# Solvability by radicals is independent of the choice of splitting fields

I am trying to prove the following exercise:

If $E/k$ and $E'/k$ are splitting fields of $f(x)\in k[x]$ and there is a radical extension $K_t/k$ with $E\subset K_t$, prove that there is a radical extension $K_r'/k$ with $E'\subset K_r'$.

Since splitting fields are isomorphic, I know I need to use an isomorphism between $E$ and $E'$ to construct a radical extension for $E'$ from the radical extension of $E$. But, I have no idea how to proceed. Any hints?

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You want to use the fact that you can extend an automorphism of $E$ to $K_r$, then apply that automorphism to $K_r$. It will send $E\to E'$ by construction, and the rest is history. – Adam Hughes Jul 3 '14 at 12:45
@AdamHughes Why an extended automorphism of $K_r$ will send $E$ to $E'$? I am really confused. Could you please clarify your hints a little bit more? – Y. Fan Jul 3 '14 at 13:15
You're specifically extending an automorphism of $E$ which sends it to $E'$. Sorry that wasn't clear in the original hint. – Adam Hughes Jul 3 '14 at 13:19
@AdamHughes Sorry, I am still confused. Are you saying that I should extend an isomorphism $f:E\to E$ to an isomorphism $f^*:K_t\to K_t$, and then $f^*$ maps $E$ to $E'$? Or I should extend an isomrophism $f:E\to E'$? I really need a clearer way to proceed. – Y. Fan Jul 3 '14 at 13:36
Reread my last comment. I specifically indicate the isomorphism sends $E\to E'$. – Adam Hughes Jul 3 '14 at 13:37