Error in proof that submodules of f.g. modules are f.g. All modules are over a ring $R$ which is a PID. I wanted to show that every submodule of a finitely generated module is finitely generated. A key to this problem seems to be the following theorem (not too hard to prove by induction) :
Every submodule $M$ of $R^n$ has a basis with at most $n$ elements. That is, $M \cong R^m$ for $m\leq n.$
Now, my latest attempt at a proof goes like this: Let $M$ be our finitely generated submodule, generated by $x_1, \cdots, x_n.$ Then the R-module homomorphism $\phi: R^n \to M$, $ (r_1, \cdots, r_n) \mapsto r_1 x_1 + \cdots + r_n x_n$ is surjective. Suppose $N$ is a submodule of $M$; we want to show that $N$ is finitely generated. The preimage of $N$ under $\phi$ is a submodule of $R^n$, so by our theorem, $\phi^{-1}(N)$ has a basis $y_1, \cdots, y_m$ for some $m\leq n.$ Now, $N=\phi( \phi^{-1}(N))$ so N is generated by $\phi(y_1),\cdots \phi(y_m).$
That proof seems to be fine, but my first "proof" was somewhat different:
1) $R^n$ is semi-simple: Every submodule $N$ of $R^n$ is isomorphic to $R^m$ for some $m\leq n$ so letting $N'= R^{n-m}$ we have $R^n= N \oplus N'$. 
2) If $N$ is a submodule of $M$ then there is another submodule $N'$ such that $M= N \oplus N'$ so $M/N \cong N',$ which allows us to identify submodules and quotient modules.
The argument goes like this: 
a) Suppose $M$ is a finitely generated $R$ module, generated by say $n$ elements, so $M \cong R^n/\ker \phi$ where $\phi$ is the same homomorphism in the new proof above.
b) By 1) and 2) $R^n/\ker \phi$ is isomorphic to some submodule of $R^n$, and that submodule is isomorphic to $R^m$ for some $m\leq n$ by our theorem.
c) Joining these up, we get $M\cong R^m$ *for some $m\leq n.$ Now by our theorem, any submodule $N$ of $M$ is isomorphic to $R^k$ for $k\leq m$ and thus finitely generated. 
Now, the claims seem ridiculous to me starting around where I put the asterisk. In particular, it would appear to prove that any finitely generated module is free. I can't think of a counterexample over a PID right now, but that seems like it can't be true. What have I done wrong?
 A: Not every finitely generated module is free. For example, $\mathbb Z / n \mathbb Z$ is finitely generated by $1$ over $\mathbb Z$ but it does not have a basis since any set of elements of it is linearly independent. (Do you see why?)
As for the proof that every submodule of a f.g. $R$-module is again finitely generated: see for example here for a proof.
In your proof it seems to me that you are confusing free and finitely generated. Could that be the case?
I might add the outline of the argument later. Hope this helps.
Edit
Regarding your proof: you cannot write "Let $M$ be our finitely generated submodule..." since that is not true. It is a random assumption. You don't know that $M$ is finitely generated. The statement you want to show is: for any submodule $M$ of $R^n$ you have $M \cong R^m$. Hence your proof would have to start "Let $M$ be any submodule of $R^m$...".
A: Your "preliminary fact" 1) is not correct: modules (free or not) over PIDs are not usually semi-simple. For example, the $R=\mathbb Z$-module $M=\mathbb Z$ has submodule $N=2\mathbb Z$ which is "uncomplemented" in the sense that there is no other submodule $N'$ such that $M=N\oplus N'$. The failure of this "fact" is what begins your troubles.
A: Another approach to the problem your wrote in bold would be to combine these facts:


*

*A module is Noetherian iff all of its submodules are finitely generated.

*A PID is a Noetherian ring.

*A finitely generated module over a Noetherian ring is a Noetherian module.
I'm betting you might be willing to take one or two for granted at this point. But even if that's not the case, they are all good facts to know, so it would be worth your time to work them out.
(Has this appeared above? The text is rather dense...)
