most recent 30 from http://mathoverflow.net 2013-05-18T23:55:06Z http://mathoverflow.net/feeds/question/30066 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/30066/cardinality-of-maximal-linearly-independent-subset ashpool 2010-06-30T15:54:33Z 2010-07-22T19:05:16Z

<p>M a finitely generated module over a commutative ring A. I can't think of an example of two maximal linearly independent subsets of M having different cardinality. I know that they all have the same cardinality if A is integral domain. Any suggestions are welcome!</p>

http://mathoverflow.net/questions/30066/cardinality-of-maximal-linearly-independent-subset/30070#30070 Arturo Magidin 2010-06-30T16:27:08Z 2010-06-30T16:45:46Z

<p>If you consider rings that are not necessarily commutative, here's an example: let $V$ be a countable dimensional vector space over a field $F$, and let $A$ be the ring of all endomorphisms of $A$. I claim that $A\cong A\oplus A$ (as left $A$-modules); if so, then using (and iterating) this isomorphism you can find maximal linearly independent subsets of any finite cardinality. </p> <p>To see that $A\cong A\oplus A$, it suffices to exhibit a two-element $A$-basis for $A$. Let $e_1,e_2,\ldots$ be a basis for $V$. Let $f_1\in A$ be the endomorphism that maps $e_2,e_4,e_6,\ldots$ to $e_1,e_2,e_3,\ldots$, respectively, and maps every odd-indexed basis element to $0$; let $f_2\in A$ be the endomorphism that maps $e_1,e_3,e_5,\ldots$ to $e_1,e_2,e_3,\ldots$, and maps the even-indexed basis elements to $0$. Then $f_1,f_2$ spans $A$: if $\varphi \in A$, then we can write $\varphi$ as $\varphi=gf_1+hf_2$, where $g(e_i)=\varphi(e_{2i})$ and $h(e_j)=\varphi(e_{2j-1})$. To see that $f_1$ and $f_2$ are $A$-linearly independent, suppose that $af_1+bf_2=0$; evaluating at the odd indexed $e_i$ shows that $b(e_j)=0$ for all $j$, and evaluating at the even indexed $e_i$ shows $a(e_j)=0$ for all $j$. Thus, $f_1,f_2$ is also a basis for $A$, which gives an isomorphism $A\cong A\oplus A$. Being bases, they are certainly maximal linearly independent sets. </p>

http://mathoverflow.net/questions/30066/cardinality-of-maximal-linearly-independent-subset/30369#30369 ashpool 2010-07-03T03:26:49Z 2010-07-22T19:05:16Z

<p>I found an old paper by Lazarus (Les familles libres maximales d'un module ont-elles le meme cardinal?, Pub. Sem. Math. Rennes 4 (1973), 1-12) which contains the the following result: Let A be a commutative ring with unit and M an A-module. In the following situations, maximal linearly independent subsets of M have the same cardinality:</p> <ol> <li><p>If M is a free A-module of infinite rank.</p></li> <li><p>If A is reduced and has only finitely many minimal primes (e.g. integral domain, reduced Noetherian ring)</p></li> <li><p>If A is Noetherian and M is a free A-module.</p></li> <li><p>If A is Noetherian and M is a submodule of a free A-module of finite rank.</p></li> <li><p>If A is Noetherian and M has an infinite linearly independent subset.</p></li> <li><p>If A is Noetherian and M is a submodule of a flat A-module.</p></li> <li><p>If A is Artin local and the zero ideal $(0)\subset A$ is irreducible.</p></li> </ol> <p>And the examples given in the paper of modules not satisfying this same cardinality property are highly nontrivial.</p>