Let $R$ be a nonzero commutative ring with $1$, such that all finite matrices over $R$ have a *Smith normal form*. Does it follow that $R$ is a principal ideal domain?


If this fails, suppose we additionally suppose that $R$ is an integral domain?


What can we say if we impose the additional condition that the diagonal entries be unique up to associates?



Does Smith normal form imply PID?
The implication is false without the assumption that R is Noetherian, because finite matrices don't detect enough information about infinitely generated ideals.


For example, let R be the ring
$$
\bigcup\_{n \geq 0} k[[t^{1/n}]]
$$
where $k$ is a field (an indiscrete valuation ring). Any finite matrix with coefficients in R comes from a subring $k[[t^{1/N}]]$ for some large $N$, and hence can be reduced to Smith normal form within this smaller PID.


However, the ideal $\cup (t^{1/N})$ is not principal.


