1

This question was part of my abstract algebra assignment and I was unable to solve 1 part among it. So, I am posting it here as I need help.

Suppose [F : K] is finite. Then the following conditions are equivalent : (i) F is Galois over K ; (ii) F is separable over K and a splitting field of a polynomial f $\in$ K[x] ; (iii) F is a splitting field over K of a polynomial $f \in K[x]$ whose irreducible factors are separable.

I am having problem only in the part in which I ahve to assume (iii) and prove (i) holds.

Can you please help with that? I have to prove that $Aut_K F = K $ itself. But i don't see any reason why it should hold.

Do you mind giving some hints.

1 Answers1

0

I assume your definition of 'Galois-extension' in the finite case is $F^{Aut_KF}=K$. Hints:

  1. First conclude from the given assumptions in (iii) that $F/K$ is normal and separable.
  2. Then show that in general for a normal extension $F/K$ we have that $F^{Aut_KF}$ is purely inseparable over $K$ by showing that there is only one $K$-homomorphism $F^{Aut_KF}\to C$ where $C$ is some algebraic closure of $F$.
  3. Conclude under the additional assumption $F/K$ separable that $F^{Aut_KF}=F$ holds.
leoli1
  • 6,979
  • 1
  • 12
  • 36