Let $D$ a PID, $F$ a free module rank $n$, $N$ a submodule of $F$.

I want to prove (or find a counterexample) of: $\text{rank}(F/N)=0 \Longleftrightarrow\text{rank}(F)=\text{rank}(N)$

  • $\text{rank}(F/N)=0\Rightarrow F/N=0 \Rightarrow F=N \Rightarrow \text{rank}(F)=\text{rank}(N)$
  • I know it to the left is false for a ring (I take $d$ a non-zero divisor and I prove that is $(d)$ is not free. I think taking $\mathbb{Z}$ and $2\mathbb{Z}$ can work, because $\{1\}$ is a basis of $\mathbb{Z}$ and $\{2\}$ appears to be a basis of $2\mathbb{Z}$, so their rank is 1 but I can't prove $\text{rank}(\mathbb{Z}/2\mathbb{Z})$ is non-zero (basically because I think it is not free).

Does this counterexample work?

  • $\begingroup$ @user26857 Over a PID, $M=T(M)\bigoplus F$, with $F$ a free module. Then $\text{rank}(M):=\text{rank}(F)$ $\endgroup$ – user203327 Jan 5 '15 at 18:29

Hint: $\text{rank}(F/N)=0\Rightarrow F/N$ is a torsion module, not $\{ 0\}.$ But the conclusion is true. Use Structure Theorem for Finitely Generated Modules over a PID

Since $N$ is submodule of $F,$ there is a basis $y_1, y_2, \cdots , y_n$ of $F$ and $d_1, d_2, \cdots , d_r \in D$ with $d_1| d_2| \cdots |d_r$ such that $d_1y_1, d_2y_2, \cdots , d_r y_r$ is a basis of $N$ (rank $N = r$) and $F/N \cong D^{n-r} \times D/d_1D \times \cdots \times D/d_rD.$

In this case, rank$(F/N) = 0 \Rightarrow r = n.$ On the other hand, let $r = n.$ Then $F/N$ is a torsion module and hence is or rank zero.

  • $\begingroup$ I understand $F/N$ is a torsion module, but I can not get the conclusion. Also, why that basis of $N$ exists? $\endgroup$ – user203327 Jan 5 '15 at 17:13
  • $\begingroup$ this a standard fact. see any algebra book (e.g. Dummit & Foote). you can also see Theorem 2.3 in math.iitb.ac.in/atm/final1.pdf $\endgroup$ – Krish Jan 5 '15 at 17:26
  • $\begingroup$ But why $F/N$ torsion module implies that the ranks are equal? $\endgroup$ – user203327 Jan 5 '15 at 17:58
  • $\begingroup$ If I am not mistaken, it is basically: $F/N\simeq D^{n-r}\bigoplus T(F/N)$. So $0=\text{rank}(F/N)\Rightarrow n-r=0 \Rightarrow n=r$; and $\text{rank}(F)=\text{rank}(N)=n \Rightarrow F/N\simeq D^0\bigoplus T(F/N)=T(F/N)\Rightarrow \text{rank}(F/N)=0$ $\endgroup$ – user203327 Jan 5 '15 at 18:44
  • $\begingroup$ @user203327: yes!! you got it. :) $\endgroup$ – Krish Jan 5 '15 at 19:05

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