Rings in this question are assumed to be commutative. I am asking "natural" examples of rings of weak dimension $\le1$ which are not semihereditary. It would be better if there are integral domains as examples. (see Update)

There are several equivalent characterizations of rings of weak dimension $\le1$ in [SP, Tag 092S]. One is the following: a ring is of weak dimension $\le1$ if every finitely generated ideal is flat.

Another slightly stronger concept is that of semihereditary rings. When talking about integral domains, they are also called Prüfer. A ring is semihereditary if every finitely generated ideal is projective.

Since a module is finite projective if and only if it is flat of finite presentation, we see that a ring of weak dimension $\le1$ is semihereditary if and only if it is coherent. However, I want to see some examples of non-coherent rings of weak dimension $\le1$, especially those which appear "naturally" in commutative algebra or algebraic geometry. A bit googling does not lead me to anything.

Update: There is no such example for integral domains: every finitely generated flat module over an integral domain is necessarily projective. More generally, there is a criterion for semihereditary rings in

Vasconcelos, W. V. (1969). On Finitely Generated Flat Modules. Transactions of the American Mathematical Society, 138, 505–512. https://doi.org/10.2307/1994928

Theorem 4.2. For a commutative ring $R$, the following are equivalent:

  1. $R$ is semihereditary;
  2. $R$ has weak dimension at most one and the annihilator of each element is finitely generated.

This theorem suggests that there are examples of rings of weak dimension $\le1$ but not semihereditary. However, I do not find any example in the literature. On the other hand, it seems slightly counterintuitive that, rings of weak dimension $0$ (aka. absolutely flat rings, or von Neumann regular rings) are semihereditary, since every finitely generated ideal is a direct summand of the ring itself, thus projective.

  • 2
    $\begingroup$ I don't think a Prüfer ring is the same as a semihereditary ring. (And, less seriously, the first sentence "All rings are commutative" is false :-) ). Von Neumann regular rings (i.e., rings for which all modules are flat) satisfy what you're asking for, but I'm not sure those are the kind of examples you're interested in. $\endgroup$ Jan 11, 2023 at 20:18
  • 1
    $\begingroup$ @JeremyRickard Yes, sure, although it would be better if there are integral domains as examples (where the word Prüfer applies). In retrospect, von Neumann regular rings are also covered in the same section stacks.math.columbia.edu/tag/092A $\endgroup$
    – Z. M
    Jan 11, 2023 at 21:40
  • $\begingroup$ @JeremyRickard Sorry, it seems to me that von Neumann regular rings are semi-hereditary, since finitely generated ideals are direct summands of the ring itself (Exercise 2.27 in Atiyah–Macdonald), thus projective? $\endgroup$
    – Z. M
    Jan 12, 2023 at 9:55
  • $\begingroup$ Yes, that's true. Sorry. $\endgroup$ Jan 12, 2023 at 10:05
  • $\begingroup$ I'd recommend to not trust in anything written on Stacks that involves non-Noetherian rings. There are too many, I'd say, blunders on this site resulting from users blindly inserting words into template statements without realising which sections have "all rings are Noetherian" implicitly written somewhere. $\endgroup$
    – Denis T
    Nov 15, 2023 at 3:24

1 Answer 1


(NB: all rings mentioned below are commutative.)

The most natural example of a ring which has weak global dimension equal to 1, but is NOT semihereditary, is $B[[X]]$, where $B$ is $$\frac{\prod_{\omega} \Bbb F_2} {\bigoplus_{\omega} \Bbb F_2 }.$$ Proof is a bit involved; I'll update the answer in few days to incorporate it.

Archimedean version of the above is $C(\beta \Bbb N \setminus \Bbb N)$, or $C(\beta \Bbb R_{\geq 0} \setminus \Bbb R_{\geq 0})$, the latter one being in some sense "generic ring of wdim 1". See [1] for context and proofs (...some of which are left as excercises).

Another example is due to Jensen. Consider a subring $$C \subset \prod_{i \in \Bbb N} k[X_i],$$ which consist of sequences with finitely many non-constant entries; then the ring $\langle C, u \rangle$, generated by $C$ and an element $u = (0, X, 0, X, 0,..)$ will have wdim equal to 1.

Also there are two (folklore, as far as I know) results which are probably relevant for geometric applications.

  1. $R$ is vN regular $\Leftrightarrow$ $R[X]$ is semihereditary.

  2. $R$ is semihereditary $\Leftrightarrow$ finitely presented modules over $R[X]$ have pdim not exceeding two.

  3. ? (...I'd like to have a continuation of this sequence, but the direction in which it should go is unclear to me. This feels a bit like a "half-integral global dimension hierarchy"; I have tried formulating this condition properly and have failed several times.)

[1]: L. Gillman, M. Jerison, Rings of Continuous Functions


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.