Elementary divisor rings (Smith normal form) 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.)
 A: 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.
