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I want to prove the following statement:

Let $M, L$ be $R$ - modules. If for every $(N, \phi: N \to L)$, where $N$ is a finitely generated submodule of $M$, $\phi$ can be extended to a map from $M \to L$, then any arbitrary submodules of $M$ satisfy this as well.

My strategy: This reminds me of the Baer's criterion for injective module. Take an arbitrary submodule $N$, and $g: N \to L$, the idea for the proof is to consider all pairs $(\tilde N, f: \tilde N \to L)$, where $N \subset \tilde N$ is a submodule of $M$ and $f$ is the extension of $g$. Then use Zorn's lemma, we can find a maximal such element. Need to show this maximal element is $M$. Then I get stuck. Is this doable?

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This is false. For instance, let $k$ be a field $R=k\oplus V$ where $V$ is an infinite-dimensional $k$-vector space, with multiplication on $R$ defined by $(a,v)\cdot(b,w)=(ab,aw+bv)$. Let $M=R$ and construct $L$ as follows. Start with $L_0=V$, which is an ideal of $R$. For each finite dimensional subspace $W\subset V$ and each $R$-linear homomorphism $f:W\to L_0$, pick a linear complement $W^\perp$ to $W$ inside $V$, adjoin a generator to $L_0$ which is annihilated by $W^\perp$ and which is acted on by $W$ according to $f$. Let $L_1$ be the module obtained by adjoining all these new generators. Repeat the same process, adjoining generators to $L_1$ to obtain a module $L_2$, and so on. Let $L$ be the direct limit of the $L_n$.

By construction, if $N$ is any finitely generated submodule of $M$, any homomorphism $N\to L$ extends to $M$. Indeed, if $N$ is not $0$ or all of $M$, then $N$ is a finite-dimensional subspace of $V$, and then the image of our homomorphism $N\to L$ is contained in some $L_n$ and by construction we can extend it to a homomorphism $M\to L_{n+1}$. On the other hand, note that every finitely generated submodule of $L$ is finite-dimensional over $k$ (by construction, every element of $L_0$ and every new generated we adjoined is annihilated by a complement of some finite-dimensional subspace of $V$ and thus generates a finite-dimensional submodule). Thus the identity map $V\to V=L_0\subset L$ cannot be extended to a homomorphism $M\to L$, since if it were then its image would be a cyclic submodule of infinite dimension.


Here's another way to get a counterexample. Let $R$ be any von Neumann regular ring which is not semisimple (for instance, an infinite Boolean ring). Then every finitely generated ideal in $R$ is a direct summand, but there are non-finitely generated ideals which are not. So, taking $M=R$ and $L$ to be a non-finitely generated ideal, every homomorphism from a finitely generated submodule of $M$ extends to all of $M$, but the identity map $L\to L$ does not extend to $M$.

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