# Proving that $\mathbb C[C_7]\cong\bigoplus_{i=1}^7\mathbb C$

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?

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$e \mathbb{C}$ is isomorphic to $\mathbb{C}$. The isomorphism sends $e$ to $1$. –  Qiaochu Yuan Feb 14 '12 at 0:12
Ah, indeed, they are. Thank you! –  user23211 Feb 14 '12 at 9:32

Another way to look at it, is that you have an epimorphism $\mathbb{C}[x]\to\mathbb{C}[\mathbb{Z}/7\mathbb{Z}]$ and it's not hard to see that the kernel of this map is $(x^7-1)$. Thus, as rings, $\mathbb{C}[x]/(x^7-1)\cong\mathbb{C}[\mathbb{Z}/7\mathbb{Z}]$ and the rest follows from CRT.

EDIT: OF course, this easily generalizes to show that $\mathbb{C}[\mathbb{Z}/n\mathbb{Z}]\cong\mathbb{C}^n$ as $\mathbb{C}$-algebras. Here's a bit of a thought experiment for you. Let's see if we can see, theoretically, how everything got set up. It's easy to see (by cyclicity!) that $\mathbb{C}[\mathbb{Z}/n\mathbb{Z}]$ is a singly-generated commutative $\mathbb{C}$-algebra which is what guarantees us the $\mathbb{C}$-algebra epimorphism $\mathbb{C}[x]\to\mathbb{C}[\mathbb{Z}/n\mathbb{Z}]$. It seems fairly interesting that the kernel of this map (which, of course, is $(x^n-1)$) is the "presentation" of $\mathbb{Z}/n\mathbb{Z}$. One then may start to wonder if, in general, given a group $G$ with presentation $\left\langle S\vert R\right\rangle$ and commutative ring $k$ if $k[G]$ is nothing more than $k\left\langle S\right\rangle/(R)$ where $k\left\langle S\right\rangle$ is the free (non-commutative) $k$-algebra over $S$. It's a good exericse to see if, in general, this is true, and see to what extent we can (if it's wrong) fix it.

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Thank you! I think it's neater than the idempotents. –  user23211 Feb 14 '12 at 10:56
@ymar: that may be so, but it doesn't generalize to noncommutative groups whereas the approach using idempotents does. –  Qiaochu Yuan Feb 14 '12 at 16:36

If you know about Wedderburn's theorem, you can use the fact that your group algebra is artinian, semisimple and commutative over the algebraically closed field $\mathbb C$ and of dimension $7$.

Alternatively, you need seven idempotents. If $\lambda$ is a primitive $7$th root of unity and $g$ generates your group, then for each $j\in\{0,\dots,6\}$ define $$e_j=\frac{1}{7} \sum_{i=0}^6\lambda^{ij}g^i.$$ You can then prove that the $e_j$ are central orthogonal idempotents, by direct computation. Since they are $7$, they must be a basis.

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The definition I know of semisimple implies Artinian (every module is a direct sum of simple modules) and is easy to verify for group algebras of finite groups. Which definition are you using (semiprimitive, perhaps)? –  Qiaochu Yuan Feb 14 '12 at 0:10
Thank you very much, Mariano! –  user23211 Feb 14 '12 at 10:56
@QiaochuYuan, I never remember the precise definitions of these things :) By semisimple I meant indeed that modules are semisimple, as in your parenthesis. –  Mariano Suárez-Alvarez Feb 14 '12 at 19:49

This can be understood in terms of discrete Fourier analysis with $N:=7$. Put ${\mathbb C}[C_7]=:R$, ${\mathbb Z}/(7{\mathbb Z})=:Z_7$, and let $V\simeq{\mathbb C}^7$ be the space of functions $f:\ Z_7\to{\mathbb C}$. Finally put $e^{2\pi i/7}=:\omega$.

Then one defines the Fourier transform $$\Phi:\quad R\to V\ ,\qquad a=\sum_j a_j g^j\ \mapsto \hat a$$ of elements $a\in R$ by $$\hat a(k)\ :=\ \sum_j a_j\omega^{jk}\qquad(k\in Z_7)\ .$$ The DFT inversion formula $$a_j={1\over7}\sum_k \hat a(k)\ \omega^{-jk}$$ shows that $\Phi$ is a bijective linear map from $R$ onto $V$, and the calculation $$\hat a(k)\ \hat b(k)=\sum_{j,l} a_j b_l \omega^{(j+l)k} =\sum_r\Bigl(\sum_{j+l=r} a_j b_l\Bigr) \omega^{rk} =\widehat{ab}(k) \qquad(k\in Z_7)$$ proves that this map is even a isomorphism of rings.

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+1: Join us and spread the gospel (to EE guys at least): Discrete Fourier analysis is just representation theory of finite cyclic (or abelian) groups! –  Jyrki Lahtonen Feb 14 '12 at 18:37