# 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.

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I think you mean "both $\gamma$ and $\gamma^2+1$" in the second paragraph... – Arturo Magidin Apr 30 '12 at 2:30
@ArturoMagidin, yes thank you. – Edison Apr 30 '12 at 2:33

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 $K(\gamma^2+1) = K(\gamma^2)$.

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Arturo, I guess that by "Dedekind's Product Theorem" you mean what I know as "The Tower Law". I had never seen that name for it before. Do you have a reference where it is called in that way? Or maybe a history book where the result is attributed to Dedekind? – Adrián Barquero Apr 30 '12 at 2:46
@AdriánBarquero: As I recall, there's a book of Stillwell's that attributes it to Dedekind; I'll look it up in my office tomorrow. (I hope I haven't been misattributing it!) (A friend of mine calls it the "Royal Dutch Airline Theorem", because he always states it with fields called $K$, $L$, and $M$.) – Arturo Magidin Apr 30 '12 at 2:49
Thanks Arturo.  – Adrián Barquero Apr 30 '12 at 3:03
Arturo, you don't need to look for it now. I just found it in google books. I found it in Section 5.5 in page 81 of Stillwell's book Elements of Algebra: Geometry, Numbers, Equations. – Adrián Barquero Apr 30 '12 at 3:08
@AdriánBarquero: Yup, that's the one I was going to check. For a second there, I was afraid I had misremembered and it was attributed to Dirichlet... Glad to know I've been attributing it correctly. – Arturo Magidin Apr 30 '12 at 15:14