Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 8 down vote accepted

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.