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.
Best Answer
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.