Example of a projective module which is not a direct sum of f.g. submodules? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T06:03:54Z http://mathoverflow.net/feeds/question/55704 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55704/example-of-a-projective-module-which-is-not-a-direct-sum-of-f-g-submodules Example of a projective module which is not a direct sum of f.g. submodules? Pete L. Clark 2011-02-17T06:45:31Z 2011-02-18T17:18:30Z <p>This semester I am teaching a graduate course in commutative algebra, and I have been taking the occasion to try to look at the proofs of some the results in my basic source material (Matsumura, Eisenbud, Bourbaki...) which I skipped as a little too complicated the first $N$ times around. </p> <p>Recently I got the chance to read and understand I. Kaplansky's big theorem on projective modules, i.e., that a(n even infinitely generated) projective module over a local ring is free. En route to establishing this, he proves another result which is interesting but rather technical:</p> <p>Theorem (Kaplansky, 1958): Every projective module is a direct sum of countably generated projective submodules.</p> <p>For my take on this result, see $\S 3.10$ <a href="http://math.uga.edu/~pete/integral.pdf" rel="nofollow">of these notes</a>. In particular, it raises two natural questions:</p> <blockquote> <p>Question 1: Is there a ring $R$ and an $R$-module $M$ which is <em>not</em> a direct sum of countably generated submodules?</p> <p>Question 2: Is there a ring $R$ and a projective $R$-module $P$ which is <em>not</em> a direct sum of finitely generated submodules?</p> </blockquote> <p>I was able to look up that the answer to Question 1 is "yes". In particular, I found work of L. Fuchs which says that for every infinite cardinal $\kappa$ there is an indecomposable (i.e., not expressible as a nontrivial direct sum) commutative group $G$ of cardinality $\kappa$. I would however be interested in hearing other examples or other takes on Question 1.</p> <p>My real question is Question 2: presumably the answer is either <em>yes</em> or <em>unknown</em>, or people would mention the stronger result when Kaplansky's Theorem is discussed. A theorem of Bass that M. Reyes pointed out to me in his answer to another recent question of mine on modules is relevant in this regard: obviously an affirmative answer to Question 2 must involve an infinitely generated projective module, and if $R$ is Noetherian and connected then every infinitely generated projective module is free, hence a direct sum of singly generated submodules! </p> http://mathoverflow.net/questions/55704/example-of-a-projective-module-which-is-not-a-direct-sum-of-f-g-submodules/55706#55706 Answer by Mariano Suárez-Alvarez for Example of a projective module which is not a direct sum of f.g. submodules? Mariano Suárez-Alvarez 2011-02-17T06:59:44Z 2011-02-17T06:59:44Z <p>[Warfield, Robert B., Jr. Rings whose modules have nice decompositions. Math. Z. 125 1972 187--192. <a href="http://www.ams.org/mathscinet-getitem?mr=MR0289487" rel="nofollow">MR0289487</a> (44 #6677)] shows that over commutative Artinian rings with non-principal ideals, there exist indecomposable modules which are not countably generated. This answers (1).</p> http://mathoverflow.net/questions/55704/example-of-a-projective-module-which-is-not-a-direct-sum-of-f-g-submodules/55707#55707 Answer by Gjergji Zaimi for Example of a projective module which is not a direct sum of f.g. submodules? Gjergji Zaimi 2011-02-17T07:06:34Z 2011-02-17T07:06:34Z <p>For question two the example that is given most frequently seems to be that of the ring $R$ of continuous real valued functions on $[0,1]$ and the ideal of all functions $f$ which vanish on some interval $[0,\epsilon(f)]$ where $\epsilon(f)\in (0,1)$. This ideal is countably generated and projective but not a direct sum of finitely generated submodules. You might also want to take a look at the article "When every projective module is a direct sum of finitely generated modules" by W. McGovern, G. Puninski and P. Rothmaler.</p> http://mathoverflow.net/questions/55704/example-of-a-projective-module-which-is-not-a-direct-sum-of-f-g-submodules/55880#55880 Answer by Pete L. Clark for Example of a projective module which is not a direct sum of f.g. submodules? Pete L. Clark 2011-02-18T17:18:30Z 2011-02-18T17:18:30Z <p>I just wanted to add a little bit to the argument given in Lam's book (as communicated by Gjergji Zaimi). Let $R = C([0,1],\mathbb{R})$ be the ring of real-valued continuous functions on the closed unit interval, and let $I$ be the ideal of all functions $f$ which vanish identically on some neighborhood of $0$. Lam shows that $I$ is projective (a nice application of the Dual Basis Lemma) and also not free: indeed, he remarks that every element $f$ of $I$ has a nontrivial annihilator -- namely any nonzero function with support contained in the zero set of $f$ -- whereas for a free $R$-module $\bigoplus_{i \in I} R$ any standard basis element $e_i$ clearly has zero annihilator.</p> <p>What Lam does not address -- as far as I can see -- is why $I$ is moreover not a direct sum of finitely generated submodules. But here is a nice argument for this using Swan's Theorem: we are asking whether the projective module $I$ is a direct sum of finitely generated projective modules. But every finitely generated projective module over $R$ corresponds to a vector bundle over $[0,1]$. However, since $[0,1]$ is contractible, every vector bundle over $[0,1]$ is trivial, and thus every finitely generated projective $R$-module is free. Thus, if $I$ were a direct sum of finitely generated submodules, it would itself be free, which we previously saw is not the case.</p> <p>I'm sure there's also a purely algebraic proof of this, but I am very fond of Swan's Theorem...</p>