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!

  • $\begingroup$ $L=Q(\alpha)$ 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
  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.