3
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ The forward direction of your proof would only hold if $M$ has free rank $0,$ unless I'm missing something. $\endgroup$
    – Aurel
    Apr 12, 2018 at 0:11
  • $\begingroup$ 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. $\endgroup$ Apr 12, 2018 at 0:28

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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 $0$. $\endgroup$ Apr 12, 2018 at 18:10
  • $\begingroup$ 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 $r.$ $\endgroup$
    – Aurel
    Apr 12, 2018 at 19:41

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.