If $G \cong \mathbb Z/3\mathbb Z$, then $\mathbb R[G] \cong \mathbb R \times \mathbb C$

Question

Let $G = \{1,g,g^2\}$ be the cyclic group of order three. Consider the group ring $\mathbb R[G]$, then $\mathbb R[G] \cong \mathbb R \times \mathbb C$ with the isomorphism $$ \varphi(1) = (1,0), \quad \varphi(g) = (1,\omega), \quad \varphi(g^2) = (1,\overline{\omega}) $$ this is written in this document on page 7. But I am unsure, first $$ (1,0)(1,\omega) = (1, 0) \ne (1,\omega) $$ so $(1,0)$ does not behave like an identity. Further in $\mathbb R[G]$ we have $(1+g)(1+g) = 1 + 2g + g^2$, but $$ ((1,0) + (1,\omega))((1,0)+(1,\omega)) = (2,\omega)(2,\omega) = (4,\overline \omega) \ne (1,0) + 2(1,\omega) + (1,\overline \omega) $$ The first objection might be fixed by setting $\varphi(1) = (1,1)$, but then the second still does not works out.

So how does this isomorphism work? Or have I used the wrong multiplication, isn't it defined componentwise?

Answer 1

The document only says that $$\phi(g)=(1, \omega)$$

As you observed you must have $$\phi(1)=(1,1)$$

Fixing this, the second relation becomes:

$$((1,1) + (1,\omega))((1,1)+(1,\omega)) = (2,1+\omega)(2,1+\omega) = (4,1+2\omega+\bar{w}) \\ =(1,0) + 2(1,\omega) + (1,\overline \omega)$$

Answer 2

So $\phi$ is defined by $\phi(g) = (1,\omega)$. Hence we must have $$ \phi(1) = \phi(g^3) = (1, \omega)^3 = (1,1) $$ (Perhaps $(1,0)$ is a typo). For your second objection, note that $1+g$ mapsto $(2, 1+\omega)$, and $$ \phi(1 + g)^2 = (2,1+\omega)^2 = (4, 1 + 2\omega + \bar\omega) $$ On the other hand \begin{align*} \phi(1 + 2g + g^2) &= (1, 1) + (2, 2\omega) + (1, \bar\omega)\\ &= (4, 1 + 2\omega + \bar\omega)\\ &= \phi(1+g)^2 \end{align*}