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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.

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  • $\begingroup$ A sufficient condition is that $M$ is flat, because of the isomorphism $\mathrm{Hom(-,\mathrm{Hom(M,E)})}\cong \mathrm{Hom(-\otimes M,E)}$. I don't think you can say more. $\endgroup$
    – abx
    Commented Apr 3, 2016 at 6:31
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    $\begingroup$ also asked math.stackexchange.com/questions/1725640/… $\endgroup$
    – user 1
    Commented Apr 3, 2016 at 9:03

1 Answer 1

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  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. 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.

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.

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