operatorname{Supp}M=\operatorname{Supp}M'\cup \operatorname{Supp}M''$. Furthermore can we obtain that $\dim M=\dim M'+\dim M''$? In fact, I want to know that given a module $M$ with $\dim M=n$, is it possible to construct an exact sequence as above with $\dim M'=0$ and $\dim M''=n-1$?
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$\begingroup$ I've heard of the Krull dimension of a ring, but what is the Krull dimension of a module? $\endgroup$– Dylan WilsonCommented Aug 3, 2010 at 6:56
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$\begingroup$ Or did you mean the projective dimension? $\endgroup$– Dylan WilsonCommented Aug 3, 2010 at 6:57
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$\begingroup$ For the second question, dimM' should be 1. $\endgroup$– TmobiusXCommented Aug 3, 2010 at 8:05
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$\begingroup$ -1. This is perhaps a research level topic, but not a research level question. $\endgroup$– user2035Commented Aug 3, 2010 at 9:41
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The dimension of $M$ is the maximum of the dimensions of $M'$ and $M''$, so what you are asking for cannot happen (at least not in the noetherian case). To Dylan: the dimension of a module is the dimension of its support.
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$\begingroup$ Thanks! Is it called the "Krull dimension" though? That's what confused me... $\endgroup$ Commented Aug 3, 2010 at 7:50
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$\begingroup$ Is there a reference for this result? Thanks! $\endgroup$– TmobiusXCommented Aug 3, 2010 at 8:04
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$\begingroup$ $dim(X \cup Y) = max(dim(X),dim(Y))$ is a an easy exercise in general topology. $\endgroup$ Commented Aug 3, 2010 at 8:45
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2$\begingroup$ To the OP: I think this should be proved in any book that discusses dimensions of modules. In case as Martin says it is a very easy exercise. To Dylan: the dimension of the spectrum of a ring A is the Krull dimension of A, and the definition for modules is a useful generalization. It can also be interpreted as the Krull dimension of the ring modulo the annihilator of the module (I am always referring to finite modules over noetherian rings). $\endgroup$– AngeloCommented Aug 3, 2010 at 9:25
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$\begingroup$ Does the equality hold when M is not finite? $\endgroup$– TmobiusXCommented Aug 4, 2010 at 1:50