MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let (A,m) be a local ring and M be a finitely generated A-module contained in a free module F of rank r with length(F/M) < $\infty$. Then I have the following question : Is the statement "M doesn't have a non-trivial free summand if and only if M$\subset$mF " true? I was trying around Nakayama's lemma

share|cite|improve this question
Dear Bala, perhaps you mean. ``$M$ doesn't contain a non-trivial free summand of $F$ if and only if $M \subseteq m F$''? – Karl Schwede Nov 17 '12 at 13:10
Dear Karl, thank you for your post. I was not aware of the concept of "M contains free summand of F". It was helpful and the proof of course. However I was looking for M containing a free summand(without condition "of F") i.e. there exists a non zero free module G such that M\congruent G\oplus H for some module H. From the second para of your proof it seems one direction of my question is correct. Isn't it? – Bala Nov 19 '12 at 6:18
Dear Bala, of course you are right. If $M$ contains a free summand of $F$ then it certainly contains a free summand. – Karl Schwede Nov 19 '12 at 17:04

The answer to your question is: no, it is not true. Here is a counterexample.

$(A,\mathfrak m)=(\mathbb Z_p,p\mathbb Z_p)$ is the (local Noetherian commutative) ring (PID) of $p$-adic integers (you can construct this ring in many ways... probably the standard one is to complete $\mathbb Z$ with respect to the $p$-adic topology -a base for this topology is given by the powers of the maximal ideal $p\mathbb Z$- the result is that $\mathbb Z_p$ is the inverse limit of the groups of the form $\mathbb Z/p^n$ with the canonical transition maps).

Now, take $F=A$ and $M=p^2 A$. It follows essentially by the definition of $\mathbb Z_p$ that I gave you, that $F/M\cong \mathbb Z/p^2$ that has composition length $2$, in particular this is finite. Anyway, $M\cong A$ as modules (just consider the morphism $A\to M$ such that $x\mapsto p^2x$... then surjectivity is obvious and injectivity follows by the fact that $A$ is a domain and so multiplication by $p^2$ has trivial kernel). Thus $M$ has not only free summands but it is itself free. Finally, notice that $pA\supsetneq p^2A=M$ so $M$ is properly contained in $\mathfrak m F$.

share|cite|improve this answer

Ok, I'm going to answer a different question for which the answer is "yes".

Question: The module $M$ contains no summand of $F$ if and only if $M \subseteq m F$.

Proof: Suppose that $M$ did contain an $F$-summand. In other words, there is a surjective map $\phi : F \to R$ such that $\phi|_M$ is also surjective (note if $\psi : M \to R$ is surjective, sending $x \mapsto 1$, then the composition $R \xrightarrow{1 \mapsto x} M \xrightarrow{\psi} R$ gives us a summand). Thus $$m = mR = m \phi(F) = \phi(mF) \supseteq \phi(M) = R$$ which is impossible.

Conversely, suppose that $M$ is not contained in $m F$. Choose an element $x \in M$, not in $mF$. If $e_1, \dots, e_n$ is a basis for $F$, then $x = \sum a_i e_i$ for some $a_i \in R$, at least one of the $a_i$ not in $m$. The projection onto that $i$th summand then clearly sends $x$ to a unit, and so by rescaling, I have a map $\phi : F \to R$ which sends $x$ to $1$. In particular, $x$ generates a summand of $F$. And so $M$ contains an $F$-summand. (You can also easily do this part with Nakayama's lemma).

share|cite|improve this answer
I guess it's also worth noting here that $F/M$ doesn't have to be finite length for this to work. Also, this fact is essentially the reason why the Betti numbers of a finite module over a local ring (which are defined in terms of Tor) also always occur as the ranks of the free modules in any minimal free resolution of the module. – Neil Epstein Nov 18 '12 at 13:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.