# Group Ring calculation

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 $\omega = 1.$ –  Geoff Robinson Dec 31 '11 at 10:45
@GeoffRobinson hmm what is the primitive cube root of unit? is it just the complex roots? –  simplicity Dec 31 '11 at 10:49
@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. –  simplicity Dec 31 '11 at 10:56
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. –  Jyrki Lahtonen Dec 31 '11 at 13:39

It looks like the task is to decompose this group ring into a product of rings. I hope to convey that the decomposition is far from arbitrary: it's completely determined by idempotents.

In short, $R=S\oplus T$ iff there are central idempotents $e$ and $f$ such that $e+f=1$ and $ef=fe=0$. These are called orthogonal idempotents. If the factorization has more factors, then you just have more central idempotents adding up to 1, and they all have pairwise product 0. So if you determine the idempotents, you know everything about the decompositions.

A quick calculuation with $(\alpha+\beta\sigma+\gamma\sigma^2)^2=\alpha+\beta\sigma+\gamma\sigma^2$ yields the system of equations: $\alpha^2+\beta\gamma=\alpha$; $\gamma^2+\alpha\beta=\beta$; and $\beta^2+\alpha\gamma=\gamma$. Using something to solve this system (doing it by hand is a little tedious) you can find that over the complex numbers there are 8 solutions to the system including the $a,b,c$ you had above.

If the characteristic is not 3, then Maschke's theorem says $K[C_3]$ is semisimple, so we can expect a lot of idempotents. Since this group algebra is clearly commutative, we can expect all of the idempotents to be central.

If we explore $K=\mathbb{F}_3$ for a moment, then we have the additional relation that $\delta^3=\delta$ for all $\delta \in K$. Using this relation, you can show that the above system of equations yields only 0 and 1 as potential idempotents. So, $\mathbb{F}_3[C_3]$ is indecomposable.

P.S.: What isomorphism do you have in mind when you say $K[x]/(x^3)\cong K[C_3]$ for char($K$)=3?

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