1
$\begingroup$

Let $M$ be a module over a commutative ring (with unity) $R$. Let $\phi : M \to M$ be an $R$-module homomorphism. Then we have a dual map $\phi^* : M^* \to M^*$ given by $\phi^*(f)=f\circ \phi, \forall f \in M^*$. My questions are:

(1) If $\phi $ is surjective, then $\phi^*$ is injective. Under what conditions on $R$ or $M$, can we say that $\phi$ is injective $\implies \phi^*$ is surjective ?

(2) If $M$ is torsion-less, i.e., $\bigcap_{f\in M^*}\ker f=0$, then $\phi^*$ surjective $\implies \phi$ is injective. Under what conditions on $R$ or $M$, can we say that $\phi^*$ is injective $\implies \phi$ is surjective ?

[NOTE: Here, $M^*:=\mathrm{Hom}_R(M,R)$]

$\endgroup$
  • 1
    $\begingroup$ I am guessing this is not the sort of condition that you're interested in (hence why this is only a comment), but if $M$ is Artinian and finitely generated, then any injective endomorphism is necessarily an isomorphism. This gives one situation where (1) holds. $\endgroup$ – Alex Wertheim May 16 at 23:16
  • $\begingroup$ @AlexWertheim: sure ... but that is too strong ... I am looking for some condition to deal with taking the dual ... $\endgroup$ – user102248 May 16 at 23:23
  • 1
    $\begingroup$ The question is too broad. For example, if $R\to R$ is multiplication by a non-zero divisor which is a non-unit, then it is is injective, but the dual is not surjective. So, unless you have some restrictions, it is difficult to answer in this open ended fashion. $\endgroup$ – Mohan May 16 at 23:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.