# $\operatorname{Hom}(\mathbb{Z}G,I) \cong I$

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

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Hi @Amy: could you please write down what've you done here? Also, what's the background for the question, is it from a book...? –  DonAntonio Jun 14 '12 at 14:52
What is $\mathbb ZG$? What is the relevance of $\zeta\ne 2$ -- 2 cannot be a root of unity in the first place (assuming characteristic 0, but otherwise what is $\mathbb Z$ doing there?) –  Henning Makholm Jun 14 '12 at 14:54
Hi, yes it is from a book, where the author wants to proof, that $Ext_{\mathbb{Z}G}^1 (\mathbb{Z},I) \cong I/(\zeta -1)I$. In the proof he uses the sentence, that the congruence in the title is clear. I'm not sure, if $p\neq 2$ is important, but I wanted to mention it. $G$ is a group, that is isomorphic to the dihedral group with $2p$ elements and $\mathbb{Z}G$ the groupring –  james B. Jun 14 '12 at 15:04
@HenningMakholm : sorry, I meant $p\neq 2$ –  james B. Jun 14 '12 at 15:13
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 $f_m(X) = Xm$
@JyrkiLahtonen good point - the universal property of a free module of rank one is exactly it, or the universal property of $\mathbb{Z}_{\langle e \rangle} \uparrow ^G$. –  mt_ Jun 15 '12 at 10:44