Signature function in field extension is onto

Question

I need help with this problem.
Suppose $K/\mathbb{Q}$ is a separable field extension. We know that there exists $\theta$ such that $K=\mathbb{Q}(\theta)$. We also know that if $\theta_1,\ldots,\theta_n$ are the complex roots of $\min(\mathbb{Q},\theta)$ then the immersions $\sigma_i:K\to \mathbb{C}$ are given by $\sigma_i(\theta) = \theta_i$.
Suppose that $\sigma_1,\ldots,\sigma_r$ are real immersions and define $$ \operatorname{sign}: K^\times \to \{ \pm 1 \}^r $$ where $\operatorname{sign}(\alpha) = (\operatorname{sgn}(\sigma_1(\alpha)),\ldots,\operatorname{sgn}(\sigma_r(\alpha)))$.

I need to prove that this $\operatorname{sign}$ function is onto, i.e, all of the $2^r$ signs are taken. Can you help me with some guidance or hint? I am really confused with this. I understand the roots of the minimal polynomial are permuted by the $\sigma_i$ and I believe I need to associate this problem with the permutation sign (the permutation of the roots).

Answer 1

You need some approximation result. If $K$ has $r$ real and $2s$ imaginary embeddings, then the image of $K$ under the natural map $\Phi$ from $K$ to $\Bbb R^r\times \Bbb C^s$ is dense. Given this, there will be elements of $K$ such that $|\sigma_j(\alpha)-\epsilon_j|<1$ for all $j$ where the $\epsilon_j$ are your favourite choices from $\{\pm1\}$. This $\alpha$ will have the right signs.

To see this density result, note that the closure of the image of $\Phi$ will certainly be a real vector subspace of $\Bbb R^r\times \Bbb C^s$. If is is a proper subspace, then there will be a linear dependence between the embeddings of $K$ into $\Bbb C$ and that would contradict Dedekind's theorem on linear independence of characters.