# Does Smith normal form imply PID?

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?

• Doesn't Smith normal form hold for a principal ideal ring, possibly with zero divisors? I am thinking of $\mathbb{Z}/n\mathbb{Z}.$ – Victor Protsak Jul 10 '10 at 6:42
• For a summary of what is known about this problem, see Theorem 2.1 of my paper math.mit.edu/~rstan/papers/snf_survey.pdf. – Richard Stanley Feb 25 at 14:09

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.

• @Tyler: I don't think it affects the rest of your argument, but: to get a valuation ring, don't you want $k[[t^{\frac{1}{n}}]]$ instead of $k[t^{\frac{1}{n}}]$? – Pete L. Clark Jul 10 '10 at 18:20
• Yes, you are correct - I added that sentence at the last minute. It is simply a ring with an indiscrete valuation. – Tyler Lawson Jul 10 '10 at 19:36

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 module, 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 :-)

• For R a domain, it implies that R is one-dimensional regular, hence Dedekind, so every nonzero ideal is a product of maximal ideals, therefore principal itself. – user2035 Jul 10 '10 at 7:13
• Thanks a-fortiori: each localization at a maximal ideal is a local ring of height at most one, so $R$ has Krull dimension $\le 1$. – Robin Chapman Jul 10 '10 at 7:19
• A commutative noetherian ring whose maximal ideals are principal is indeed a principal ideal ring (even if it is not domain). See Theorem 12.3 of Kaplansky's article "Elementary divisors and modules," Trans. Amer. Math. Soc. 66 (1949), 464-491. – Manny Reyes Jul 10 '10 at 14:55

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

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