I am told that $K_p = \frac{\mathbb{Q}[x]}{(\Phi_p(x))}$ where $\Phi_p(x) = x^{p-1} + ... + x + 1 $.
Subfield L is then defined such that: $\mathbb{Q} \subset L \subset K_p$ where $[L:\mathbb{Q}] = 2$.
I am asked to prove that every automorphism $\phi \in Gal(K_p, \mathbb{Q})$ preserves this subfield L.
Clearly, every automorphism in the Galois group preserves $\mathbb{Q}$, but I am unsure why this should be true for L.
Am I correct to say that all elements in L take the form $q_2 x^2 + q_1 x + q_0$ for some $q_i \in \mathbb{Q}$ ?
I am quite new to this stuff. Any help at all would be very much appreciated!
This is best approached in much greater abstraction. Suppose that $L/K$ is Galois with Galois group $G$, let $H$ be a subgroup of $G$, and let $F\subset L$ be the fixed field of $H$.
Exercise: Given $\sigma\in G$, what is the subgroup $U$ of $G$ that fixes the subfield $\sigma F$ of $L$?
The significance of this exercise in your context is that Galois theory tells you that $\sigma F =F$ if and only if $U=H$. If you have done the exercise correctly, then you should find that this holds for all $\sigma\in G$ if and only if $H$ is normal in $G$. From this, you should easily be able to deduce your exercise, but this more general statement is extremely useful in much wider contexts.