When does a torsion $\mathbb{Z}_p$-module have unique maximal submodule?

Solution 1:

As said in coments, it is not clear what exactly you are asking. Maybe the following reminder helps:

For a unital commutative ring $R$, every cyclic $R$-module is isomorphic to $R/I$ for some ideal $I$ of $R$.

The only ideals of $\mathbb Z_p$ are $(0)$ and all $(p^n)$ for $n \in \{0,1,2,...\}$.

Hence up to isomorphism, the only cyclic $\mathbb Z_p$-modules are $\mathbb Z_p$ itself and $\mathbb Z_p/(p^n)$ for $n \in \{0,1,2,...\}$. It is well-known that $\mathbb Z_p/(p^n) \simeq \mathbb Z/(p^n)$.

If you have a $\mathbb Z_p$-module $M$ which is torsion and has finite cardinality ($p^c$, say), then in particular it is finitely generated, and hence by the structure theory of f.g. torsion modules over PIDs (or an easier version thereof in this case of a DVR), it is a direct sum of cyclic modules. So here:

$$M \simeq \bigoplus_{n \ge 0} (\mathbb Z/p^n)^{i_n} \qquad \qquad (*)$$

where $(\mathbb Z/(p^n))^{i_n}$ is short notation for $\underbrace{\mathbb Z/(p^n) \oplus ... \oplus \mathbb Z/(p^n)}_{i_n \text{ times}}$, and only finitely many $i_n$ are $\neq 0$. Actually, for $M$ to have cardinality $p^c$, we need $\prod_{n \ge 0} n \cdot i_n = c$.

Obviously the cyclic ones among these are the ones where $i_n=1$ for $n=c$ and $i_n=0$ for all $n \neq c$.

Or in other words, if the module has cardinality $p^c$ and is cyclic, it must be $\simeq \mathbb Z/p^c$. So: If we know the module has cardinality $p^c$, then for it to be cyclic it suffices to show, for example, that it has an element of order $p^c$ (which is nearly trivially equivalent to being cyclic, if you think about it).

As regards torsion, if we know that $M$ has exact torsion $p^t$ (i.e. $p^t \cdot m=0$ for all $m$ but there is $m \in M$ with $p^{t-1} \cdot m \neq 0$), then in $(*)$ this only imposes the restriction that $i_t \neq 0$ and $i_{t+1}=i_{t+2}= ... =0$. As said above, such module is cyclic iff $i_t=1$ and all other $i_n=0$ i.e. iff $M =\simeq \mathbb Z/(p^t)$ i.e. iff it has cardinality $p^t$. (It would suffice to show its cardinality is $\le p^t$.)

As said in comments, I find it hard to imagine how you are given a module $M$ for which you can decide at all whether it is cyclic, but without that decision being kind of immediate via any of the criteria above.