Trace form and totally real number fields

Question

$\newcommand\Q{\mathbb{Q}}$ Let $K$ be a number field then there is a quadratic form over the $\Q$ vector space $K$ given by $$\tau: K\rightarrow \Q \qquad y\mapsto\mathrm{Tr}_{K/\Q}(y^2)$$ which is also known as the trace-form of $K$. As far as I can recall, the trace form is positive definite iff $K$ is totally real. Where can I find the proof of this statement? Or is it rather trivial to show this?

Answer 1

Tensoring up with $\Bbb R$, the trace form extends to a real quadratic form on $K\otimes\Bbb R$, given by the same formula. We have the usual decomposition $K\otimes\Bbb R\cong \Bbb R^{r_1}\oplus\Bbb C^{r_2}$ as $\Bbb R$-algebras. If there is a copy of $\Bbb C$ present, then on this the trace form is $z\mapsto z^2+\overline z^2$ which takes positive, negative and zero values, so is indefinite.