Construct $K=\{ (a+bi)+(c+di)\sqrt{2} \}$ with Extension Fields. where $i=\sqrt{-1}$. Can take it for granted that K is a field where degree is 4. construct K and confirm its degree
we need irreducible polynomials in $\mathbb{Q}$ where the solution for x is $i,\sqrt{2}$.
For $i$, $x^2+1=0$. For $\sqrt{2}$, $x^2-2$
First Extension] Now, let $F=\mathbb{Q}[x]/(x^2+1)$. The elements of F are polynomials of degree at most 1 that is $$F=\{[a+bx]:a,b\in\mathbb{Q} \}=\{[a+bi]:a,b\in\mathbb{Q} \}$$ since $[x]=[i]$. Also $[F:K]=deg(x^2+1)=2$
Second Extension] Consider $S=F/(x^2-2)$
$\vdots$
Missing Cant really explain it in words
$$S=\{(a+bi)+(c+di)\sqrt{2} :a,b,c,d \mathbb{Q} \}$$ $[S:F]=deg(x^2-2)=2$.
Also $[S:K]=[S:F][F:K]=2*2=4$ which confirms the degree
Have the intuition in missing steps. But could use some help.
The field is given by $K=\mathbb{Q}(i,\sqrt{2})$. We can construct it from $\mathbb{Q}$ by a quadratic extension $L=\mathbb{Q}(\sqrt{2})$ with minimal polynomial $x^2-2$, and then again a quadratic extension $K=L(i)$ with minimal polynomial $x^2+1$. Since clearly $i\notin\mathbb{Q}(\sqrt{2})$, we have $$ [\mathbb{Q}(\sqrt{2},i):\mathbb{Q}(\sqrt{2}]=2. $$ Now we can use the dimension formula: $$ [\mathbb{Q}(\sqrt{2},i):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2},i):\mathbb{Q}(\sqrt{2}]\cdot [\mathbb{Q}(\sqrt{2}):\mathbb{Q}] =2\cdot2=4. $$ No more intuition is needed, I think.