Can a module be an extension in two really different ways?

Question

^{(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 $Aut(N)$.

Answer 1

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 $I=0$).

Answer 2

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)

Answer 3

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.