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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!

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  • $\begingroup$ $L=Q(\alpha)$ 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. $\endgroup$ – Ayoub Mar 16 at 15:24
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    $\begingroup$ 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. $\endgroup$ – reuns Mar 16 at 15:46
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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.

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