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Let, $C_3=\langle\sigma\mid\sigma^3=1\rangle$, $a=\frac{1}{3}( 1+ \sigma +\sigma^2)$, $b=\frac{1}{3}(1+w \sigma+ w^2 \sigma^2)$ and $c=\frac{1}{3}(1+w^2\sigma+ w\sigma^2)$ where $w$ is the primitive cube root of 1.

Characteristic is $0$. Need to decompose the group ring $KG$ (where $K$ is a field) into $KG \cong aK \oplus bK \oplus cK$. Personally, I don't see why there are two decomposition from $KG$ into $aK \oplus bK \oplus cK$(when char $K=0$) and $KG \cong K[x]/(x^3)$ (when char $K=3$). Seems do arbitrary.

In the notes it says you need to calculuate $a^2=a$, e.t.c show they are all idempotents.

However, it says that you need to show all crossproduct of $a,b,c$ are $0$.

However, when I do $ab=\frac{1}{3}(1+w+w^2+(1+w+w^2)\sigma+(1+w+w^2)\sigma^2)$. This doesn't equal $0$ when you take $w=1$.

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    But $\omega$ is supposed to be a primitive cube root of unity, and that excludes the possibility $\omega = 1.$2011-12-31
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    @GeoffRobinson hmm what is the primitive cube root of unit? is it just the complex roots?2011-12-31
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    @GeoffRobinson Thanks, yeah see $w^1=1$ so 1 can't be a primitive cube root of unit. So then it must be ab=0. Thanks a lot, would never figure that simple thing out myself.2011-12-31
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    Apparently you though of $\omega$ as a variable as opposed to a constant. Ok, live an learn :-) But if $\omega=1$, then $a=b=c$, and that should have tipped you off!? Also, there's a lingering typo. It should read $KG\simeq K[x]/(x^3)$, when $char K=3$. The case $char K=2$ is not a problem. As long as $K$ contains the field of 4 elements, it will have a primitive cube root of 1. OTOH the case $char K=3$ will give you a headache, because you cannot divide by 3. Therefore you cannot even define $a,b,c$ with these formulas.2011-12-31

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