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Why is $KG \cong K[x]/(x^p-1)$, for $K=\mathbb{Z}/p \mathbb{Z}$ and $G = \langle g | g^p=1 \rangle$ a ring isomorphism?

If I take $g \mapsto [x]$, I'd have an group homomorphism. And than an algebra(?) homomorphism f: $KG \rightarrow K[x]$, $a_g g \mapsto a_g[x]$, with $a_g \in K$. Now I need $\ker(f)=(x^p-1)$, not?

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    Dear Jyrki, as usual you display a fine sense of fair play, but in my opinion you shouldn't have deleted your answer.2012-07-05

2 Answers 2

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You should map the other way, namely consider $K[x] \to K[G]$ taking $x\mapsto g.$ Then it should be easy to see it's a homomorphism, and clear that $\operatorname{Ker}(f)=(x^p-1).$

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    Thank you for help!! Now it works2012-07-05
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If $R$ is a ring and $G$ is a cyclic group of order $n$, then for every ring $A$ we have $\hom(R[G],A) \cong \hom(R,A) \times \hom(G,A^*) \cong \hom(R,A) \times \{a \in A : a^n = 1\} \cong \{\phi \in \hom(R[x],A) : \phi(x)^n=1\} \cong \hom(R[x]/(x^n-1),A)$ Yoneda implies $R[G] \cong R[x]/(x^n-1)$. We have used the universal properties of group rings, polynomial rings, and quotients.