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(Edit: 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.)

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

\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}

Must there be an isomorphism between these two short exact sequences?


Alternative formulation

$Ext^1(M,N)$ parameterizes extensions of $M$ by $N$ modulo isomorphims of extensions. Suppose I'm interested in parameterizing extensions of $M$ by $N$ modulo abstract isomorphisms (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.

Does the quotient set $Aut(M)\backslash Ext^1(M,N)/Aut(N)$ parameterize extensions of $M$ by $N$ modulo abstract isomorphism?

Note: I'm not 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 there exists some isomorphism between them which is generated by $Aut(M)$ and $Aut(N)$.

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There are currently three answers to this question, and I want to accept all of them! –  Anton Geraschenko Oct 15 '10 at 18:07
    
Are you going to let little things like rules and ethics stop you? Go for it! Gerhard "Ask Me Not About Propriety" Paseman, 2010.10.15 –  Gerhard Paseman Oct 15 '10 at 21:24

3 Answers 3

up vote 12 down vote accepted

It is worth noting some very interesting cases when the answer is yes. An amazing result by Miyata 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!

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.

Added: in fact, this paper by Janet Striuli 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 $I=0$).

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I see ... my $E$ represents two different elements $E_1$ and $E_2$ of $Ext^1(M,N)$ which happen to be abstractly isomorphic. The abstract isomorphism induces an isomorphism between the trivial extension and the "Yoneda quotient" (difference in $Ext^1$) of $E_1$ and $E_2$. Applying Miyata's result, you get that the "Yoneda quotient" of $E_1$ and $E_2$ splits, and this splitting induces the isomorphism of exact sequences I was looking for. This is great! Thank you for this answer! –  Anton Geraschenko Oct 15 '10 at 18:03
    
Wait a second. Greg Muller's example $R=\mathbb{C}[x]$ is commutative and noetherian, and the modules are finitely generated. Am I misunderstanding Miyata's result, or am I misunderstanding Greg's example? –  Anton Geraschenko Oct 16 '10 at 6:51
    
@Anton: Miyata result only applies for $E=M\oplus N$. It does not prove it for all $E$ over Noetherian commutative ring. –  Hailong Dao Oct 16 '10 at 7:06
    
@Anton: to be precise, I don't think your first comment work. The fact that the middle modules of $\alpha, \beta$ are isomorphic does not imply that $\alpha -\beta$ is isomorphic to the trivial extension. If that is the case we can concludes that $\alpha = \beta$ already! –  Hailong Dao Oct 16 '10 at 7:22
2  
Ah, I think I see the problem. You have answered my original question (I think my first comment really does work). Greg has answered my "alternative formulation". The problem is that the two formulations are not equivalent after all. I overlooked the fact that it's possible to have an automorphism of $E$ as an extension ... an automorphism which induces the identity maps on both $M$ and $N$. So you (and Miyata) are correct: Greg's extension does split. Greg is also correct: the isomorphism with the direct sum is not induced by isomorphisms of $M$ and $N$. –  Anton Geraschenko Oct 18 '10 at 1:28

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 $$ M = \left[\begin{array}{cc} 0 & 1 \\\ 0 & 0 \end{array}\right], N = [0] $$ be modules of dimension 2 and 1, respectively. Then extensions of $N$ by $M$ correspond block diagonal matrices of the form $$ \left[ \begin{array}{cc} N & C \\\ 0 & M \end{array}\right] $$ 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 $C$.

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.

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1  
This is a wonderful example! Thanks. –  Anton Geraschenko Oct 14 '10 at 18:10

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 $A^\infty\oplus E^\infty\oplus B^\infty$.

(Here $(\mathord-)^\infty$ denotes the countable direct sum of its argument)

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This is also a wonderful example! –  Anton Geraschenko Oct 14 '10 at 18:13

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