When Does Such a Short Exact Sequence Exist?

[caffeinemachine]

$\newcommand{\Z}{\mathbb Z}$.
Question: Let $m$ and $n$ be positive integers. What are all the abelian groups $A$ such that there is a short exact sequence $0\to \mathbb Z/p^m\mathbb Z\to A\to \mathbb Z/p^n\mathbb Z\to 0$.

It is clear that any such abelian group $A$ has cardinality $p^{m+n}$. Thus by Fundamental Theore of Finitely Generated Abelian Groups we have $A\cong \Z/p^{k_1}\Z\oplus \cdots\oplus \Z/p^{k_r}\Z$ for some $k_1\leq \cdots \leq k_r$. Also, since $A$ contains a copy of $\Z/p^m\Z$, a cyclic group of order $m$, we must have $k_r\geq m$. I am unable to see what to do next.

Also, I am particularly interested in the solution given here in the form of Proposition 0.1 on pg 2 :http://www.math.ku.dk/~moller/f03/algtop/opg/S2.1.pdf
On the first line of the solution it uses a notion of "pull-back" which I do not understand. Can somebody shed some light on this solution?

Thanks.

 

[johng1]

CaffeinMachine,
I read the proposition and proof in your link. The following contains an alternate proof and some discussion of the "pull back" idea.

 

[caffeinemachine]

CaffeinMachine,
I read the proposition and proof in your link. The following contains an alternate proof and some discussion of the "pull back" idea.

Hello johng,

Sorry for the late reply. I am not getting much time because of the courses I am doing.

I have one question.

I do not understand the part where you said that $\ker$ is a free abelian group. Before that I understand everything. Can you please elaborate on that point.

Thanks.

 

[johng1]

CaffeineMachine,
I don't understand the source of your misunderstanding. The fundamental theorem on finitely generated abelian groups says (among other things) that such a group is the direct sum of a finite group and a free abelian group. Presumably you followed why ker has no elements of finite order. So ??

 

[caffeinemachine]

CaffeineMachine,
I don't understand the source of your misunderstanding. The fundamental theorem on finitely generated abelian groups says (among other things) that such a group is the direct sum of a finite group and a free abelian group. Presumably you followed why ker has no elements of finite order. So ??

I thought about it for some time. I have the following proposition:

Hypothesis: Let $M$ be a finitely generated module over a PID $R$ (in our case it is $\mathbb Z$). So we can write $M=M^{\text{tor}}\oplus F$, where $M^{\text{tor}}$ is the torsion submodule of $M$ and $F$ is any maximal free submodule of $F$. Let $N$ be a submodule of $M$. We know that $N$ is also a finitely generated module over $R$ and $N=N^{\text{tor}}\oplus F'$ (where $F'$ is a maximal free submodule of $N$).

Proposition: If $N\cap M^{\text{tor}}=0$, then $N$ is free.
Proof: Suppose $N$ is not free. Then $N^{\text{tor}}\neq 0$. But any nonzero member of $N^{\text{tor}}$ is also in $M^{\text{tor}}$ and we have $M^{\text{tor}}\cap N\neq 0$, contradicting our assumption. Thus $N$ is free.

Is the above alright. Also, we can use this to settle that $\ker$ is free. Also, we do not need the fundamental theorem of finitely generated abelian groups if the above argument is correct, for the decomposition $M=M^{\text{tor}}\oplus F$ can be proved to hold by an elementary reasoning.

 

[johng1]

I agree with everything you said in your last post. Basically, the fundamental theorem on finitely generated abelian groups is just a special case of the general proposition on finitely generated modules over a PID. If you are now happy, so am I.

 

[caffeinemachine]

I agree with everything you said in your last post. Basically, the fundamental theorem on finitely generated abelian groups is just a special case of the general proposition on finitely generated modules over a PID. If you are now happy, so am I.

I think I understand your proof now. My vote for next MHB algebra award has been sealed.

 

[Deveno]

I think I understand your proof now. My vote for next MHB algebra award has been sealed.

johng always writes impeccably clean expositions. Your vote would be well-deserved.