Galois finite extension Let $K/ \mathbb{Q}$ be a finite Galois extension, $K \otimes_{\mathbb{Q}} \mathbb{R} \simeq \mathbb{R}^s \oplus \mathbb{C}^t$. 
Prove that either $s=0$ or $t=0$. 
 A: The automorphisms of $K/\Bbb Q$ are continuous, so they extend to a completion. If we denote an infinite place by $\mathfrak{p}$, we know one completion is $K_{\mathfrak{p}}$. As all other completions are given by $\sigma(K)_{\sigma(\mathfrak{p})}$ and $\sigma(K)=K$ (since we are dealing with a Galois extension) we use the fact that $\text{Gal}\left(K/\Bbb Q\right)$ permutes the places transitively to conclude all completions are isomorphic as $\Bbb R$ vector spaces, in particular they have the same dimension.
Now as
$$K\otimes_{\Bbb R}\Bbb R \cong\prod_{\mathfrak{p}\text{ real}}K_{\mathfrak{p}}\times\prod_{\mathfrak{p}\text{ complex}}K_{\mathfrak{p}}$$
we conclude the result.
A: [Addendum] It is known that here $s$ is the number of distinct real embeddings of $K$, and $d$ is the number of pairs of complex conjugate pairs of embeddings of $K$. I set out to prove this form of the claim. IOW I seek to prove that either all the embeddings of $K$ are real or none is. [/Addendum]
As $\Bbb{C}$ is algebraically closed we can assume that $K$ is a subfield of $\Bbb{C}$. Let $[K:\Bbb{Q}]=n$. It is known that there are exactly $n$ embeddings $\sigma_1,\sigma_2,\ldots,\sigma_n$ from $K$ to $\Bbb{C}$. But we also know that $K$, being Galois, has $n$ distinct automorphisms. It follows that the above embeddings are exactly the automorphisms of $K$, and thus have the property $\sigma_i(K)=K$. As $K$ either is or is not a subset of the reals the claim follows from this.
