# $M$ is a cyclic module if and only if it has a single invariant factor

Suppose $R$ is a Euclidean domain (not necessarily a field) and let $M$ be a finitely generated $R$-module. Prove $M$ is a cyclic module if and only if it has a single invariant factor.

If $M$ has a only has invariant factor $d_1$, it seems right away that we can use the Structure theorem for finitely generated modules over a PID to get $M \cong R/(d_1)$. Thus $M$ is cyclic.

For the other direction, suppose $M$ is cyclic, i.e. $\exists m \in M$ so that $M = (m) = \{rm : r\in R\}$. We need to show there is only one invariant factor. I don't know how to go about this. Thoughts?

• The forward direction of your proof would only hold if $M$ has free rank $0,$ unless I'm missing something. Apr 12, 2018 at 0:11
• Yes you're right. It'd be $M \cong R/(d_1) \oplus R^k$. I really have no idea how to approach this, it doesn't seem like we have any other tools besides the structure theorem. Apr 12, 2018 at 0:28

It is true that if $M = (m)$ is a cyclic module over a PID $R$ then it has a single invariant factor. To see this note that there is a canonical R-epimorphism $R\twoheadrightarrow M$ given by $r \mapsto rm.$ Then by definition, we have that kernel of this map is the annihilator, $\text{Ann}(m) = \{r \in R; rm = 0\}$ of $m$ (prove that this is an ideal of R). Moreover, since $R$ is a PID, we have that $\text{Ann}(m) =(d)$ for some $d \in R.$ We can conclude the proof by the first isomorphism theorem.
The converse, on the other hand, is only true if the rank of $M$ is $0.$ An counterexample for the nonzero rank case is $\mathbb{Z} \oplus \mathbb{Z}/2$ considered as a $\mathbb{Z}$-module.
• Thank you. I still have one question: what does rank $0$ mean? Isn't that like having a basis of $0$ elements? How does that work? What condition on $M$ would make it rank $0$. Apr 12, 2018 at 18:10
• Rank 0 means that in the decomposition $M = R^r \oplus_{i =1}^n R/(d_i),$ we have that $r = 0.$ In general the rank of $M$ is the value of $r.$ Apr 12, 2018 at 19:41