I am looking for a proof of the following result: Let $\mathfrak{g}$ be a Lie algebra and $I$ an injective $\mathfrak{g}$-module. Then $\mathrm{H}^q(\mathfrak{g},I)=0$ $\forall q>0$. More precisely, I am looking for a proof in a textbook so that I can cite it. The result shows up in Weibel's book (Exercise 2.5.1), but only as an exercise (no proof).
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2$\begingroup$ Usually one defines $H^q(\mathfrak{g},M)$ as $\operatorname{Ext}^q_{U(\mathfrak{g})}(k,M) $, so the vanishing when $M$ is injective is obvious. To see that this is equivalent with the definition in terms of the Chevalley-Eilenberg complex, see Cartan-Eilenberg, chapter 13. $\endgroup$– abxCommented Oct 25, 2019 at 14:14
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If you define Lie algebra cohomology via injective resolutions, then your claim is trivially true, because you can take the injective resolution which is I in degree 0 and 0 in higher degrees.
To see that your preferred definition agrees with that via injective resolutions, you have to check two things:
any two injective resolutions are chain homotopy equivalent (this is a variant of the fundamental lemma of homological algebra and should have the same proof) and thus yield the same cohomology groups
your preferred definition uses an injective resolution
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$\begingroup$ I use the definition via the Chevalley-Eilenberg complex, $\mathrm{C}^q(\mathfrak{g},I)=\text{Hom}_k(\wedge^q\mathfrak{g},I)$. Is there a way to get the result directly also with this definition? $\endgroup$– SleipnirCommented Oct 25, 2019 at 16:36
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$\begingroup$ It should be possible to write down the 0-homotopy directly. But this will be the same chain homotopy equivalence that one constructs in the proof of the fundamental lemma. $\endgroup$– ThiKuCommented Oct 25, 2019 at 16:45