Baer's Criterion for injectiveness of modules says: "An $R$-module $E$ is injective iff for all ideals $I$ of $R$, every homomorphism $f\colon I \to E$ can be extended to $R$." I wonder if there is a graded version for this? I mean:

Let $R$ be a graded ring. Let $M$ be a graded $R$-module.

If $Ext_R^1(R/I,E)=0$ for all homogeneous ideals $I$ of $R$ then $Ext_R^1(M,E)=0$ for every graded $R$-module $M$?

  • $\begingroup$ Can you not just copy the usual proof inserting the word "graded" everywhere? $\endgroup$ Feb 20 '15 at 8:25
  • 1
    $\begingroup$ @Eric: No - cf. my answer. But in some way also Yes - using the right interpretation of "graded Ext" and "graded R"... $\endgroup$ Feb 20 '15 at 12:12

To get the correct graded version you have to observe all possible shifts. More precisely:

Let $G$ be an abelian group, let $R$ be a $G$-graded ring, and let $M$ be a $G$-graded $R$-module. Then, the following statements are equivalent:

(i) $M$ is injective;

(ii) For every $g\in G$, every monomorphism $v\colon N\rightarrow R(g)$ and every morphism $w\colon N\rightarrow M$ there is a morphism $u\colon R(g)\rightarrow M$ such that $u\circ v=w$.

(Note that in the above, "morphism" means morphism in the category of $G$-graded $R$-modules.)

In order to proof this you have to observe that the category of $G$-graded $R$-modules is an AB5-category and that $\bigoplus_{g\in G}R(g)$ is a generator of this category, and then you can apply Lemma 1 in Section 1.10 of Grothendieck's Tohoku paper (Link).

  • $\begingroup$ I added a link to the Tohoku paper (where you can also find a definition of AB5). $\endgroup$ Feb 20 '15 at 14:20
  • $\begingroup$ I do not know of any translation. But do not fear - it is not french but maths! $\endgroup$ Feb 20 '15 at 14:55
  • 1
    $\begingroup$ Michael Barr has an English translation available here: math.mcgill.ca/barr/papers/gk.pdf $\endgroup$ Feb 21 '15 at 9:08

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.