Descent of the two nontrivial characters of C₃

A rational root $r$ of $p$ would satisfy $(r+1\mathrm{/}2{)}^2+3\mathrm{/}4=0$, impossible in the ordered field $\mathbb{Q}$. A reducible quadratic over a field has a linear factor and thus a root, so $p$ is irreducible and $(1,\zeta )$ is a rational basis of $E$. We have ${\zeta }^3=1$ and $\zeta \mathrm{\ne }1$; the two roots are $\zeta ,{\zeta }^2$, and they are distinct since equality would force $\zeta =1$. Both lie in $E$. An embedding of this quadratic field is determined by a root, so the identity and $\zeta \mapsto {\zeta }^2$ are its two automorphisms. Equivalently it is a finite normal separable extension, hence Galois, with this two-element group.

Multiplication sends $1\mapsto \zeta$ and $\zeta \mapsto -1-\zeta$, giving $M$. Direct multiplication gives $M^2=\left(\begin{array}{cc}-1& 1\\ -1& 0\end{array}\right)$, $M^2+M+I=0$ and $M^3=I$. Thus it defines a $C_3$-action. For $0\mathrm{\ne }x\in S$, the vectors $x,\zeta x$ are rationally independent, because $(a+b\zeta )x=0$ in the field implies $a=b=0$. Any nonzero invariant rational subspace contains such $x$ and $\zeta x$, and hence equals $S$. This proves simplicity.

Put $t=1+g+g^2$. Since every group element occurs three times in its square, $t^2=3t$. Thus $e=1-t\mathrm{/}3$ is central and $e^2=e$. Also $te=0$, whence $e+ge+g^{2}e=0$. The two coefficient vectors $e=(2-g-g^2)\mathrm{/}3$ and $ge=(-1+2g-g^2)\mathrm{/}3$ are independent: $ae+bge=0$ gives $2a-b=-a+2b=0$, so $a=b=0$. They span the ideal by the relation just found. Evaluation $g\mapsto \zeta$ sends $e\mapsto 1$ and $ge\mapsto \zeta$, so restricts to an algebra isomorphism $\mathbb{Q}[C_3]e\to E$, preserving the block unit. The full evaluation map has kernel $\mathbb{Q}t$: if $a+b\zeta +c{\zeta }^2=0$, then $(a-c)+(b-c)\zeta =0$, so $a=b=c$.

For either $\lambda =\zeta$ or ${\zeta }^2$, set $v_{\lambda }=(1,-\lambda {)}^{\mathsf{T}}$. Then $M{v}_{\lambda }=(\lambda ,1+\lambda {)}^{\mathsf{T}}=\lambda v_{\lambda }$. The determinant of $(v_{\zeta },v_{{\zeta }^2})$ is $\zeta -{\zeta }^2\mathrm{\ne }0$, so these form an $E$-basis of $E{\otimes }_{\mathbb{Q}}S$. Consequently both eigenline modules occur exactly once. Conjugating coefficients interchanges the two eigenvectors and their distinct eigenvalues; the lines give nonisomorphic one-dimensional modules since an intertwiner between them would force $\zeta ={\zeta }^2$.

To check the splitting hypothesis for the whole algebra, evaluation at $1,\zeta ,{\zeta }^2$ gives $E[C_3]\to E^3$. Its inverse sends the $\lambda$th coordinate vector to ${\mathrm{\ell }}_{\lambda }(g)$, where ${\mathrm{\ell }}_{\lambda }(X)={\prod }_{\mu \mathrm{\ne }\lambda }(X-\mu )\mathrm{/}(\lambda -\mu )$ and the product ranges over the other two roots. All denominators are nonzero, and ${\mathrm{\ell }}_{\lambda }(\mu )={\delta }_{\lambda \mu }$ proves the inverse identities on evaluations; a degree at most two polynomial vanishing at three distinct roots is zero, by successive division by $X-\mu$. Thus the algebra is split. The characteristic-zero specialization of F1 now identifies the orbit in step 3.1 with the simple rational module in step 2.1, with multiplicity $1$ as computed.

Finally $\mathrm{tr}(I)=2$, $\mathrm{tr}(M)=-1$ and $\mathrm{tr}(M^2)=-1$. On either eigenline the traces are its scalar values; adding gives $\zeta +{\zeta }^2=-1$ at $g$ and ${\zeta }^2+{\zeta }^4=-1$ at $g^2$. This checks the character and the identity value directly. [F2, step 1.1, step 2.1, step 3.1, algebra] QED