When one can write $M=\mathbb{Z} \cdot m$ for some $m \in M$?

Let $M$ be a finitely generated $\mathbb{Z}$-module such that $p^n \cdot m=0$ for all $m \in M$.

Assume also that $M$ is a free $\mathbb{Z}/p^n\mathbb{Z}$-module.

Question:

When can we write $M=\mathbb{Z} \cdot m$ for some $m \in M$ ?

or in other word, when is $M$ a cyclic $\mathbb{Z}$-module ?

My effort:

If the rank of $M$ as a free $\mathbb{Z}/p\mathbb{Z}$-module be $1$, then we have $$M \cong \mathbb{Z}/p^n\mathbb{Z}.$$ In the above case $M$ being torsion-free $\mathbb{Z}/p^n\mathbb{Z}$ and $\mathbb{Z}/p^n\mathbb{Z}$ is a PID, so we conclude $M$ is a cyclic $\mathbb{Z}/p^n\mathbb{Z}$-module and hence $$M=\mathbb{Z}/p^n\mathbb{Z} \cdot m, ~~~~~~~~~~~~(1)$$ for some $m \in M$.

But I want show $M=\mathbb{Z} \cdot m$ for some $m \in M$.

As the action of $\mathbb{Z}$ on $M$ factors through the quotient $\mathbb{Z}/p^n\mathbb{Z}$, we have $$p^n\cdot m=0 \Rightarrow z \cdot m=(z+p^n \mathbb{Z}) \cdot m,$$ which says we can view the elements of $\mathbb{Z}$ as belonging to $\mathbb{Z}/p^n\mathbb{Z}$.

From relation $(1)$, we have $$M=\mathbb{Z}/p^n\mathbb{Z} \cdot m \cong \mathbb{Z}/p^n\mathbb{Z} \cdot m \oplus p^n \mathbb{Z} \cdot m \cong \mathbb{Z} \cdot m$$ for some $m \in M$.

Any help is appreciated. Thanks

Solution 1:

There are a lot of misconceptions in this question. I will only address those that are relevant for answering the question (and formulating it properly).

Let $M$ be a $\mathbb{Z}$-module such that $p^n \cdot m=0$ for all $m \in M$.

Equivalently $M$ is an abelian group, in which the order of every element divides $p^n$. In particular $M$ is a $p$-group.

Since $\mathbb{Z}$ is a PID, by structure theorem $M=F\oplus T$, where $F$ is free $\mathbb{Z}$-module and $T$ is torsion $\mathbb{Z}$-module.

This is not true. The structure theorem holds for finitely generated modules over a PID.

Therefore $M$ is a free $\mathbb{Z}/p^n\mathbb{Z}$-module.

This simply does not follow, even if $M$ were finitely generated. I have no idea what your line of thought could be. For a very simple counterexample, take $p=2$ and $M=\Bbb{Z}/2\Bbb{Z}$ and $n=2$. Be sure to look up the definition of a free module.

Question:

When can we write $M=\mathbb{Z} \cdot m$ for some $m \in M$ ?

You already note that the action of $\Bbb{Z}$ on $M$ factors over $\Bbb{Z}/p^n\Bbb{Z}$, and so $$\mathbb{Z} \cdot m=(\Bbb{Z}/p^n\Bbb{Z})\cdot m.$$ By definition $M=(\Bbb{Z}/p^n\Bbb{Z})\cdot m$ for some $m\in M$ if and only if $M$ is a cyclic $\Bbb{Z}/p^n\Bbb{Z}$-module. Be sure to look up the definition of a cyclic module.