ramification of infinite place of ${Q}$

Question

Can any one explain me "The infinite place $v=\infty$ of ${Q }$ is unramified in ${Q(\sqrt{2})} $ but is ramified in ${Q(\sqrt{i})}$

Answer 1

There are two possibilities to extend the archimedean absolute value $| \cdot |_\infty$ of $\Bbb Q$ to $\Bbb Q(\sqrt{2})$: the two real embeddings give two different absolute values, as for example $|1+\sqrt{2}|\neq|1-\sqrt{2}|$. This means that action of the Galois group on the set of extensions has trivial stabilizers. Thus the infinite place is unramified.

But there's only one extension of $|\cdot |_{\infty}$ to $\Bbb Q(i)$, as complex conjugation preserves the Archimedean absolute value, so the Galois group acts trivially on the extensions.