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](http://people.brandeis.edu/~igusa/Math101bS07/Math101b_notesD.pdf) 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.

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