Injective dimension $n$ implies $Ext^n$ does not vanish with an injective Let $M$ be a finitely generated module and suppose that the injective dimension of $M$ is $n$. I want to show that there exists an injective module $I$ such that $Ext^n(I,M)\neq 0$ (and if the projective dimension of $M$ is $n$, then there exists a projective $P$ such that $Ext^n(M,P)\neq 0$, but I suspect the proof will be similar). An idea is to use the fact that there exists a free module $F$ such that $Ext^n(M,F)\neq 0$ (see this question), then use the fact that $F=P\oplus P'$ for some $P$ projective and then use additivity of the $Ext$ functor, but this seems to me like too much work and moreover only works for the projective case. I suspect that there should be a simpler way...
 A: In fact, a proof dual to the one you linked to works.  Let's spell it out.
Let $\ 0\to M\to I_0\to \ldots \to I_{n-1} \stackrel{f}{\to} I_n \to 0$ be a minimal injective resolution of $M$.  The module $I_n$ is non-zero, since $M$ has injective dimension $n$.
Take $I=I_n$.  To compute $Ext^n(I_n,M)$, we apply the functor $Hom(I_n,-)$ to the above resolution.  We get the following complex:
$$ Hom(I_n,I_0) \to Hom(I_n,I_1) \to \ldots Hom(I_n,I_{n-1}) \stackrel{f_*}{\to} Hom(I_n,I_n) \to 0. $$
Then $Ext^n(I_n, M) = Hom(I_n, I_n)/Im(f_*)$.  Thus to prove that $Ext^n(I_n, M)\neq 0$ it suffices to prove that there is a morphism in $Hom(I_n, I_n)$ that is not in the image of $f_*$.
Consider the identity morphism $id_{I_n}\in Hom(I_n, I_n)$.  If this morphism were in the image of $f_*$, then there would be a morphism $g\in Hom(I_n, I_{n-1})$ such that $f\circ g = id_{I_n}$.  Since $I_n$ is injective, this would imply that $g$ is a section, so $I_{n-1}\cong I_n\oplus \ker(f)$.  Thus
$$0\to M\to I_0\to \ldots \to I_{n-2}\to \ker(f)\to 0  $$
would be an injective resolution of $M$, contradicting the fact that its injective dimension is $n$.
Therefore $id_{I_n}$ is not in the image of $f_*$, and $Ext^n(I_n, M)\neq 0$.
