# Inverse of the Structure Theorem for Finitely Generated Modules over PID

We know that for a PID $$R$$, any finitely generated module is of the form $$\frac{R}{(a_1)} \oplus \dots \oplus \frac{R}{(a_s)}$$. I was wondering if the converse of this statement is true, that is, is it true that for a domain $$R$$, if any f.g. module is isomorphic to $$\frac{R}{I_1} \oplus \dots \oplus \frac{R}{I_s}$$ for ideals $$I_i \triangleleft R$$, must $$R$$ be a PID?

I know that a counter example to the above statement must be non Noetherian (and infact, it must be that any non-principal ideal is not f.g.) because if $$I = (a_1, ..., a_n)$$ is a non principal f.g. ideal, that it is a torsion free $$R$$ module, but it isn't free, since $$x_1, ... x_m$$ are not linearly independent if $$m > 1$$ since $$x_2 x_1 + (-x_1)x_2 = 0$$, hence they do not form a basis.

PS: I would generally like $$R$$ to be commutative unitial and an integral domain, but I would love to know even for other cases.

• An íntegral domain in which every finitely generated ideal is principal is called a Bezout domain. It is apparently am open question of the elementary divisor theorem os true in all Bezout domains see alpha.math.uga.edu/~lorenz/Bezout.pdf – Benjamin Steinberg Jan 24 at 21:27
• @AnweshRay Yes, and this is what I stated in the second paragraph. The question regards weirder rings then, where the only non-principal ideals are those which are not f.g. and that there is at least one such ideal. – Adi Ostrov Jan 24 at 23:20
• As Benjamin Steinberg has hinted, a reasonable conclusion here should be that $R$ is a Bezout domain, rather than a PID. One of the first examples of a non-PID Bezout domain, $k[x^{1/p^{\infty}}]$, is an elementary divisor ring that is not a PID, hence provides a counterexample to the statement. – Victor Protsak Jan 25 at 4:14
• Possible duplicate: mathoverflow.net/questions/31275 – Victor Protsak Jan 25 at 6:30
• @VictorProtsak I don’t think that possible duplicate is the same question. I don’t see why having Smith normal form implies anything about the structure of finitely generated, but not finitely presented, modules. – Jeremy Rickard Jan 25 at 8:20

There seems to be quite some literature about rings with this property, sometimes under the name "FGC domains".

From Googling, not personal knowledge:

In Theorem 14 of Kaplansky, Irving, Modules over Dedekind rings and valuation rings, Trans. Am. Math. Soc. 72, 327-340 (1952). ZBL0046.25701, it is proved that an almost maximal valuation domain has this property.

"Maximal" means that a system $$x\equiv a_i\pmod{I_i}$$ of congruences has a solution if each pair of the congruences has a solution, and "almost maximal" is the same, but only for such systems of congruences with $$\bigcap_iI_i\neq0$$. I'm not sure what the simplest example is that is not a PID, but I guess the completion of $$R=k[x^\alpha\mid 0<\alpha\in\mathbb{Q}]$$ with respect to the set of ideals $$x^\alpha R$$ is a maximal valuation domain.

Osofsky constructed an example of an FGC domain that is neither a PID nor a valuation domain, and work of many people, culminating in Brandal, Willy; Wiegand, Roger, Reduced rings whose finitely generated modules decompose, Commun. Algebra 6, 195-201 (1978). ZBL0368.13005, led to the theorem that FGC domains are precisely the almost maximal Bezout domains.

And there's a whole Springer Lecture Notes volume by Brandal on the not-necessarily-domain case:

Brandal, Willy, Commutative rings whose finitely generated modules decompose, Lecture Notes in Mathematics. 723. Berlin-Heidelberg-New York: Springer-Verlag. 116 p. (1979). ZBL0426.13004.