Can a module be an extension in two really different ways? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T05:58:42Z http://mathoverflow.net/feeds/question/42174 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42174/can-a-module-be-an-extension-in-two-really-different-ways Can a module be an extension in two really different ways? Anton Geraschenko 2010-10-14T16:56:01Z 2010-10-18T01:37:58Z <p><sup>(<strong>Edit:</strong> I've realized that there was an error in my reasoning when I was convincing myself that these two formulations are equivalent. Hailong has given a beautiful affirmative answer to my first question in the case of finite type modules over a noetherian commutative ring. Mariano has given a slick negative answer to the question for non-finite-type modules. Greg has given a beautiful negative answer to my "alternative formulation" even in the finite type case over a noetherian commutative ring. I'm accepting Hailong's answer since that's the one I imagine people will be most immediately interested in if they find this question in the future.)</sup></p> <p>Suppose we're working the category of modules over some ring $R$. Suppose a module $E$ is an extension of $M$ by $N$ in two different ways. In other words, I have two short exact sequences</p> <p> \begin{array}{ccccccccc} 0&\to &N&\xrightarrow{i_1}&E&\xrightarrow{p_1}&M&\to &0\\ & & \wr\downarrow ?& & \wr\downarrow ?& & \wr\downarrow ?\\ 0&\to &N&\xrightarrow{i_2}&E&\xrightarrow{p_2}&M&\to &0 \end{array} </p> <blockquote> <p>Must there be an isomorphism between these two short exact sequences?</p> </blockquote> <hr> <h2>Alternative formulation</h2> <p>$Ext^1(M,N)$ parameterizes extensions of $M$ by $N$ modulo isomorphims <em>of extensions</em>. Suppose I'm interested in parameterizing extensions of $M$ by $N$ modulo <em>abstract isomorphisms</em> (which don't have to respect the submodule $N$ or the quotient $M$). One obvious thing to note is that there is a left action of $Aut(M)$ on $Ext^1(M,N)$, and that any two extensions related by this action are abstractly isomorphic. Similarly, there is a right action of $Aut(N)$ so that any two extensions related by the action are abstractly isomorphic.</p> <blockquote> <p>Does the quotient set $Aut(M)\backslash Ext^1(M,N)/Aut(N)$ parameterize extensions of $M$ by $N$ modulo abstract isomorphism?</p> </blockquote> <p>Note: I'm <strong>not</strong> asking whether all abstract isomorphisms are generated by $Aut(M)$ and $Aut(N)$. They certainly aren't. I'm asking whether for every pair of abstractly isomorphic extensions <em>there exists</em> some isomorphism between them which is generated by $Aut(M)$ and $Aut(N)$.</p> http://mathoverflow.net/questions/42174/can-a-module-be-an-extension-in-two-really-different-ways/42179#42179 Answer by Greg Muller for Can a module be an extension in two really different ways? Greg Muller 2010-10-14T17:22:31Z 2010-10-14T17:28:19Z <p>I believe this is a counter example. Let $R=\mathbb{C}[x]$, and consider finite-dimensional modules (ie, f.d. vector spaces equipped with a distinguished endomorphism). For convenience, I will identify a module with a matrix, implicitly choosing a basis. Let $$M = \left[\begin{array}{cc} 0 &amp; 1 \\ 0 &amp; 0 \end{array}\right], N = [0]$$ be modules of dimension 2 and 1, respectively. Then extensions of $N$ by $M$ correspond block diagonal matrices of the form $$\left[ \begin{array}{cc} N &amp; C \\ 0 &amp; M \end{array}\right]$$ where $C$ is some $1\times 2$-matrix. Since the automorphisms of $M$ and $N$ act as conjugation by the appropriate matrix, we see that they preserve the rank and nullity of $C$. </p> <p>Now, note the two extensions $$C =\left[ 0 \; 0 \right],\;\; C' = \left[ 0 \; 1\right]$$ give isomorphic extensions (ie, conjugate matrices), but $C$ and $C'$ have different ranks.</p> http://mathoverflow.net/questions/42174/can-a-module-be-an-extension-in-two-really-different-ways/42184#42184 Answer by Mariano Suárez-Alvarez for Can a module be an extension in two really different ways? Mariano Suárez-Alvarez 2010-10-14T17:50:47Z 2010-10-14T18:36:28Z <p>Silly example: pick any non-split extension $$\mathcal E:0\to A\to E\to B\to0$$ and consider the boring extension $$\mathcal F:0\to A^\infty\oplus E^\infty\oplus B^\infty\to A^\infty\oplus E^\infty\oplus B^\infty\to 0\to 0$$ whose non-zero map is an identity. Then the sequence $\mathcal E\oplus\mathcal F$ is not split, yet the modules which appear in it are the same ones that appear in the split extension of $B$ by $A^\infty\oplus E^\infty\oplus B^\infty$.</p> <p>(Here $(\mathord-)^\infty$ denotes the countable direct sum of its argument)</p> http://mathoverflow.net/questions/42174/can-a-module-be-an-extension-in-two-really-different-ways/42309#42309 Answer by Hailong Dao for Can a module be an extension in two really different ways? Hailong Dao 2010-10-15T17:09:43Z 2010-10-15T20:59:36Z <p>It is worth noting some very interesting cases when the answer is yes. An amazing <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.kjm/1250524308" rel="nofollow">result by Miyata</a> states that if $R$ is Noetherian and commutative, $M,N$ are finitely generated and $E \cong M\oplus N$, any exact sequence $0 \to M \to E \to N \to 0$ must split! </p> <p>This holds true slightly more generally, when $R$ is (not necessarity commutative) module-finite over a Noetherian commutative ring. Also, the statement holds for finitely generated pro-finite groups, see Goldstein-Guralnick, J. Group Theory 9 (2006), 317–322. </p> <p>Added: in fact, this <a href="http://www.math.unl.edu/~jstriuli2/research/module2.pdf" rel="nofollow">paper by Janet Striuli</a> may be useful for you. She addressed the question: if two elements $\alpha, \beta \in \text{Ext}^1(M,N)$ give isomorphic extension modules, how close must $\alpha, \beta$ be? Her Theorem 1.2 extend Miyata's result (let $I=0$). </p>