Is the ring $R={\mathbb Z}/n{\mathbb Z}$ an elementary divisor ring? Can you provide an easy proof from well-known results, or a reference to this result?

(Recall that an elementary divisor ring $R$ is one for which every matrix over $R$ is equivalent to a diagonal matrix. The existence of Smith normal form shows that when $R$ is a principal ideal domain, it is an elementary divisor ring. Of course, when $n$ is composite then $R={\mathbb Z}/n{\mathbb Z}$ is not even a domain, although it is a principal ideal ring.)


This follows from the fact that Smith normal forms over $\Bbb{Z}$ exist.

If $\overline{A}$ is matrix over $\Bbb{Z}_n$ then we can, arbitrarily, lift the entries to integers and form a matrix $A$ with integer entries. By the existence of Smith normal forms over $\Bbb{Z}$ there exists invertible (i.e. determinant $=\pm1$) integer matrices $P$ and $Q$ such that $PAQ=D$ is a diagonal matrix.

Reducing that matrix equation modulo $n$ gives $$ \overline{D}=\overline{P}\overline{A}\overline{Q}. $$ Here $\overline{D}$ is obviously a diagonal matrix, and because $\overline{P}$ and $\overline{Q}$ have determinants $=\pm\overline{1}$ they are invertible over $R$ (alternatively you can reduce their inverses in a matrix ring over $\Bbb{Z}$ modulo $n$). The claim follows from this.

You may lose the divisibility relations among the diagonal entries. I do suspect the invariant factors to survive in some form (think: finitely generated abelian groups), but divisibility arguments in the presence of zero divisors need a little bit of extra care.

  • $\begingroup$ Thanks! Actually, I guess the divisibility relations are also induced in ${\mathbb Z}/n{\mathbb Z}$ from the fact that they hold in ${\mathbb Z}$ $\endgroup$ – user6584 Sep 14 '18 at 13:17
  • $\begingroup$ True, @user6584. It's less clear in which form the uniqueness survives. We do get some uniqueness because the "Smith form" describes the structure of a certain $\Bbb{Z}_n$-module. That is still unique as an abelian group, so even that will survive in some form. $\endgroup$ – Jyrki Lahtonen Sep 14 '18 at 14:44
  • $\begingroup$ I just noticed this discussion which is relevant: mathoverflow.net/questions/44576/… $\endgroup$ – user6584 Sep 27 '18 at 17:05

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.