Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $R$ be a quasi-Frobenius ring (so $R$ is self-injective and left and right noetherian). I want to prove that $lD(R)=0$ or $\infty$, where $lD(R)$ denotes the left global dimension.

I'm unsure about how to go about proving this; the only thing I can think of is to somehow show that if we assume that some $R$ module $A$ has a finite projective resolution then it is in fact projective and hence has global dimension $0$. I tried to do this using the fact that a module over a quasi Frobenius ring is projective if and only if it is injective, but I didn't get far.

share|cite|improve this question
up vote 4 down vote accepted

Suppose you have a module $M$ with finite projective dimension, say $n$: $$0\to P_n\to \dots\to P_0\to M\to 0.$$ Look at the injection $0\to P_n\to P_{n-1}$. You have that $P_n$ is injective, hence $P_n$ is a direct summand of $P_{n-1}$. But then you could leave out the summand $P_n$ in both $P_n$ and $P_{n-1}$, contradiction to projective dimension $n$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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