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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.

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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 $\mbox{length}(\mathcal{I}^{\bullet})\leq\mbox{length}(\mathcal{Q}^{\bullet})$.

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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

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