$\newcommand\Q{\mathbb Q}$I have a series of questions related to sum of $k$-powers of algebraic numbers (it is actually only one question that I try to weaken/strengthen the conditions). Suppose we know the following $$\sum_{i=1}^n a_i^k \in \Q$$ for algebraic number $a_i$'s for which not all of them are in $\Q$. Then I ask

  1. If $n=k=2$ and $\Q(a_1,a_2)$ is real then is it true that $[\Q(a_1,a_2):\Q]$ is even?
  2. Same question as 1. except we assume the general condition $n\geq 2$ and $\Q(a_1,\dots,a_n)$ is real
  3. Same question as 1. except we assume any $k\geq 2$ and ask if $[\Q(a_1,a_2):\Q]$ is divisible by $k$
  4. Same question as 2. except we assume any $k\geq 2$ and ask if $[\Q(a_1,\dots, a_n):\Q]$ is divisible by $k$

5-8. We ask the same questions as 1-4. without assuming $\Q(a_1,\dots,a_n)$ is real.

Probably the answers are trivial (for instance if 1. is already not true). And I think 5-8. is generally easier than 1-4 (I feel that for 5-8. all of them are generally false).


$\newcommand\Q{\mathbb Q}$For sums of squares the answer is no. Take any real number field $K$ that has odd degree over $\Q$. Take a non rational number $x\in K$. Then you can find a natural number $m$ such that $m>x^2$ with respect to any ordering of $K$. By Artin's theorem it follows that $m-x^2$ is a sum of squares over $K$ (this is the smallest preordering of $K$). So $m$ is the sum of squares in $K$ for which one of the summands is $x^2$ with $x$ not in $\Q$ but $[K:\Q]$ is odd.


(1) seems to be false in quite a strong way. Suppose $n = k = 2$ and you even insist that the sum is equal to $1$, e.g.

$$x^2 + y^2 = 1.$$

Then you can make $K = \mathbf{Q}(x,y)$ be any field you want, and certainly of odd degree and totally real. Here is the construction. Given any number field $K$, by the primitive element theorem one can write $K = \mathbf{Q}(t)$ for some $t$. Then write

$$x = \frac{2t}{t^2+1}, \qquad y = \frac{t^2-1}{t^2+1}.$$

Then $x^2 + y^2 = 1$, and $x,y \in \mathbf{Q}(t)$, so $\mathbf{Q}(x,y) \subseteq \mathbf{Q}(t)$. Finally, note that

$$t = \frac{1+x+y}{1+x-y} \in \mathbf{Q}(x,y),$$

and so $K = \mathbf{Q}(t) = \mathbf{Q}(x,y)$.

  • $\begingroup$ for a while I was confused with $x,y,t$ thinking they were indeterminates/transcendental over $\mathbb Q$ $\endgroup$ – quantum Aug 30 '18 at 8:55
  • $\begingroup$ @quantum I answered your questions and you did not award me the green. I curse you and your descendants to the end of time. $\endgroup$ – Baby Monkey Oct 1 '18 at 23:50

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