Let $L/K$ be a normal extension of number fields and let $\sigma: $K$ \rightarrow \mathbb{C}$ be an embedding of $K$. We know that there are $[L:K]=n$ extensions of $\sigma$ to L.
$\sigma$ is a real embedding if $\sigma(K) \subseteq \mathbb{R}$ and a complex embedding otherwise.
If $\sigma$ is a complex embedding, then, any extension of $\sigma$ to $L$ must also be a complex embedding. But if $\sigma$ is a real embedding, the extensions of $\sigma$ to $L$ might not be all real. If $\sigma$ is a real embedding, is it true that the extensions to $L$ must be all real or all complex? I have worked some examples and it seems to be true (e.g.: $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt[4]{2}) \subseteq \mathbb{Q}(\sqrt[8]{2})$) but I do not know how to prove it in general.
If it is not true, is there a way to know how many of the extensions are real and how many of them are complex?
Thank in advance.
Best Answer
All extensions of $\sigma \colon K \to \mathbf C$ to an embedding of $L$ have the same image as a subfield of $\mathbf C$, so all extensions are real or all extensions are complex.
Let $\sigma' \colon L \to \mathbf C$ and $\sigma'' \colon L \to \mathbf C$ be extensions of $\sigma$, so $\sigma'(L)/\sigma(K)$ is a normal extension and $\sigma''(L)/\sigma(K)$ is a normal extension. To show $\sigma'(L) = \sigma''(L)$, we'll show $\sigma'(L) \subset \sigma''(L)$. The reverse containment then follows by swapping the roles of $\sigma'$ and $\sigma''$.
Pick $\gamma \in \sigma'(L)$, so $\gamma = \sigma'(\alpha)$ for some $\alpha \in L$. We want to show $\gamma \in \sigma''(L)$.
Let $m(x)$ be the minimal polynomial of $\alpha$ over $K$, with $n = \deg m(x)$. Applying $\sigma'$ to both sides of the equation $m(\alpha) = 0$ in $L$ tells us $(\sigma m)(\gamma) = 0$, where $(\sigma m)(x)$ is the polynomial whose coefficients are $\sigma$ applied to the coefficients of $m(x)$. Since $(\sigma m)(x)$ is irreducible in $\sigma(K)[x]$ with a root $\gamma$ in $\sigma'(L)$, and $\sigma'(L)$ is a normal extension of $\sigma(K)$, $(\sigma m)(x)$ splits completely in $\sigma'(L)[x]$: $$ (\sigma m)(x) = \prod_{j=1}^n (x - r_j) $$ where $r_1, \ldots, r_n \in \sigma'(L)$ and $r_1 = \gamma$. (Some $r_j$'s may be repeated.)
Now apply $\sigma''$ to both sides of the equation $m(\alpha) = 0$ to get $(\sigma m)(\sigma''(\alpha)) = 0$ in $\sigma''(L)$. Since $\sigma''(L)$ is a normal extension of $\sigma(K)$, by the same reasoning as above $(\sigma m)(x)$ splits completely in $\sigma''(L)[x]$: $$ (\sigma m)(x) = \prod_{j=1}^n (x - s_j) $$ where $s_1, \ldots, s_n \in \sigma''(L)$. Thus we have factored $(\sigma m)(x)$ into monic linear factors in $\mathbf C[x]$ in two ways. By unique factorization in $\mathbf C[x]$, these linear factorizations must be the same (up to the order of multiplication). Thus $\gamma = r_1$ must be some $s_j$, so $\gamma = s_j \in \sigma''(L)$.