0
$\begingroup$

Suppose you are given a bounded chain complex $M$ over a commutative ring $R$.

Is there a clear relation between homological dimensions of $M$ and homological dimensions of its cohomologies?

For example, suppose I know that $injdim(H^i(M))<\infty$ for all $i$, does this imply that $injdim(M)<\infty$? what about the converse? what about projective and flat dimension?

Any reference for this?

Improve this question
$\endgroup$
4
  • $\begingroup$ What do you mean by the homological/injective dimension of a chain complex? $\endgroup$
    – Fernando Muro
    Commented Aug 24, 2015 at 15:28
  • 1
    $\begingroup$ If you mean the minimum length of an injective complex that is quasi-isomorphic to the original complex, then I think the answer is yes, by writing the complex up to quasi-isomorphism as an iterated extension of its homology groups and applying the mapping cone to construct an injective resolution of the complex from the injective resolution of the homology groups. $\endgroup$
    – Will Sawin
    Commented Aug 24, 2015 at 15:42
  • $\begingroup$ @WillSawin do you mean complexes of injective modules? By contrast, injective complexes are contractible, I think. $\endgroup$
    – Fernando Muro
    Commented Aug 24, 2015 at 22:24
  • $\begingroup$ @FernandoMuro Yes, I mean precisely that. $\endgroup$
    – Will Sawin
    Commented Aug 25, 2015 at 2:01

1 Answer 1

Reset to default
3
$\begingroup$

Consider the ring $R = \mathbb Z/4$. Then $\mathbb Z/4$ is both injective and projective over itself, whereas $\mathbb Z/2$ has infinite projective and injective dimension.

We could have the chain complex: $ 0 \to \mathbb Z/4 \to^{\cdot 2} \mathbb Z/4 \to 0$ with the homologies having infinite dimension and the terms of the complex not. Or we could have $0 \to \mathbb Z/2 \to^{\cdot 1} \mathbb Z/2 \to 0$ with the homologies having having $0$ dimension, but the terms of the complex not.

(As Fernando asked, am I interpreting your question correctly?)

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .