Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

$A$ a Noetherian local ring, $M\neq 0$ a finite $A$-module. I'm not quite sure about the relation between finiteness of projective and injective dimensions of $M$. Does the finiteness (or infiniteness) of one necessarily imply the finiteness (or infiniteness) of another?

share|improve this question

2 Answers 2

up vote 7 down vote accepted

To complement Mariano's answer: If finite projective dimension implies finite injective dimension for any module $M$, then $R$ better have finite injective dimension (the converse is also quite easy).

The local rings $R$ which have finite inj. dim. over themselves are also known as Gorenstein rings. In fact, a theorem by Foxby says that $R$ possesses a module of both finite proj. and inj. dim. if and only if $R$ is Gorenstein.

share|improve this answer
Thanks. How do I see that if $R$ has a finite injective dimension then finite projective dimension of $M$ implies finite injective dimension of $M$? –  ashpool Aug 6 '10 at 16:10
You can use induction on the length of the min. free res. of $M$, for instance. –  Hailong Dao Aug 6 '10 at 17:23
I'm sorry, I still have no idea how to proceed. To begin with, if $A$ has a finite injective dimension and $M$ is projective over $A$, why does $M$ have a finite injective dimension? If you could point to any reference that would be great, too. –  ashpool Aug 7 '10 at 3:34
@kwan: break the res. into short exact sequences and use the fact that if 2 modules have fi. inj. dim., so is the third one. For ref (without proof), look at "Cohen-Macaulay rings" 1st ed by Bruns-Herzog, Section 3.1, esp. exer. 3.1.25. –  Hailong Dao Aug 7 '10 at 18:08
Thanks. The same method (of using minimal free resolution) doesn't seem to work to show the converse, if $A$ is Gorenstein and $M$ has a finite injective dimension then $M$ has a finite projective dimension. I tried to use finite injective resolution instead but it doesn't seem to work either. I would appreciate any suggestions. –  ashpool Aug 7 '10 at 20:33

No. For example, there are rings $A$ for which the residue field $k$ is of infinite projective dimension but of finite injective dimension. On the other hand, if $k$ is of finite projective dimension, then $A$ is of finite global dimension, so $k$ (and everything else) has finite injective dimension.

For a small non-commutative example, consider the algebra $A$ which is the quotient of the path algebra of the quiver


modulo the ideal generated by $\beta\alpha$ and $\beta^2$. Then the simple module supported on the vertex $1$ is an injective of infinite projective dimension. By duality, the opposite algebra has a projective module of infinite injective dimension.

share|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.