# Extending maps from a submodule

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?

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 $$M$$.