# If $\operatorname{rk}(N)=1$, and $M/N$ is torsion, why is $\operatorname{rk}(M)=1$?

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!

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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 $n=1$.

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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.

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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 math.uga.edu/~pete/integral.pdf (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.

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