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So this was a question I sort of stumbled on and realised I was quite stumped. Suppose we have two finitely generated $R$-modules $M, N$ (I have the group ring $R=\mathbb{Z}[G]$ in mind) which appear in short exact sequences

$$0\to X\to M\to Y\to 0,$$

$$0\to X\to N\to Y\to 0.$$

I know that $M$ and $N$ are indecomposable and I know that ${\rm Ext}^1(Y, X)\cong\mathbb{Z}/2 \Bbb Z$. Are $M$ and $N$ isomorphic?

If I can find a homomorphism between them which fits into a commutative diagram, then obviously this is true. But must such a homomorphism exist?

Improve this question
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    $\begingroup$ @LSpice yes, sorry. I should have thought more on the wording of this question. First time posting, so I didnt think. My own stupid fault. $\endgroup$
    – DJWilliams
    Commented Jun 13, 2020 at 19:20
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    $\begingroup$ So $M$ and $N$ are indecomposable. In particular these short exact sequences are nonsplit, so they correspond to non-zero elements of $\operatorname{Ext}^1(Y,X)$. If $\operatorname{Ext}^1(Y,X) \cong \mathbb{Z}/2\mathbb{Z}$, this means that these two sequences are equivalent, and in particular $M$ and $N$ are isomorphic. $\endgroup$
    – spin
    Commented Jun 13, 2020 at 19:36
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    $\begingroup$ Well, $x \mapsto -x$ is the identity map on $\mathbb{Z}/2\mathbb{Z}$, so being in inverse classes is the same as being in the same class! Perhaps I'm misunderstanding what you mean? The important point is that $M$ and $N$ being indecomposable implies that neither extension corresponds to the identity element in $\text{Ext}^1$, and there is only one non-identity element in $\mathbb{Z}/2\mathbb{Z}$. Small note: this implication isn't entirely trivial, you need to use the fact that if $X = 0$ or $Y = 0$ then we would have $\text{Ext}^1(Y,X) = 0 \neq \mathbb{Z}/2\mathbb{Z}$. $\endgroup$
    – diracdeltafunk
    Commented Jun 14, 2020 at 6:24
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    $\begingroup$ Ah, yes, the assumption that $\text{Ext} \cong \mathbb{Z}/2$ is very strong and it is what allows us to conclude that any two non-split extensions are equivalent. If $\text{Ext} \cong \mathbb{Z}/3$ this will not be the case, however, any two non-split extensions will still be isomorphic, as you suspected! This is because if $0 \to X \xrightarrow{f} Z \xrightarrow{g} Y \to 0$ represents an element $a \in \text{Ext}^1(Y,X)$, then $-a$ is represented by $0 \to X \xrightarrow{-f} Z \xrightarrow{g} Y \to 0$ (check this by computing the Baer sum of these two extensions and verifying it splits) $\endgroup$
    – diracdeltafunk
    Commented Jun 14, 2020 at 21:20
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    $\begingroup$ (cont.) However, for $\lvert \text{Ext}^1 \rvert > 2$ we can't use this argument anymore, because even after taking the quotient of $\text{Ext}$ by $a \sim -a$ we will have more than one nontrivial class. In general, though, I agree that $\text{Ext}$ is a spookily powerful tool: that's why it's so popular and deeply studied! Anyway, the implication I was talking about was "if $M$ and $N$ are indecomposable then neither extension corresponds to the identity element of $\text{Ext}^1$". This is true, but only because we had also assumed $\text{Ext}^1 \neq 0$. Still very simple, but not immediate. $\endgroup$
    – diracdeltafunk
    Commented Jun 14, 2020 at 21:32

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