Let $w$ be a primitive fifth root of unity, and let $L=\mathbb Q(w)$. Show that $\sigma:w\mapsto w^2$ is an automorphism of $L$ and calculate its fixed field.
I know that the minimal polynomial of $w$ is $x^4+x^3+x^2+x+1$, but how can I show that $\sigma$ is an automorphism? Can I just say that automorphisms of $L$ permute the roots of the minimal polynomial of $w$, so then $\sigma$ is an automorphism?
Since $w\mapsto w^2$, we must have that $w^2\mapsto w^4$, $w^3\mapsto w$, and $w^4\mapsto w^3$, but I'm not sure how to use this to find which field $\sigma$ fixes. I'm not even sure of which intermediate fields there are between $\mathbb Q$ and $\mathbb Q(w)$. Is there a set way to do this?
Best Answer
$L=\mathbb Q(\omega)$
There is a unique automorphism defined by $\sigma:w\mapsto w^2$.
Any automorphism leaves the base field $\mathbb Q$ fixed, so we know $\sigma(r) = r$ when $r \in \mathbb Q$. Furthermore the equation $\sigma:w\mapsto w^2$ tells us the behavior of $\sigma$ on any power of $w$. By linearity of field automorphisms we know the unique result that any element $\sum_i c_i w^i \in L$ maps to.
Consider an arbitrary element $l = c_0 + c_1 w + c_2 w^2 + c_3 w^3 \in L$.
$\begin{align} \sigma(l) =&\, c_0 + c_1 w^2 + c_2 (- 1 - w - w^2 - w^3) + c_3 w \\ =&\ (c_0 - c_2) + (c_3 - c_2) w + (c_1 - c_2) w^2 - c_2 w^3 \end{align}$.
and to have $l = \sigma(l)$ it will require $c_2 = 0$ and matching up the coefficients of $w$:
which all together implies $c_1 = c_2 = c_3 = 0$. So the fixed field is just $\mathbb Q$.