RG is a group ring and R is a ring. Let $f:RG\rightarrow R$ be a homomorphism such that $f(\sum r_ig_i) = \sum r_i$ . Prove that the kernel of $f$ is a principal ideal.
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What are R and G? Rings? Are mentions of "i" meant to be subscripts? – hardmath Jul 03 '14 at 02:17
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I think $RG$ is supposed to be a group ring. – Nishant Jul 03 '14 at 02:17
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I forgot to indicate that it was RG and R, RG is a group ring and R is a ring – Alex Pozo Jul 03 '14 at 02:23
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2This kernel is the augmentation ideal is generated by ${g-1 : g\in G}$. I don't think it is principal in general, but perhaps it is under some specific conditions on $R$ and $G$. – Prahlad Vaidyanathan Jul 03 '14 at 02:44
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1Why do you think this is true? – Thomas Andrews Jul 03 '14 at 02:49
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Let $R$ be a (commutative) domain, and let $G = \mathbb{Z}^2$, so that $R[G] = R[X, X^{-1}, Y, Y^{-1}]$. The kernel $I = \ker f = (1 - X, 1 - Y)$. Suppose $I$ is principal, so that $I = (\alpha)$ for some $\alpha\in R[X, Y]$. Assume without loss of generality that $X, Y\!\not|\, \alpha$. Then $\alpha | (1 - X) X^n Y^m$ and $\alpha | (1 - Y)X^{n'} Y^{m'}$ in $R[X, Y]$ for some $n, m, n', m'$. Since $R[G]/I = R$ is a domain, $I$ is prime. Thus $\alpha |(1 - X)$ and $\alpha | (1 - Y)$ in $R[X, Y]$, forcing $\alpha\in R$. But then $I$ contains $X\in R[G]$, which is clearly not in $I$.
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