# Sums of squares in global fields (Reference Request)

There is a result due to Siegel that, for a number field $K$, any totally positive element of $K$ is the sum of four squares of $K$. This is discussed in another question (sum of squares in ring of integers).

Is the result true in the more general setting of global fields? That is, for an algebraic function field $F$ is a totally positive element the sum of four squares?

Of course, in the number field terminology totally positive requires embeddings of $K$ into $\mathbb{R}$, which we do not have in the function field setting. So (I believe) we say an element $f \in F$ is totally positive if $f$ is a square in $F_{\infty}$, the completion of $F$ at the nonarchimedian infinite place $\infty$.

• Usually a global (function) field is the field of rational functions of an algebraic curve over a finite field. Is that what you have in mind? Then could you explain what is the "nonarchimedian infinite place $\infty$"? – abx Aug 9 '18 at 15:06
• @abx Yes, an algebraic field extension of $\mathbb{F}_q (t)$ where $\mathbb{F}_q$ is a finite field of characteristic $p$ and $q = p^n$ elements, and $t$ is transcendental over $\mathbb{F}_q$ is the $F$ I have in mind. In which case the prime at infinity is the ideal of $F$ generated by $\frac{1}{t}$ with corresponding valuation equivalence class denoted by $\infty$. This place $\infty$ is nonarchimedian and infinite. It takes value $v_{\infty}(f) = -\deg(f)$ for $f \in F$. – user221330 Aug 9 '18 at 17:31

Corollary 1.5 on page 379 of Lam's Introduction to quadratic forms over fields (in the Sums of Squares section) states that if $F$ is any nonreal global field then every element of $F$ is a sum of four squares. Since you say you are interested in finite extensions of $\mathbb F_q(t)$, and all fields of characteristic other than $0$ are necessarily nonreal, it follows that the answer to your question "yes".