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!

• $L=Q(\alpha)$ for some $\alpha$ that is root of a degree 2 polynomial p with coefficients in Q. Any automorphism of $K_p$ either fixes $\alpha$ (and thus L) or sends it to the other root of $p$. Either way, $L$ is preserved. – Ayoub Mar 16 at 15:24
• The problem is the word preserved. Then didn't mean $\forall a \in L, \phi(a) = a$ but $\forall a \in L , \phi(a) \in L$ ie. $\phi|_L \in Gal(L/\mathbb{Q})$. Since $K_p/\mathbb{Q}$ is a normal extension this is equivalent to saying $L/\mathbb{Q}$ is a normal extension. – reuns Mar 16 at 15:46

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.