Let $R$ be the group ring $\mathbb C[C_7],$ where $C_7=\{1,g,\ldots,g^6\}$ is a cyclic group. I would like to prove that
$\mathbb C[C_7]\cong\bigoplus_{i=1}^7\mathbb C.$
I was thinking that I could use the following theorem:
If $R$ is a commutative unital ring and $\{e_i\}_{i=1}^n$ are orthogonal idempotents in $R$, such that $\sum_{i=1}^ne_i=1,$ then $R\cong\bigoplus_{i=1}^ne_iR.$
That would mean I have to find seven orthogonal idempotents $\{e_i\}_{i=1}^7$ with
$e_1+e_2+e_3+e_4+e_5+e_6+e_7=1,$
such that $e_i R\cong \mathbb C.$
Is this the right way to do it? I've managed to find one idempotent $e\in R$ such that $eR = e\mathbb C.$ That is, I take
$e=\frac 17 (1+g+\cdots+g^6).$
I have
$e^2=\frac 17 \cdot \frac 17(1+g+\cdots+g^6)(1+g+\cdots+g^6)=\frac 17\cdot \frac 17\cdot 7(1+g+\cdots+g^6)=e,$
so $e$ is idempotent. For $\sum_{i=0}^6 r_ig^i\in R$, I have
$ \begin{eqnarray} e\sum_{i=0}^6 r_ig^i&=&\frac 17(1+g+\cdots+g^6)(r_0+r_1g+\cdots+r_6g^6)\\ &=&\frac 17(r_0+r_1g+\cdots+r_6g^6\\ &\,& \;\;\,+r_6+r_0g+\cdots+r_5g^6 \\ &\,& \;\;\,\vdots\\ &\,& \;\;\,+r_1+r_2g+\cdots+r_0g^6)\\ &=&e\sum_{i=0}^6 r_i\in e\mathbb C, \end{eqnarray} $ (where $\mathbb C$ denotes the natural copy of the field $\mathbb C$ contained in $R$). Every element of $e\mathbb C$ can be written as $e\sum_{i=0}^6 r_i$, so $eR=e\mathbb C.$
But I need $eR\cong \mathbb C.$ $e\mathbb C$ is not isomorphic to $\mathbb C,$ is it? The former doesn't have a unity, right?
But even if I'm missing something very simple here and $eR\cong \mathbb C,$ then I still don't have seven idempotents -- just the one. Could you please help me with this?