Showing two field extensions are the same Consider the finite extension $F/K$, [$F:K$]=15. Suppose for some $\gamma\in F$, $F=K(\gamma)$. I want to show that $F=K(\gamma^{2}+1)$.
Since the extension is finite, in particular it is algebraic. Therefore I think as long as I show that both $\gamma$ and $\gamma^{2} +1$ satisfy the same minimal polynomial the conclusion will follow. Is this correct? Also, I do not see the significance that the degree of the extension is playing, I would appreciate a hint about this as well.
 A: It would be true, indeed, that if $\gamma$ and $\gamma^2+1$  satisfy the same minimal polynomial, then $K(\gamma)\cong K(\gamma^2+1)$ are isomorphic, hence have the same degree over $K$. Since $K(\gamma^2+1)\subseteq K(\gamma)$, the equality of degree would suffice to show equality of fields.
But that is not always the case.  For example, if $K=\mathbb{Q}$ and $\gamma=\sqrt[15]{2}$, then notice that $\sqrt[15]{4}+1$ does not satisfy the same minimal polynomial as $\gamma$ (neither does $\sqrt[15]{4}$).
Instead, note that $K(\gamma)$ is an extension of $K(\gamma^2+1)$, since $\gamma^2+1\in K(\gamma)$. Now notice that $\gamma$ satisfies the polynomial $x^2 -(\gamma^2+1) +1 \in K(\gamma^2+1)[x]$, so the degree of $K(\gamma)$ over $K(\gamma^2+1)$ is at most $2$. Can it be equal to $2$?
More generally:

Proposition. Let $K$ be a field, and let $u$ be algebraic over $K$. If $[K(u):K]$ is odd, then $K(u)=K(u^2)$.

(I know two proofs: a slick one using Dedekind's Product Theorem, and a direct computational one.)
You can deduce what you want after noting that $K(\gamma^2+1) = K(\gamma^2)$.  
