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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 $M$?

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  • $\begingroup$ Can you not just copy the usual proof inserting the word "graded" everywhere? $\endgroup$ Feb 20 '15 at 8:25
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    $\begingroup$ @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
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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).

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  • $\begingroup$ 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
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    $\begingroup$ Michael Barr has an English translation available here: math.mcgill.ca/barr/papers/gk.pdf $\endgroup$ Feb 21 '15 at 9:08

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