It's well known that when a ring $R$ is a PID, every submodule of a free $R$-module is free. I'm interested in cases when the converse holds -- that is, in rings $R$ which have the property that every submodule of a free $R$-module is free. It's easy enough to see that such a ring cannot have any zero divisors and that every ideal of the ring must be principal. I would like to know if there are any "noncommutative PIDs", that is, if there are any rings which (1) have no zero divisors, (2) have only principal ideals, and (3) are noncommutative. Is there any sort of structure theorem for such rings? For instance, Hungerford proved in this paper that every principal ideal ring (a ring whose ideals are all principal) which is commutative is a direct sum of homomorphic images of PIDs.

EDIT: Robert Lewis reminded me that one class of examples of the ring I am looking for is division rings. Are there any examples that are not division rings? (After all, the ideal structure of a division ring is not very interesting.)

  • $\begingroup$ So for rings where commutativity isn't assumed, we usually make a left/right distinction. If you are only interested in principal two-sided ideals, then simple domains will do, but I doubt they have the property on free modules that you mentioned. $\endgroup$
    – rschwieb
    Mar 24, 2014 at 10:58
  • 1
    $\begingroup$ In noncommutative domains it is not true that a free right ideal has to be principal. E.g. $xR + yR$ in the free algebra $R=k\langle x,y\rangle$ over a field $k$. A ring for which every right ideal is free is called a right fir (fir=free ideal ring). For right firs every submodule of a free right module is free. So while your question about PIDs has been answered, left and/or right firs are precisely the rings that satisfy the property that submodules of free right/left modules are free. P.M. Cohn has a lot of work on firs. See Free Ideal Rings and Localization in General Rings. $\endgroup$
    – moonlight
    Sep 11, 2015 at 10:44

3 Answers 3


Fred Byrd's suggestion of the quaternions does the trick, but if you want there to be some irreducible elements, you can use $\Bbb H[x]$, which is a left-and-right PID by the same argument you use for polynomial rings over fields. It is definitely not a division ring.

As for a structure theorem, I believe the one you're looking for is for hereditary Noetherian prime rings, of which a left-and-right principal ideal domain is an example. Unfortunately I do not know this one by heart and don't have the resources handy.

If I were you I'd check out the work by Robson, Eisenbud, Griffith, Connell and Jategaonkar on the topic of hereditary Noetherian rings. I would guess the conclusion is probably available in Rowen's ring theory books and maybe Goodearl and Warfield's book on noncommutative noetherian rings.

I remember the theorem going something like this:

Hereditary Noetherian rings decompose into an Artinian ring and a finite product of hereditary Noetherian prime rings.

I also managed to find in Faith's Rings and things p 115 that proper homomorphic images of HNP rings are all Artinian serial rings. I highly suspect that these are exactly the Artinian rings that appear in the above decomposition in the Artinian part.

Let's compare this to Hungerford's result. (Principal ideal ring=PIR, principal ideal domain-PID, and special principal ideal ring=SPIR.)

Theorem 1: Every PIR is a finite direct sum of homomorphic images of PIDs.

Lemma 10: Every PIR is a (finite) direct sum of PIDs and SPIRs.

Corollary 11: Every SPIR is a homomorphic image of a PID.

Now a SPIR with a unique maximal ideal is an Artinian ring whose ideals are linearly ordered. A finite product of SPIRs is an example of an Artinian serial ring. So, that is why So, I sincerely believe that the hereditary Noetherian ring theorem is the proper extension of Hungerford's theorem.


The quaternions, I believe, will do the trick.

As for a general structure theorem, I'm afraid that's a bit out of my depth.

  • $\begingroup$ Why are all the ideals of the quaternions principal? $\endgroup$
    – user134824
    Mar 24, 2014 at 5:30
  • $\begingroup$ @user134824: Is it not the case that the quaternions form a division ring? Then the only ideals, left, right, or two-sided are $\{ 0 \}$ and the quaternions themselves. $\endgroup$ Mar 24, 2014 at 5:36
  • $\begingroup$ Of course! No nontrivial ideals. Thanks. $\endgroup$
    – user134824
    Mar 24, 2014 at 5:40

The universal enveloping algebra of $\mathfrak{sl}_2(\mathbb{C})$ is a noncommutative PID which is not a division ring. See for example Catoiu's Ideals of the enveloping algebra $U(sl_2)$ p.145 and the references cited there. Siciliano and Usefi give more examples in Enveloping algebras that are principal ideal rings.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .