Let $G=\langle \sigma \rangle $ be a cyclic group of order $3$ (or if you prefer $G=C_3$) .
- We would like to prove that $\Bbb C G$ is a direct sum of fields.
- We would like to prove that $\Bbb Q G$ is also a direct sum of fields.
- If $K$ is a field of characteristic $3$, we would like to prove that $$KG \cong K[X]/\langle X^3 \rangle,$$ as rings.
Proof. 1. We define the map $\theta: \Bbb C[X] \longrightarrow \Bbb C G,\ f(X) \longmapsto \theta (f(X)):= f(\sigma).$ Then, this is a ring homomorphism with image $\mathrm{Im} \theta= \Bbb C G$, so $\theta$ is in fact a surjective homomorphism. We now claim that $\mathrm{Ker} \theta = \langle X^3-1 \rangle$. For the $\supseteq$, we have $$\theta(X^3-1)= \sigma ^3 -1 =0 \implies X^3-1 \in \mathrm{Ker} \theta \implies \langle X^3-1\rangle \subseteq \mathrm{Ker} \theta.$$
For the $\subseteq$, take $f(X)\in \mathrm{Ker} \theta$. Then, by the Division Algorithm we have $$f(X)=q(X)(X^3-1)+r(X),\ \text{ where } r(X)=0 \text{ or } \deg r(X) \leq 2.$$ Then, $0=\theta (f(X))=\theta(r(X)) \iff r(\sigma)=0 \iff a\sigma^2+b\sigma+1=0 \iff a=b=0$, where $a,b\in \Bbb C$ So, $r(X)=0$ and $\mathrm{Ker} \theta \subseteq \langle X^3-1 \rangle$. Finally, we obtain the equality. So, by the 1st Iso Theorem for rings, we have $$\Bbb C C_3 \cong \Bbb C[X]/ \langle X^3-1 \rangle.$$ Since $X^3-1=(X-1)(X^2+X+1)=(X-1)(X-\alpha)(X-\alpha^2)$, be the Chinese Remainder Theorem, we have $$\Bbb C C_3 \cong \Bbb C[X]/ \langle X^3-1 \rangle \cong \Bbb C[X]/\langle X-1 \rangle \times \Bbb C[X]/\langle X-\alpha \rangle \times \Bbb C[X]/\langle X-\alpha^2\rangle,$$ and we are done.
- Following the same technique, since $X^3-1=(X-1)(X^2+X+1)$ is the factorisation into irreducibles, we obtain $$\Bbb Q C_3 \cong \Bbb Q[X]/ \langle X^3-1 \rangle \cong \Bbb Q[X]/\langle X-1 \rangle \times \Bbb Q[X]/\langle X^2+X+1 \rangle,$$ and we are done.
Questions.
1) Are the proofs of 1. and 2. complete and correct?
2) How do we cope with 3. ?
Thank you.