Real and complex embeddings of a Galois extension

Question

I need to show that if $K$ is Galois over $\mathbb{Q}$ and $r_1(K)>0 (resp. r_2(K)>0)$, then $r_1(K)=[K:\mathbb{Q}]$(resp. $r_2(K)=[K:\mathbb{Q}]/2$)

I am lost at the very first step. I am guessing we need to show that $r_1 = 0$ or $r_2=0$ for a Galois extension. I know what a Galois extension is and that $r_1 + 2r_2=n$ where $n=[K:\mathbb{Q}]$ but I am not able to piece together all concepts for the proof. What am I missing? I would like someone to be give me a hint or point me in the right direction. I am not a math student so I am really bad at proving things but I would like some direction so that I can work it out myself. Thanks.

Answer 1

Hint: If $K$ is Galois over $\mathbb{Q}$, then all embeddings $K\to\mathbb{C}$ have the same image.

Answer 2

@Eric Wofsey: Is this what you meant? Here is what I think you meant.

Since all the embeddings from $K\rightarrow\mathbb{C}$ have the same image if K is Galois then either $r_1=0$ or $r_2=0$. If former, then $r_2=n/2$ and similarly $r_1=0$

Is this right? If yes, do we have to show the embeddings indeed have same image? OR is it implicit from the definition of Galois extensions?