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?

  • 1
    $\begingroup$ Take $R = \Bbb{Z}$, $n = 1$. Then take $m=1$, and $R = 2\Bbb{Z} \cong \Bbb{Z}$. Then $\Bbb{Z}/2\Bbb{Z}$ is the quotient and not $\Bbb{Z}^{1-1}$. $\endgroup$
    – user38268
    Commented Nov 10, 2012 at 14:18
  • $\begingroup$ @BenjaLim Thanks for your quick example. I realized that 1) and 2) don't quite patch together the way I thought they did to imply that false statement. All we can conclude is that $R^n/R^m \cong R^l$ for some $l\leq n.$ Can you see why the "proof" I have is incorrect though? $\endgroup$ Commented Nov 10, 2012 at 14:24
  • $\begingroup$ Why is $M$ finitely generated? $\endgroup$
    – user38268
    Commented Nov 10, 2012 at 14:29
  • $\begingroup$ @BenjaLim That is the assumption. $\endgroup$ Commented Nov 10, 2012 at 14:29
  • $\begingroup$ No, you want to prove that if $M$ is a submodule of $R^n$ then $M$ is finitely generated, free of rank less than or equal to $n$. Where in the assumption above is $M$ finitely generated? $\endgroup$
    – user38268
    Commented Nov 10, 2012 at 14:31

3 Answers 3


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.

  • $\begingroup$ It appears I'm not appreciating when honest equality is important and when isomorphism is good enough. If I replace $R^n = N \oplus N'$ by $R^n \cong N \oplus N'$ then it seems to be still be a valid result. Is the moral here that to be a semi simple module, every submodule must have a complementary submodule such that their direct sum is equal to the whole module, and mere isomorphism is not good enough? It seems odd that the property of semi-simplicity is not isomorphism invariant. $\endgroup$ Commented Nov 13, 2012 at 14:08
  • $\begingroup$ Well, yes, being isomorphic to the whole module is not enough for submodules to be equal to it. Being off by finite index in the case of $2\mathbb Z\subset \mathbb Z$ is the deal-killer her. Semi-simplicity is isomorphism-invariant, but $\mathbb Z^n$ is not semi-simple. Very few modules over non-field PIDs are semi-simple. The structure theorem implies that only $\mathbb Z$-modules of the form $\mathbb Z/p_1\oplus\ldots\oplus \mathbb Z/p_n$ with distinct primes $p$ are semi-simple, among finitely-generated $\mathbb Z$-modules. $\endgroup$ Commented Nov 13, 2012 at 16:02
  • $\begingroup$ Ok I think I understand everything now. I always had it in my head that for all algebraic intents and purposes, isomorphism was as good as equality, but obviously the whole definition of semi-simplicity isn't even really meaningful with just isomorphism so that is a good example to contradict that principle. Thank you very much for your answer! I've accepted your answer but the system tells me I can only award the bounty 9 hours from now, so I'll come back later to do that. $\endgroup$ Commented Nov 14, 2012 at 3:23

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.


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 $R^m$...".

  • $\begingroup$ Thanks for your example (the "why" is because torsion elements can't be part of a basis). There is a decent chance I have confused some things in my argument, but I'm having trouble seeing exactly where. Each little step seems solid to me, but the final result is ridiculous. $\endgroup$ Commented Nov 10, 2012 at 14:32
  • $\begingroup$ @KatieDobbs I don't know what torsion elements are, I'm sorry. Could you state the explanation without torsion elements? (It's a (simple) statement of the definition of linear independence of basis elements.) $\endgroup$ Commented Nov 10, 2012 at 14:34
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    $\begingroup$ Sorry, I'm quite new learning this so I don't know what terminology in my book is common or not. Basically, every element satisfies a non-trivial R linear dependence relation, since $nm=0$ for any $m\in \mathbb{Z}/n\mathbb{Z}.$ $\endgroup$ Commented Nov 10, 2012 at 14:40
  • $\begingroup$ @KatieDobbs Example of torsion free but not finitely generated: $\Bbb{Q}$ as an abelian group. Example of finitely generated but not free: any finite abelian group. $\endgroup$
    – user38268
    Commented Nov 10, 2012 at 14:41
  • 1
    $\begingroup$ @KatieDobbs Yes, excellent, that's what I wanted to hear. $\endgroup$ Commented Nov 10, 2012 at 14:41

Another approach to the problem your wrote in bold would be to combine these facts:

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

  2. A PID is a Noetherian ring.

  3. 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...)

  • $\begingroup$ I think I've managed to prove the fact I want in another way. When I first attempted this problem I thought of a certain proof, but it seems to be dodgy because it proves something too strong, and I want to know at which step it fails. I understand the text is a bit dense, you can understand my question if you only read from "That proof seems to be fine, but my first "proof" was somewhat different...". There are two preliminary facts, 1) and 2), that are external to the problem at hand (but would still like checking on them) and then I use these facts in my argument in steps a), b) and c). $\endgroup$ Commented Nov 10, 2012 at 17:44
  • $\begingroup$ @KatieDobbs That's OK, I understand :) I thought the other posts had probably addressed that, so I wanted to throw out another approach that might shed some light on other approaches. Good luck! $\endgroup$
    – rschwieb
    Commented Nov 11, 2012 at 0:25

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