Let $F \subset L$ be two fields, and define $K = \{\alpha \in L\mid [F(\alpha): F] \text{ is a power of 2} \}$. Our problem is to prove that $K$ is a field.

Closure under reciprocation is easy (because $F(\alpha) = F(\alpha^{-1})$). We ran into trouble proving closure under addition and multiplication. Our initial attempt was to prove that for any $\alpha, \beta \in L$, $[F(\alpha, \beta): F]$ divides $[F(\alpha): F] [F(\beta): F]$, but this relies on a result that $[F(\alpha, \beta): F(\alpha)] = \deg_{F(\alpha)} (\beta)$ divides $[F(\beta), F] = \deg_F (\beta)$, which (as far as we can tell) is not necessarily true.

EDIT: As pointed out below, the problem is unsolvable; the professor who assigned it agrees.

  • 2
    $\begingroup$ I suspect there are examples where a sum of two elements of degree 4, say, is an element of degree 6. $\endgroup$ – Gerry Myerson Feb 6 '14 at 8:16
  • $\begingroup$ Apparently not, if the problem is correct.... $\endgroup$ – Greg Martin Feb 6 '14 at 8:19
  • 2
    $\begingroup$ @Greg, that's a BIG if.... $\endgroup$ – Gerry Myerson Feb 6 '14 at 8:24
  • 2
    $\begingroup$ I wonder if finite fields were meant? $\endgroup$ – user14972 Feb 7 '14 at 0:51

Let $\alpha$ and $\beta$ be two of the roots of an $S_4$ quartic; then they both have degree 4, but their sum has degree 6.

A thorough discussion of degrees of sums, and products, of algebraic numbers, can be found here.

EDIT: With a little help from Wolfram Alpha, and possibly with many mistakes of my own, I present this example. The $S_4$-quartic $x^4-x-1$ has two real roots, which add up to $$y=\sqrt{\left({9+\sqrt{849}\over18}\right)^{1/3}-4\left({2\over3(9+\sqrt{849})}\right)^{1/3}}$$ and the minimal polynomial for $y$ is $y^6+4y^2-1$.

The two real roots of the quartic are of the form $(1/2)y\pm z$, where $z$ is a radical expression about twice as complicated as the expression for $y$. Go to Wolfram, and see for yourself.

EDIT: The dead link pointed to this paper: Paulius Drungilas, Arturas Dubickas, Chris Smyth, A degree problem for two algebraic numbers and their sum, Publ. Mat. 56 (2012), 413--448. This link should work (for now).

| cite | improve this answer | |
  • $\begingroup$ For anybody wanting to find the relevant passage in the paper: look at Proposition 29 (ii) on page 14, plugging in $n=4$. $\endgroup$ – J.R. Feb 6 '14 at 8:44
  • $\begingroup$ It would be even nicer to have an explicit example - say, an $S_4$ quartic with particularly accessible roots. Of course, the $S_4$ condition makes it quite hard to write down these roots.... $\endgroup$ – Greg Martin Feb 6 '14 at 18:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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