Why does $Ext_{R}^{n}(I,B)\neq 0$, $I$ is injective with $pd(I)=n$, & $B$ the image of a projective $P$, imply $Ext_{R}^{n}(I,P)\neq 0$?

Question

I've see a proof that say that as $Ext_{R}^{n}(I,B)\neq 0$, where $I$ is an injective module with projective dimension $n$, and as $B$ is the image of a projective module $P$ $Ext_{R}^{n}(I,P)\neq 0$ but I don't understand quite why this is the case. I'm guessing it's to do with the projective resolutions but I'm not quite sure how. Can anyone explain this?

Answer 1

$I$ has projective dimension $n$; so let $0\to P_n\to ...\to P_0\to I$ be a projective resolution.

Then applying $\hom_R(-,M)$ to this for an $R$-module $M$ yields that $\mathrm{Ext}^n_R(I,M) = \hom_R(P_n, M)/\mathrm{Im}\hom_R(P_{n-1}, M)$ : the point is that the differential is $0\to P_n$ so the kernel is the whole $\hom_R(P_n,M)$

But now for any epimorphism $M\to N$, $\hom_R(P_n,M) \to \hom_R(P_n, N)$ is also an epimorphism : that's the definition of projectivity !

Hence by the above formula $\mathrm{Ext}^n_R(I, M)\to \mathrm{Ext}^n_R(I,N)$ is also an epimorphism.

Now if $B$ is the image of the projective module $P$, this precisely means there is an epimorphism $P\to B$, and so there is an epimorphism $\mathrm{Ext}^n_R(I,P)\to \mathrm{Ext}^n_R(I,B)$. So if $\mathrm{Ext}^n_R(I,B)\neq 0$ then $\mathrm{Ext}^n_R(I,P) \neq 0$