Question about example 7.1.7 Alaca/Williams. So they are trying to prove that given $K=\mathbb{Q(\sqrt{2}+i)}$:
$O_K= \mathbb{Z}+\mathbb{Z}\sqrt{-1} + \mathbb{Z}\sqrt{2} + \mathbb{Z}(\frac{1}{2}(\sqrt{2}+\sqrt{-2}))$
You have that $K$'s subfields are $\mathbb{Q},\mathbb{Q}(\sqrt{-1}),\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{-2})$
Now,given $\alpha \in O_K$, if $\alpha$ belongs to any of the subfields, you can easily show that $\alpha \in \mathbb{Z}+\mathbb{Z}\sqrt{-1} + \mathbb{Z}\sqrt{2} + \mathbb{Z}(\frac{1}{2}(\sqrt{2}+\sqrt{-2}))$
Suppose that $\alpha$ does not belong to any of the subfields:
Now here is what I do not get, they state that since $\alpha \in K$, you ca express it as:
$\alpha = a + b i+ c\sqrt{2}+ d\sqrt{2}i$ with $a,b,c,d \in \mathbb{Q}$
(Why? Is this a general thing? Is it because the extension is Galois?)
And then they proceed with my second doubt, they state that the conjugates of $\alpha$ are:
$\alpha' = a - b i+ c\sqrt{2}- d\sqrt{2}i$
$\alpha'' = a + b i-c\sqrt{2}- d\sqrt{2}i$
$\alpha''' = a - b i- c\sqrt{2}+ d\sqrt{2}i$
I don't get why this is? I mean in this particular case the conjugate are directly related to the subfields, but I don't get how you get to those equations for the conjugates. This could just be a lot of operating though, this book has a habit of not showing any of the work (Although I get why, some of the work ommited is not at all trivial, for example, finding the subfields was a hassle, they just invoke them. And I am trying to do the work for this examples so when I get to do excercises I will have a better idea on how to solve them)

Okay so the first step is to convince yourself that $K = \Bbb Q(\sqrt{2},i)$, if that hasn't been done in the mentioned argument. This isn't hard, but it's far from obvious if you haven't done it before, so let's delve into it.
Note that since $\sqrt{2}+i \in \Bbb Q(\sqrt{2},i)$, we already have $K \subset \Bbb Q (\sqrt{2},i)$. To achieve the equality, we can show that $\sqrt{2},i \in K$. But moreover, if we show that one of these is in $K$, since their sum is already an element of $K$, we get the other for free. So, by direct computation:
$$
K \ni (\sqrt{2}+i)^3 = (1+\sqrt{2}i)(\sqrt{2}+i) = 3i
$$
hence $i \in K$ and $K = \Bbb Q(\sqrt{2},i)$.
Thus we have a tower of extensions $K = \Bbb Q(i)(\sqrt{2}) - \Bbb Q(i) - \Bbb Q$, and there's a general way to compute a $k$-basis of $E$ if it lies in a tower $E - F -k$. Namely: consider a $k$-basis $\{x_i\}$ of $F$, an $F$-basis $\{y_j\}$ of $E$, and then do all possible products $\{x_iy_j\}_{i,j}$.
Moreover, when the extension is already in the form $\Bbb Q(\alpha)$, it's not hard to show that a $\Bbb Q$-basis is $\{1,\alpha,\ldots, \alpha^{d-1}\}$ with $d = \deg m(\alpha, \Bbb Q)$.
Thus as $\Bbb Q$-vector spaces we have $\Bbb Q(i) = \Bbb Q 1 + \Bbb Q i$ and $\Bbb Q(\sqrt{2}) = \Bbb Q 1+ \Bbb Q\sqrt{2}$, hence a $\Bbb Q$ basis for $K$ consists of all possible products of basic elements
$$
B = \{1,\sqrt{2},i,\sqrt{2}i\}.
$$
That solves the first question. As for the second: via our new characterization of $K$, we know that automorphisms are determined by the images of $i$ and $\sqrt{2}$, these have quadratic polynomials, and so their images must be $\pm i$ and $\pm \sqrt{2}$ respectively.
Hence we have $4$ different admissible morphisms, each corresponds to a different conjugate.
Edit: we did not need any Galois theory but the extension is indeed Galois, as it is the splitting field of $\{X^2-2,X^2+1\}$ and its Galois group then must be the Klein four group. Explicit generators are $$\sigma(a+ib+c\sqrt{2}+di\sqrt{2}) = a+ib-c\sqrt{2}-di\sqrt{2}$$ and $$c(a+ib+c\sqrt{2}+di\sqrt{2}) = a-ib+c\sqrt{2}-di\sqrt{2},$$ i.e. the map that sends $i\mapsto i, \sqrt{2} \mapsto - \sqrt{2}$ and complex conjugation.
