Why is the equation above satisfied, if $I$ ist an Ideal in $\mathbb{Z}[\zeta]$, $\zeta$ a $p\neq 2$ root of unity and $G$ a finite group? Thank you
$\operatorname{Hom}(\mathbb{Z}G,I) \cong I $
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abstract-algebra
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0@HenningMakholm : sorry, I meant $p\neq 2$ – 2012-06-14
1 Answers
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There is a more general fact that $\hom_{\mathbb{Z}G} (\mathbb{Z}G,M)\cong M$ (as an abelian group) for any $\mathbb{Z}G$-module $M$. The isomorphism is that $m \in M$ corresponds to the map $f_m: \mathbb{Z}G \to M$ determined by $f_m(X) = Xm$
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0@JyrkiLahtonen good point - the universal property of a free module of rank one is exactly it, or the universal property of $\mathbb{Z}_{\langle e \rangle} \uparrow ^G$. – 2012-06-15