Let $R$ be a ring of infinite global dimension. A priori we can't immediately conclude that $R$ has a module of infinite projective dimension, since it could be the case that $R$ only has a sequence of modules $(M_n)$ with $\text{p.dim} \hspace{2pt} M_n \geqslant n$. However, we then have $\text{p.dim} \hspace{2pt} (\bigoplus_n M_n) = \infty$, so there in fact does exist of module of infinite projective dimension. However, such a module need not be finitely generated.

Is there a ring $R$ of infinite global dimension which does not have a finitely generated module of infinite projective dimension?

I wouldn't be surprised if there was an obvious candidate for $R$ that I've missed.

  • $\begingroup$ Edited. Thanks, $\endgroup$
    – lokodiz
    Commented Dec 31, 2015 at 10:21

1 Answer 1


A ring $R$ for which every finitely generated right $R$-module has finite projective dimension is sometimes called right regular. Such rings are discussed in McConnell, Robson, Noncommutative Noetherian Rings, Chapter 7.7. In particular, Example 7.7.2 (6.4.9) gives an example of a ring of infinite global dimension which is left and right regular. The ring is constructed as a localization of a polynomial ring in countably many indeterminates. Maybe it is interesting to add that the ring is even Noetherian.

  • $\begingroup$ Great, thanks for that reference. $\endgroup$
    – lokodiz
    Commented Sep 4, 2015 at 12:17
  • $\begingroup$ That ring is commutative, so I don't understand this "left and right". (Btw, it is a celebrated example of Nagata.) $\endgroup$
    – user26857
    Commented Dec 30, 2015 at 21:24
  • $\begingroup$ I mean to emphasize that, precisely because it is commutative, this rings works as an example for left regular as well as right regular. Of course, you may just call it regular in the special case (At the disadvantage of possible ambiguity with other notions of regular commutative rings) $\endgroup$
    – moonlight
    Commented Dec 31, 2015 at 8:28
  • $\begingroup$ I can't see any ambiguity here. It is a regular (non-local) ring in commutative algebra, too. $\endgroup$
    – user26857
    Commented Dec 31, 2015 at 9:08

You must log in to answer this question.

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