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Let $R$ be a ring and let $M$ and $N$ be $R$-modules with submodules $L$ and $P$, respectively. Assume we know $M \cong N$ as $R$-modules. Further, suppose we are given an isomorphism

$$\psi: L \to P$$

When is it true that there exists an isomorphism $\phi: M \to N$ such that $\forall m \in L$ we have

$$\phi(m)=\psi(m).$$

One instance where this holds is when there exist direct sum decompositions $M=L \oplus L'$ and $N=P \oplus P'$, i.e. when $L$ and $P$ are direct summands of their respective modules.

Is this condition actually necessary for the two isomorphisms to agree, or are there some more general conditions under which this property holds?

Exit path
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    Your condition is not sufficient: for instance, take $R$ to be a field, $M=N=L$ to be an infinite-dimensional vector space, and $P$ to be a proper subspace of $N$ of the same dimension. – Eric Wofsey Jan 31 '16 at 00:38

3 Answers3

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Take $R = M = N = L = \mathbb{Z}, P = 2 \mathbb{Z}$ for a fairly simple example where no such isomorphism exists. I don't see a useful sufficient condition.

An equivalent question can be stated using fewer modules: suppose $M$ is a module. When does $\text{Aut}(M)$ act transitively on the set of all submodules of $M$ abstractly isomorphic to a given submodule $L$? The only interesting example that comes to mind is the case that $M$ is a finite-dimensional vector space.

Qiaochu Yuan
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  • This isn't a counterexample since $\psi$ doesn't extend to an isomorphism (so it doesn't matter that $2\mathbb{Z}$ is not a direct summand of $\mathbb{Z}$). – Noah Schweber Jan 31 '16 at 00:39
  • @Noah: it wasn't meant to be a counterexample, just an example that shows why there's no reason to expect statements like this to be true. – Qiaochu Yuan Jan 31 '16 at 01:09
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It is not necessary: take $R=\mathbb{Z}$, $M=N=\mathbb{Q}$, with the usual structure; and let $L=\mathbb{Z}$, $P=2\mathbb{Z}$ with $\psi:x\mapsto 2x$ the obvious module isomorphism. Then $\psi$ extends to an appropriate automorphism of $\mathbb{Q}$, while $\mathbb{Q}$ does not decompose as a direct sum at all (exercise).

Noah Schweber
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Your condition is not even sufficient.

Consider a module $A$ such that $A\cong A\oplus A$; let $g\colon A\to A\oplus A$ be an isomorphism.

Now take $L=A$, $L'=A$, $M=L\oplus L'$, $P=A\oplus A$, $P'=0$, $N=P\oplus P'$. Define $f\colon L\to P$ as $g$. Then you have an isomorphism $f\colon L\to P$ that obviously cannot be extended to an isomorphism $M\to N$.

Note that there are (noncommutative) rings $R$ such that $R\cong R\oplus R$ as right (or left) modules, so the above counterexample can also be with finitely generated modules. For commutative rings, just take a free module over an infinite basis.

egreg
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