Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A$ be a Noetherian (not necessarily local) ring and $M$ a finitely generated $A$-moduel. Is the length of the minimal injective resolution of $M$ always equal to the injective dimension of $M$? (Just like the projective dimension and minimal free resolution.) I suspect the formula for the Bass number $$\mu_i(\mathfrak{p},M)=\mbox{dim}_{\kappa(\mathfrak{p})}\mbox{Ext}^i_{A_{\mathfrak{p}}}(\kappa(\mathfrak{p}),M_{\mathfrak{p}})$$ might hold the key, but I can't seem to go anywhere.

share|improve this question

1 Answer 1

If $$0\rightarrow M\rightarrow Q^0\rightarrow Q^1\rightarrow\cdots\equiv\mathcal{Q}^{\bullet}$$ is any injective resolution $\mathcal{I}^{\bullet}$ of $M$, there is an injective chain map $\mathcal{I}^{\bullet}\hookrightarrow\mathcal{Q}^{\bullet}$, implying that $\mbox{length}(\mathcal{I}^{\bullet})\leq\mbox{length}(\mathcal{Q}^{\bullet})$.

share|improve this answer
There is even an easier answer. If $V^n\subset I^n$ is the cosyzygy of minimal injective resolution $\mathcal{I}^{\bullet}$, then $I^n=E(V^n)$, so $$\mbox{id}(M)\leq n\implies V^n \mbox{is injective} \implies I^n=V^n\implies I^{n+1}=0\implies\ell(\mathcal{I}^{\bullet})\leq n.$$ –  ashpool Aug 2 '11 at 1:20

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.