Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

Suppose $M$ is a finitely generated torsion-free module over a PID. If $N\leq M$ is free of rank $1$, and $M/N$ is torsion, how do we conclude $M$ is free of rank $1$?

My scattered thoughts: Since $M$ is f.g., $M/N$ is as well, and since it is torsion, the structure theorem tells me it can be factored as $$ M/N=\langle \bar{v}_1\rangle\oplus\cdots\oplus\langle \bar{v}_n\rangle $$ for cyclic modules $\langle \bar{v}_i\rangle$ in $M/N$. I know that each $\langle \bar{v}_i\rangle=L_i/N$ where $L_i$ is a submodule of $M$ containing $N$. Since $M/N$ is torsion, I can send any element $m$ of a generating set for $M$ into $N$ by multiplication by some appropriate nonzero $r_m\in R$.

I wanted to show that any finite generating set of $M$ can actually be reduced down to one element somehow, but I don't have any good ideas. Any ideas of how to go forward? Thanks!

share|cite|improve this question
up vote 4 down vote accepted

First note that $M$ is free of rank $n\ge 1$. Then use the structure theorem that (in this case) says the following: there is a basis $\{x_1,\dots,x_n\}$ of $M$ and $d\in R^{\times}$ such that $\{dx_1\}$ is a basis of $N$. Then $M/N\simeq R/(d)\oplus R^{n-1}$. Since $M/N$ is torsion you get $n=1$.

share|cite|improve this answer
Thanks, I wasn't aware about that fact about bases. I'll try to work it out. – yunone Feb 23 '13 at 23:53
@yunone This follows easily if use the Smith Normal Form of matrices over a PID. – user26857 Feb 24 '13 at 0:16

Consider the short exact sequence $$0\to N\to M\to M/N\to 0.$$ Tensoring this with the quotient field $F$, which is flat, $$0\to F\otimes N\to F\otimes M\to F\otimes (M/N)\to 0.$$ Since $M/N$ is torsion, $F\otimes (M/N)=0$, so in fact we have an isomorphism $$F\otimes N\cong F\otimes M.$$ In particular, we have $$\dim_F F\otimes N= \dim_FF\otimes M.$$ As the rank of a module $Z$ is equal to $\dim_FF\otimes Z$, this is what we wanted.

share|cite|improve this answer
Note that this argument works verbatim for modules over any integral domain $R$ with fraction field $K$, where by the rank of $M$ we mean $\dim_K M \otimes_R K$. (The fact that the OP restricts to PIDs may suggest that he wants a more undergraduate-level approach using the structure theory of PIDs...So it is nice to have both answers.) – Pete L. Clark Feb 23 '13 at 23:55
@PeteL.Clark If we don't work over a PID how do we know that the functor $- \otimes_R K$ is exact? Maybe I'm confused. – user38268 Feb 24 '13 at 0:29
Localization is flat: c.f. Prop. 7.8 of (or any other reference on commutative algebra). – Pete L. Clark Feb 24 '13 at 0:40
It is much better if we reserve the term crap for other contexts, really! – Mariano Suárez-Alvarez Feb 24 '13 at 8:08
Thanks for the more general proof. – yunone Feb 26 '13 at 5:24

The other two answers have already answered your queries but here are some more general remarks. Let $R$ be a PID and consider $F = \operatorname{Frac} R$. Then for any ses of $R$ - modules

$$0 \to A \to B \to C \to 0$$

we can apply the functor $- \otimes_R F$ that is exact because over a PID a module is flat iff it is torsion free. Since $F$ is a field containing $R$ it has zero $R$ - torsion and so is flat. Thus we obtain

$$0 \to A \otimes_R F \to B \otimes_R F \to C \otimes_R F \to 0.$$

But now the point is that these are just $F$ vector spaces and so

$$\dim B \otimes_R F = \dim A \otimes_R F + \dim C \otimes_R F.$$

Hence the rank of $B$ is the sum of the rank of $A$ and $C$. Note this is a general case which includes what you need as a special case.

share|cite|improve this answer

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.