Let $A$ be any ring and consider modules on the left. For $M$ $A$-module, the trace $Tr(M,A)$ is a two-sided ideal of $A$. If $A$ is a unitary ring then:

  • $Tr(P,A)P=P$, for $P$ projective;
  • $Tr(P,A)^2=Tr(P,A)$.

Further, if $A$ is semiprimary then $I$ is an idempotent ideal of $A$ if and only if $I=Tr(P,A)$, for some finitely generated projective $P$, if and only if $I=AeA$, for some idempotent $e∈A$.

In parallel, it is easy to check that if $A$ is unitary then the reject $Rej(A,M)$ is the (left) annihilator of the module $M$.

Is there a characterisation of the reject of injective modules that is "somehow" dual to the one above?

I know that in sufficiently good contexts there is a connection between annihilators of simple modules and prime ideals... It would be nice to have something for the reject of injectives that would hold for A semiprimary (or artinian or, in the worst case scenario, an Artin algebra).

Just one more thing. In the paper "Homological Theory of Idempotent Ideals" (Auslander, Platzeck, Todorov) we can find different characterisations of $k$-idempotent ideals using $\operatorname{Ext}$. If the base ring has a self-duality, $k$-idempotent ideals may also be characterised in terms of $\operatorname{Tor}$. So...

We have that an ideal $I$ of $A$ is idempotent if and only if $\operatorname{Ext}_A^1(A/I,Y)=0$ for all $Y$ in $\operatorname{mod} A/I$ (there are many other similar characterisations).

I was trying to see if it is true that: $I=Rej(A,Q)$, for some injective $Q$, if and only $\operatorname{Tor}^A_1(A/I,Y)=0$ for all $Y$ in $\operatorname{mod} A/I$...

(Maybe the assertion "$I=Rej(A,Q)$, for some injective $Q$, if and only $\operatorname{Ext}_A^1(Y,I)=0$ for all $Y$ in $\operatorname{mod} A/I$" makes more sense.

The problem is that projectives and injectives are not quite "dual" in $\operatorname{Mod}A$. The proofs I know for the properties that relate traces and projectives use the dual basis lemma.)

  • 1
    $\begingroup$ Try to look in the first part of this book: "Equivalence and Duality for Module Categories with Tilting and Cotilting for Rings" $\endgroup$ – Simone Virili Jul 9 '13 at 19:29
  • 2
    $\begingroup$ The assertion in the parenthetical remark seems to be false. There are examples with $I=\operatorname{Rej}(A,Q)$ for some injective $Q$ and a non-split extension $0 \rightarrow I \rightarrow A \rightarrow A/I \rightarrow 0$. Take for instance the path algebra of $A_2$ and let $Q$ be the simple injective. $\endgroup$ – Dag Oskar Madsen Jul 9 '13 at 20:22
  • $\begingroup$ @Dag Oskar Madsen - Thank you very much for the counterexample. I am trying to see which properties of standard modules remain true (after dualisation) for costandard modules when the base ring does not have a self-duality. $\endgroup$ – 3 A's Jul 10 '13 at 7:10
  • $\begingroup$ In the example I gave, the sequence does split as a sequence of \emph{right} modules. Maybe there is something going on there. $\endgroup$ – Dag Oskar Madsen Jul 10 '13 at 9:06
  • $\begingroup$ @Simone Virili - Thanks very much for your suggestion. I understood this much better after realising the functor $Rej(−,Q)$, $Q$ injective, as an hereditary preradical ($Tr(P,−)$ is an cohereditary preradical if $P$ is projective). I read the first paragraphs of Chapter 2 of "Lifting Modules", Wisbauer et al, and looked up in some other places. Yet, it seems the ideal $Tr(P,A)$ has always "better" properties than $Rej(A,Q)$. I think this is because $A$ is a generator for $\operatorname{Mod}A$ (though not a cogenerator for $\operatorname{Mod}A$, in general). $\endgroup$ – 3 A's Jul 16 '13 at 10:02

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.