Third root of unity $\zeta$, extension degree of $\mathbb{Q}(\sqrt{2},\zeta)$, its $\mathbb{Q}$-Basis and extension of field morphism. Question
a) Let $\zeta$ be the third root of unity, find $[\mathbb{Q}(\sqrt{2},\zeta):\mathbb{Q}]$ and the $\mathbb{Q}$-Basis of $\mathbb{Q}(\sqrt{2},\zeta)$.
b) Let $\sigma$ be the field morphism from $\mathbb{Q}(\sqrt{2})$ to $\mathbb{Q}(\sqrt{2})$, defined by $\sigma(\sqrt{2})=-\sqrt{2}$. Find all the extensions $\tilde{\sigma}$ from $\mathbb{Q}(\sqrt{2},\zeta)$ to $\mathbb{Q}(\sqrt{2},\zeta)$ of $\sigma$.
My attempt
a) $[\mathbb{Q}(\sqrt{2},\zeta):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2},\zeta):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}]$, $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$, since the minimal polynomial of $\sqrt{2}$ over $\mathbb{Q}$ is $x^2-1$. For $[\mathbb{Q}(\sqrt{2},\zeta):\mathbb{Q}(\sqrt{2})]$ the minimal polynomial of $\zeta$ over $\mathbb{Q}(\sqrt{2})$ is $x^2+x+1$, because $\zeta$ is the third root of unity ans $\zeta\notin\mathbb{Q}(\sqrt{2})$. Therefore $[\mathbb{Q}(\sqrt{2},\zeta):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2},\zeta):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2\cdot2=4$.
Since $[\mathbb{Q}(\sqrt{2},\zeta):\mathbb{Q}]=4$ the $\mathbb{Q}$-Basis of $\mathbb{Q}(\sqrt{2},\zeta)$ has four elements: $\{1, \sqrt{2}, \zeta, \zeta\sqrt{2}\}$.
b) I know that if $\tilde{\sigma}$ is an extension of $\sigma$, then $\tilde{\sigma}(\zeta)$ is a zero for the image under $\sigma$ of $x^2+x+1$. I am not sure what this image would look like, but I know that for each of its zeroes (we can call $\beta$) has exactly one extension such that $\tilde{\sigma}(\zeta)=\beta$.
Do I have to find the image of $x^2+x+1$ under $\sigma$? Or am I approaching it wrong?
 A: Yes, this seems like the right approach. Discussing “the image of $x^2 + x + 1$ under $\sigma$” can be a bit confusing since $\sigma$ isn’t a map between polynomial rings, but what this really means is this: if $\alpha$ is a root of $x^2 + x + 1$, then $0=\tilde\sigma(\alpha^2 + \alpha + 1) = \tilde\sigma(\alpha)^2 + \tilde\sigma(\alpha) + 1$, since $\tilde\sigma$ is a field automorphism and thus must fix $\mathbb{Q}$. So $\tilde\sigma(\alpha)$ is a root of $x^2 + x + 1$ because $\sigma$ fixes the coefficients of $x^2 + x + 1$.
In general, the image of a polynomial under a field homomorphism is obtained by applying the homomorphism to the polynomial’s coefficients. If $K$ is a field, $L=K[\alpha]$ is an algebraic extension where $\alpha$ has minimal polynomial $p(x)$ over $K$, and $\sigma$ is an automorphism of $K$, then the extensions $\tilde\sigma$ of $\sigma$ to $L$ must satisfy
$
\tilde\sigma(p(\alpha)) = 0.
$
Writing $p(x) = x^n + a_{n-1}x^{n-1} + \dots + a_0$, with the $a_i$ in $K$, this means
\begin{align*}
0 = \tilde\sigma(p(\alpha))& = \tilde\sigma(\alpha)^n + \tilde\sigma(a_{n-1})\tilde\sigma(\alpha)^{n-1}+\dots+\tilde\sigma(a_0)\\
&= \tilde\sigma(\alpha)^n + \sigma(a_{n-1})\tilde\sigma(\alpha)^{n-1}+\dots+\sigma(a_0),
\end{align*}
where the final equality is because $\tilde\sigma$ agrees with $\sigma$ on $K$. So $\tilde\sigma(\alpha)$ is a root of the polynomial
$$
\sigma p(x) = x^n + \sigma(a_{n-1})x^{n-1} + \dots + \sigma(a_0).
$$
