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Let $G = \{1,g,g^2\}$ be a cyclic group of order $3$. Consider the group ring $\mathbb Q[G]$. Then we have the two $G$-invariant simple subspaces $$ I = \{ a_0 + a_1 g + a_2 g^2 : a_0 = a_1 = a_2 \} $$ and $J = \{ a_0 + a_ g + a_2 g^2 : a_0 + a_1 + a_3 = 0 \}$. The first is generated by $1+g+g^2$, the second by $1-g, 1-g^2$ and we have $$ \mathbb Q[G] = I \oplus J. $$ Another way to view the group ring is to note that $\mathbb Q[G] \cong \mathbb Q[x] / (x^3 - 1)$ and then by chinese remainder $$ \mathbb Q[G] \cong \mathbb Q[x] / (x^3 - 1) \cong \mathbb Q[x] / (x-1) \times Q[x] / (x^2+x+1) $$ so that with $\mathbb Q[x] / (x-1] \cong \mathbb Q$ and $\mathbb Q[x] / (x^2 + x + 1) \cong \mathbb Q[\omega]$ where $\omega$ denotes a primitive $3$rd root of unity we have $$ \mathbb Q[G] \cong \mathbb Q \times \mathbb Q[\omega]. $$ Now what is the relation between this isomorphism and the decomposition $Q[G] = I \oplus J$, should we have $I \cong \mathbb Q$ and $J \cong \mathbb Q[\omega]$ in some way? But I do not see how to establish such an isomophism, the "obvious" way to define $\varphi : \mathbb Q[\omega] \to J$ by $\varphi(1) = 1, \varphi(\omega) = g$ does not work out. Even to define for $\psi : \mathbb Q \to I$ by $\psi(q) = q\cdot (1+g+g^2)$ (i.e. just the coefficient with respect to the generating "vector" $1+g+g^2$) does not work, as $3+3g+3g^2 = \varphi(1)\varphi(1) \ne \varphi(1) = 1+g+g^2$. By inspection I found that for $I \cong \mathbb Q$ the mapping $\rho(q) = \frac{q}{3}(1+g+g^2)$ works out, but I don't know why (is there any theory/explanation for that...), and I do not see how to establish the isomorphism $J \cong \mathbb Q[\omega]$?

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Consider internal decomposition $\mathbb{Q}[G]=I\oplus J$. In $I$ there is an element $e=\frac{1+g+g^2}{3}$ (it is a Primitive central idempotent), and $I$ is two-sided ideal generated by this element. We can write $I=Ie$. Then $Ie$ becomes an algebra, in which additive identity is $0$ but multiplicative identity element is $e$; we can show that $Ie$ is isomorphic to $\mathbb{Q}$ as an algebra.

Next, $1-e$ is in $J$ (check). And $J$ is two-sided ideal generated by $1-e$, so write $J=J(1-e)$. Then $J$ is also an algebra with additive identity $0$ but multiplicative identity $1-e$. It becomes isomorphic to $\mathbb{Q}(\omega)$.


(Let me know if some facts are not clear, or doubtful).

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The isomorphism you have written down is just the Chinese remainder theorem. You have an isomorphism $$\phi:\mathbb{Q}[x]/(x^3-1)\to\mathbb{Q}[x]/(x-1)\oplus\mathbb{Q}[x]/(x^2+x+1),\;\;\; f+(x^3-1)\mapsto (f+(x-1),f+(x^2+x+1)).$$ I will omit the coset notation for convenience.

Under this map, we have $$\phi(x^2+x+1)=(3,0)$$ since $x^2+x+1\equiv 1^2+1+1=3$ (mod $x-1$). Hence, your map $I\to \mathbb{Q}$ should be $\frac{1}{3}(1+g+g^2)\mapsto 1$.

Now, observe that $\phi((x-1)^2)=(0,(x-1)^2)=(0,x^2-2x+1)=(0,-3x)$. Therefore, your map $\mathbb{Q}(\omega)\to J$ should be $$\omega\mapsto \frac{-1}{3}(1-g)^2=\frac{-1}{3}(1-2g+g^2).$$ As $\omega^3=1$, you have $$1\mapsto\frac{-1}{27}(1-g)^6=\frac{1}{3}(2-g-g^2).$$

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  • $\begingroup$ Thanks your isomorphism seems to work; but why you look at the image of $(x-1)^2$, whats your intuition in doing so? $\endgroup$ – StefanH Feb 4 '16 at 12:38
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    $\begingroup$ $(1-x)^2$ is $0$ mod $1-x$ and it is quadratic with lead coefficient and constant term 1. That means it is a multiple of $x$ mod $x^2+x+1$. That's what I was looking for. $\endgroup$ – David Hill Feb 4 '16 at 14:35
  • $\begingroup$ Okay, thanks for you answer. Unfortunately I can only accept one, but both of you clarified it to me from different angles! $\endgroup$ – StefanH Feb 4 '16 at 20:50

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