Does Smith normal form imply PID? - MathOverflow

Does Smith normal form imply PID?

2010-07-10T06:36:03Z

Let R be a commutative ring with a 1 different from 0, such that all finite matrices over R have a Smith normal form. Does it follow that R is a Principal Ideal Domain?

If not, what if R also has no zero divisors? (aka is an integral domain) What if additionally the diagonal entries are always unique up to associatedness?

Answer by Robin Chapman for Does Smith normal form imply PID?

2010-07-10T06:47:51Z

If every matrix has a Smith normal form, then every finitely generated $R$-submodule $M$ of $R^n$ satisfies $R^n/M$ is a finite direct sum of modules isomorphic to $R/aR$. If $R$ is Noetherian this implies that every finitely generated module is a direct sum of modules of the form $R/aR$. So if $I$ is a maximal ideal of the Noetherian $R$ then $R/I$ is a simple ideal, so if $R/I\cong R/aR$ then $I=aR$ is principal. So in a Noetherian ring with Smith normal form for all matrices, every maximal ideal is principal. Does this imply that all ideals are principal?....I'm not sure :-)

Answer by Tyler Lawson for Does Smith normal form imply PID?

2010-07-10T12:38:52Z

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.

Answer by Bill Dubuque for Does Smith normal form imply PID?

2010-07-10T15:55:39Z

Work on ring-theoretic generalizations of Hermite/Smith normal forms goes way back, but made it into the mainstream via classic papers by Helmer and Kaplansky. Nowadays such rings are called elementary divisor rings, or rings with elementary divisors (r.e.d.) or Helmer rings, etc. A search on such terms, and for citations of Kap's classic paper [1] should quickly answer all your questions and then some.

[1] I. Kaplansky, "Elementary divisors and modules," Trans. Am. Math. Soc., 66, 464-491. (1949).
http://www.ams.org/journals/tran/1949-066-02/S0002-9947-1949-0031470-3/S0002-9947-1949-0031470-3.pdf

Answer by Victor Protsak for Does Smith normal form imply PID?

2010-07-10T16:07:09Z

Such rings are apparently called elementary divisor rings. They are necessarily Bezout rings (i.e. every finitely-generated ideal is principal), but not easy to characterize completely.

The first paper giving a nontrivial sufficient condition (beyond classical case) seems to be

Helmer, Olaf The elementary divisor theorem for certain rings without chain condition. Bull. Amer. Math. Soc. 49, (1943). 225--236, MR

More complete results are in a series of papers starting with

Larsen, Max D.; Lewis, William J.; Shores, Thomas S. Elementary divisor rings and finitely presented modules. Trans. Amer. Math. Soc. 187 (1974), 231--248, MR