The rank of a not necessarily finitely generated module. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T09:20:00Z http://mathoverflow.net/feeds/question/69420 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69420/the-rank-of-a-not-necessarily-finitely-generated-module The rank of a not necessarily finitely generated module. Sándor Kovács 2011-07-03T22:11:42Z 2011-07-06T16:09:32Z <p>This question is motivated by <a href="http://mathoverflow.net/questions/69353/an-example-of-a-rank-one-projective-r-module-that-is-not-invertible" rel="nofollow">this one</a>. The main point of the question (was) to try to weaken the notion of rank. After the answers and comments, it seems this is not a good way to do it, but perhaps it is still useful for anyone trying to do the same.</p> <p><strong>Remark:</strong> Conditions added after Hailong's comment and Tom's answer: $R$ reduced and $M$ indecomposable.</p> <p>Let $R$ be a reduced noetherian ring and $M$ an indecomposable $R$-module. For any prime $p\in\mathrm{Spec} R$ with residue field $\kappa(p)$ define $$\delta_M(p)=\dim (M\otimes_R \kappa(p))$$ the local rank of $M$ at $p$.</p> <blockquote> <p><strong>Question 1:</strong> If $\delta_M(p)=1$ for all $p$, does it follow that $M_p\simeq R_p$? Or more generally, if $\delta_M$ is constant, does it imply that $M_p$ is a free $R_p$-module for all $p$?</p> </blockquote> <p><strong>Remarks</strong></p> <p><strong>1</strong> If $M$ is finitely generated, then by Nakayama lemma these questions are easy. (See Exercise II.5.8 on page 125 in [Hartshorne]).</p> <p><strong>2</strong> <a href="http://mathoverflow.net/users/6666/tom-goodwillie" rel="nofollow">Tom Goodwillie</a> points out that if $R=\mathbb Z$ and $M\subset \mathbb Q$ consists of all rational numbers $a/b$ such that $b$ is square-free, then $M_p\simeq R_p$, but it is not invertible, so that would be too much to ask.</p> <p><strong>3</strong> <a href="http://mathoverflow.net/users/14094/yves-cornulier" rel="nofollow">Yves Cornulier</a> shows <a href="http://mathoverflow.net/questions/69353/an-example-of-a-rank-one-projective-r-module-that-is-not-invertible/69414#69414" rel="nofollow">here</a> that if $M$ is projective and $M_p\simeq R_p$, then $M$ is finitely generated. In other words, if the answer to (the first part of) Question 1 is "YES" then $M$ cannot be projective. So this suggests a subquestion...</p> <blockquote> <p><strong>Question 1a:</strong> Does there exist an example of an $R$ and a non-finitely generated projective $M$ for which $\delta_M$ is constant?</p> </blockquote> <p>And let me include also a somewhat vague, but related question:</p> <blockquote> <p><strong>Question 2:</strong> Is the class of modules $M$ for which $\delta_M$ is finite for all $p$ interesting? Is there some kind of a finiteness condition they satisfy? (Other than the one that this means directly). Maybe with some additional hypetheses? (projective?)</p> </blockquote> http://mathoverflow.net/questions/69420/the-rank-of-a-not-necessarily-finitely-generated-module/69434#69434 Answer by Tom Goodwillie for The rank of a not necessarily finitely generated module. Tom Goodwillie 2011-07-04T01:35:02Z 2011-07-04T01:35:02Z <p>Question 1: No. Let $M=\oplus_p\kappa(p)$.</p> http://mathoverflow.net/questions/69420/the-rank-of-a-not-necessarily-finitely-generated-module/69504#69504 Answer by Hailong Dao for The rank of a not necessarily finitely generated module. Hailong Dao 2011-07-04T23:06:05Z 2011-07-04T23:06:05Z <p>Here are a few comments, too long to fit in the comments box. </p> <p>1) One can modify Tom Goodwillie's example as follows: for simplicity lets pick $R$ to be a local domain of dimension $1$ (so there are only $2$ prime ideals, $0$ and $\mathfrak m$, with residues $K$, the quotient field and $k$, the residue field of $R$ respectively). Then any module $M$ that fits into a short exact sequence:</p> <p>$$0 \to k \to M \to K \to 0$$ would satisfy: $\delta_M(0)= \delta_M(\mathfrak m) =1$. But $M_{\mathfrak m} =k$ is not a free $R_{\mathfrak m}=R$-module.</p> <p>Note that one can not try the exact sequence with $k, K$ swapped, since $Ext^1_R(k,K)=0$. On the other hand $Ext^1(K,k)$ is complicated, since projective resolutions of $K$ <a href="http://www.jstor.org/pss/1994890" rel="nofollow">depend on the continuum hypothesis</a>! </p> <p>2) To weed out examples like above, perhaps more relevant than indecomposability or projectivity is to require $M$ to be <em>torsion-free</em>, so $M$ injects into $M\otimes Q(R)$ ($Q(R)$ is the total ring of quotients). </p> <p>Say we assume this and the "rank" is $1$. Then immediately we know that $M$ is a submodule of $Q(R)$, and this more or less describes $M$: at every prime $p$, once we tensor with $\kappa (p)$, basically only one denominator of $M$ (that is not in $R$) is left. When $R=\mathbb Z$, one can see the example in your 2) very clear from this point of view.</p> <p>3) Finally, I am not sure we should call $\delta_M(p)$ the local "rank". When $M$ is finitely generated, $\delta_M(p)$ is the <em>minimal number of generators of $M$ locally at $p$</em>. Of course, if $M$ is locally free, it will be the rank, but we definitely do not want to assume that.</p> <p>This perhaps explains why the easy counter-examples by me and Tom Goodwillie exist: $M$ can locally have a constant number of generators, but it generally does not mean $M$ is locally free. </p>