0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ @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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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