# On the injective dimension of a module

Let $R$ be a ring and $M$ an $R$-module then

inj dim $M\leq i\in\mathbb{N}$ if and only if $\mathrm{Ext}^{i+1}(N,M)=0$ for every cyclic module $N$.

The implication from left to right is obvious, I'm finding some difficulties in proving the other implication. In case we need it I think we can suppose $M$ finitely generated and $R$ noetherian. Do you have any suggestions fot the implication I'm missing?

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This reminds me of Baer's criterion. – Jack Schmidt Oct 3 '12 at 2:02
At any rate, use Schanuel's lemma to do "dimension shifting", and it should follow from Baer's criterion. An outline is in Lam section 5C, but this specific result is not there. – Jack Schmidt Oct 3 '12 at 3:53
thank you, that helped, I posted an answer following you suggestions, hope it's right. – Chris Oct 3 '12 at 9:17

$0\rightarrow M\rightarrow I^0\rightarrow I^1\rightarrow\cdots\rightarrow I^{i-2}\rightarrow I^{i-1}\rightarrow M^\prime\rightarrow0$
where the $I^j$'s are injectives and $M^\prime=\mathrm{coker}(I^{i-2}\rightarrow I^{i-1})$. Then by dimension shifting $0=\mathrm{Ext}^{i+1}(N,M)=\mathrm{Ext}^1(N,M^\prime)$. So if $N=R/J$ then the exact sequence $0\rightarrow J\rightarrow R\rightarrow R/J\rightarrow 0$ yields an exact sequence
$0\rightarrow\mathrm{Hom}(R/J,M^\prime)\rightarrow\mathrm{Hom}(R,M^\prime)\rightarrow\mathrm{Hom}(J,M^\prime)\rightarrow0$ (because $\mathrm{Ext}^1(N,M^\prime)=0$). So by Baer's criterion $M^\prime$ is injective (because $\mathrm{Hom}(R,M^\prime)\rightarrow\mathrm{Hom}(J,M^\prime)$ is surjective for every ideal $J$).