A noetherian ring $R$ is said to be regular if every localization at a prime ideal is regular local.

On the other hand, there is another definition of regularity for non-noetherian rings:

A (commutative) ring $R$ is said to be regular if every finitely generated ideal has finite projective dimension.

In many books (e.g. 'Commutative Coherent Rings' by Sarah Glaz), it is said that these two definitions coincide for noetherian rings.

However, I can't prove it nor find any proof.

Is there any reference or proof? Thanks.

Edit: By Serre's theorem, two definitions coincide for noetherian local rings and for noetherian rings of finite dimension.

  • $\begingroup$ Could you show how some ideal in $k[t^2,t^3]$ fails to have a finite projective resolution ? $\endgroup$
    – reuns
    Jan 19, 2020 at 3:04
  • $\begingroup$ It has a cusp at the origin so it is not regular. $\endgroup$
    – Hiro Wat
    Jan 19, 2020 at 4:12
  • $\begingroup$ I see; I know that the "homological" regularity implies the "ideal-theoretic" regularity, and the converse is the problem. $\endgroup$
    – Hiro Wat
    Jan 19, 2020 at 4:16

1 Answer 1


I have proven it!

Let $M$ be a f.g. $R$-mod, and take a projective resolution

$$\cdots \overset{d_2}{\to} P_2 \overset{d_1}{\to} P_1 \overset{d_0}{\to} P_0 \to M \to 0.$$

Put $K_i := \operatorname{Ker}d_i$. It suffices to show that there exists $N \geq 0$ such that $K_N$ is projective.

For each $p \in X:= \operatorname{Spec}R$, set $d(p):= \operatorname{pd}_{R_p} M_p < \infty.$

Then $(K_{d(p)})_p$ is free, and thus there is an open neighbourhood $U_p$ of $p$ in $X$ such that $K_{d(p)}$ is free on $U_p$.

Since $X$ is quasi-compact, we can choose a finite number of primes $p_1, \cdots p_n$ so that $$X = U_{p_i} \cup \cdots \cup U_{p_n}.$$

Set $N:= \underset{i}{\operatorname{max}} d(p_i )$.

For any $p\in X$, there is an $i$ such that $p \in U_{p_i}$.

Then $(K_{d(p_i)})_p$ is free and thus $d(p) \leq d(p_i) \leq N$.

Thus $(K_N)_p$ is free. Therefore $K_N$ is projective, as wanted.


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