Let $R$ be a commutative Noetherian ring with non-zero identity, $M$ be an $R$-module
and $E$ be an injective $R$-module. When $Hom(M,E)$ is injective?  

Thanks.



When Hom(M,E) is injective?
1. $Hom(K,Hom(M,E))\cong Hom(K\otimes M,E)$
2. $F$ is flat iff $F\otimes -$ is exact.
3. Let $E$ be [injective cogenerator](https://math.stackexchange.com/questions/229168/injective-cogenerators-in-the-category-of-modules-over-a-noetherian-ring). Then $0 \longrightarrow X \longrightarrow Y \longrightarrow Z \longrightarrow 0$ is exact iff $0 \longrightarrow Hom(Z, E) \longrightarrow Hom(Y, E) \longrightarrow Hom(X, E) \longrightarrow 0$ is exact. 



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Using (1) and (2), you can see that If $M$ is flat then $Hom(m,E)$ is injective (as abx said). Using (3), you can see that if $E$ is injective cogenerator, then you have also the necessary condition.


