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Let $a>0$ be a real cyclotomic number. Is it always possible to solve in cyclotomics the equation $X\overline{X}=a$ ?

Equivalently, one might want to express $a$ as a sum of squares of two real cyclotomics. It is well-known that one square is not always enough.

(If two squares are not enough, then, is there an upper bound?).

Edit: $a$ is not only positive, but totally positive (otherwise the answer is No).

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    $\begingroup$ The number $X\bar X$ is totally positive (any of its Galois conjugates is positive). Thus any $a$ which is positive but not totally positive would be a counterexample, e.g. $a=\sqrt{2}$. $\endgroup$
    – Victor Ostrik
    Commented Apr 26, 2021 at 17:34
  • $\begingroup$ my $a=\sum_{k=1}^m b_k\overline{b}_k$, $b_k$ cyclotomic. Is $a$ totally positive? (References on the topic would be much appreciated, too) $\endgroup$
    – Dima Pasechnik
    Commented Apr 26, 2021 at 19:02
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    $\begingroup$ For the sum of squares, if $i$ is in the field, then $a=\left(\dfrac{a+1}{2}\right)^2+\left(i\dfrac{a-1}{2}\right)^2$. I'd check "A Historical View of the Pythagoras Numbers of Fields" by D. Leep $\endgroup$
    – pavl0
    Commented Apr 26, 2021 at 19:33
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    $\begingroup$ Your $a$ is totally positive, because it is the sum of totally positive elements. $\endgroup$
    – GH from MO
    Commented Apr 26, 2021 at 20:18
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    $\begingroup$ You don't need Artin-Schreier theory. If you conjugate $a$ by $\sigma$, all the $b_k$'s get conjugated by $\sigma$, because complex conjugation commutes with $\sigma$. Then use that $z\bar z$ is positive for every nonzero $z\in\mathbb{C}$. $\endgroup$
    – GH from MO
    Commented Apr 26, 2021 at 22:00

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If $a$ is a totally positive real cyclotomic number, then it is a sum of two squares of real cyclotomic numbers.

It suffices to check that the equation $x^2+ y^2 - a z^2=0$ has solutions in real cyclotomic numbers. It has solutions in a particular real cyclotomic number field $F$ if it has solutions everywhere locally. This equation has solutions locally if and only if the quaternion algebra $(a,-1)$ splits locally.

First take the field $F$ generated by $a$. If we adjoin to $F$ a sufficiently large totally real extension of the field of $2$-power roots of unity (sufficiently large depending on the set of ramified places of this quaternion algebra), producing a field $K$, then every ramified place $v$ of this quaternion algebra will not be totally split in $K$; because $K/F$ is a Galois extension of degree a power of $2$, this implies that the quaternion algebra splits over $K_v$, and so the equation has solutions locally over $K_v$, thus solutions over $K$.

How sufficiently large?

For $v$ an odd prime with residue field $q_v$, it suffices to adjoin the totally real part of the $2^n$th roots of unity for the minimum $n$ such that $q_v^2 \neq 1 \mod 2^n$. To obtain this, it suffices to have $2^n \geq q_v^2$.

An odd prime $v$ only ramifies if it divides the numerator or denominator of $a$, in which case $q_v$ divides the norm of the numerator or denominator, so it suffices to have $2^n$ at least the max of the norms of the numerator and denominator squared.

To handle the even primes $v$, we need to ensure that the image of the inertia group of $2$ acting on the totally real part of field of the $2^n$th roots of unity is strictly larger than the image of the inertia group of $2$ acting on $F$. Since the image of the inertia group acting on the real part of the field of $2^n$th roots of unity is of order the degree $2^{n-2}$, it suffices to take $2^{n-2}> \deg F$.

So it suffices to take $2^n > \max( Na_1^2, Na_2^2, 4 \deg F)) $ with $Na_1, Na_2$ the norms of the numerator and denominator. We can achieve this with $2^n \leq 2 \max( Na_1^2, Na_2^2, 4 \deg F)$, which is an extension of $F$ of degree $\leq ( Na_1^2/2, Na_2^2/2, 2\deg F)$.

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  • $\begingroup$ @GHfromMO The places themselves are not totally split, which is what makes the quaternion algebra split. $\endgroup$
    – Will Sawin
    Commented Apr 26, 2021 at 22:12
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    $\begingroup$ @GHfromMO Yeah, I was a bit lazy when writing this answer. First, there is not a unique quadratic extension, but rather more than one. For $v$ odd, we will get the unramified one. The idea is this: Frobenius at $v$ acts on the $2^n$th roots of unity by raising to the exponent $q$. Choose $n$ large enough that $q^2$ is not congruent to $1$ mod $2^n$, and it will follow that Frobenius does not act by raising to the exponent $1$ or $-1$. Since complex conjugation acts by $-1$, Frobenius is not in the group generated by complex conjugation, and thus acts nontrivially on the real part. $\endgroup$
    – Will Sawin
    Commented Apr 27, 2021 at 0:07
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    $\begingroup$ @GHfromMO Sounds good - I may try to work on my reading skills... $\endgroup$
    – Will Sawin
    Commented Apr 27, 2021 at 0:49
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    $\begingroup$ @DimaPasechnik Any irreducible representation of a finite group $G$ can be defined over $\mathbf{Q}(\zeta_{N})$ with $N = |G|$; this is a theorem of Brauer from 1945. That's also clearly best possible in general, since one can take $G = \mathbf{Z}/N \mathbf{Z}$. $\endgroup$
    – user175951
    Commented Apr 27, 2021 at 19:21
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    $\begingroup$ @DimaPasechnik My edit gives an effective bound. $\endgroup$
    – Will Sawin
    Commented Apr 27, 2021 at 19:39

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