# Equivalence of Galois groups of two different splitting fields of the same polynomial

All fields are in $\mathbb{C}$

Let $f$ be a polynomial with coefficients in the field $F$.
Let $F_1$ be a Galois extension of $F$ such that its Galois group $G(F_1/F)$ is cyclic and has prime order.

Let $K$ be the splitting field of $f$ over $F$ and $K_1$ be the splitting field of $f$ over $F_1$.

We are given that $G(K/F)$ is simple and non-abelian.

Question: Is $G(K/F) \cong G(K_1/F_1)$?

Comments: We know that $[F_1:F] = p$ for some prime $p$ and this should be the key to the proof but I can't see where to use it. And we know that $K/F$ and $K_1/F_1$ are Galois extensions because they are splitting fields over the lower fields. This question arises from the proof of proposition (9.8) of Algebra by Artin 1991 reproduced below:

• What if $F = \Bbb R$, $f(x) = x^2 + 1$, and $F_1 = \Bbb C$? Then $G(F_1/F) = \Bbb Z/(2)$, $K = \Bbb C$, and $K_1 = \Bbb C$, so $G(K/F) = \Bbb Z/(2)$ but $G(K_1/F_1) = 1$. – Stahl Mar 31 '15 at 18:51
• Very good point... I wonder if I misstated something. I'll add the full proof to my question. – Mark Mar 31 '15 at 19:04
• You forgot to include the hypothesis that $G$ is a simple nonabelian group. – Stahl Mar 31 '15 at 19:08
• In the first paragraph, he's assuming the result to be true in the case when $[F_1 : F]$ is prime: to show that it's true in this case, you need the hypothesis. The first part of the proof isn't showing that the result holds in the $[F_1 : F]$ prime case, only that the general case can be reduced to this one. – Stahl Mar 31 '15 at 19:14
• Ah! In other words I jumped ship too soon. The proof is given in the very next paragraph! Thanks – Mark Mar 31 '15 at 19:16