Galois group and trace form

Question

I am trying to prove the claim $$12\cdot Tr_{k{\zeta_3}/k}(\langle 1\rangle )=12\cdot (\langle 2\rangle +\langle 2\cdot (-3)\rangle$$ where $k$ is a field with $char k\neq 3$ and $\zeta_3$ is a third root of unity which is not in $k$. $\langle a\rangle\in GW(k)$ (or $GW(k(\zeta_3)))$ is an element of Grothendieck-Witt ring, i.e. a non-degenerate symmetric bilinear form over $k$ (or $k(\zeta_3)$) mapping $(x,y)$ to $axy$.
This is my try:
Since $Gal(k(\zeta_3)/k)=\{id,f\}$ where $f:\zeta_3\mapsto \zeta_3^2$. We apply trace formula $Tr_{k(\zeta_3)/k}:k(\zeta_3)\to k, \alpha\mapsto\sum_{\sigma\in Gal(k(\zeta_3)/k}\sigma(\alpha))$ to the bilinear form. Then from some element $\langle a\rangle\in GW(k(\zeta_3))$ we get an element $Tr_{k(\zeta_3)/k}\circ\langle a\rangle\in GW(k)$.
Explicitly, for $\langle 1\rangle\in GW(k(\zeta_3))$, we have $$Tr_{k(\zeta_3)/k}(\langle 1\rangle )(x,y)=Tr_{k(\zeta_3)/k}(xy)=id(xy)+f(xy)$$ This shows that $Tr_{k(\zeta_3)/k}(\langle 1\rangle)=\langle 1\rangle +\langle b\rangle$ where $b$ is the one corresponding to $f(xy)$ (now we leave it like this). Can someone explain to me why we have $\langle 2\rangle$ on the right hand side of claim? Any help is appreciated. Thanks!

Answer 1

Assuming that $char(k)\neq 2,3:$

Since $\zeta_3\notin k$, $k(\zeta_3)=k(\sqrt{-3})$.

Now, the $k$-basis $(1,\sqrt{-3})$ is orthogonal wrt to the trace form,since the trace of $\sqrt{-3}$ is $0$. The traces for $1^2$ and $\sqrt{-3}^2$ are respectively $2$ and $-6$. Hence $Tr_{k(\zeta_3)/k}(\langle 1\rangle)\simeq \langle 2,-6\rangle$