Let $ α = i\sqrt{2}+\sqrt{3}$ and let $K:=\mathbb{Q}(i\sqrt{2},\sqrt{3})$.
I have to find the minimal polynomial, the degree of the extension $K:\mathbb{Q}$ and thus prove that it is simple.
I started by saying that:
$a= i\sqrt{2}+\sqrt{3}$ thus $a^2 = (i\sqrt{2}+\sqrt{3})^2$ which is $a^2-1=2\sqrt{6}i$. Squaring both sides again I got: $a^4-2a^2+25=0$. Let $f(x)=x^4-2x^2+25$. Since $f(x)$ is monic, we need to show $f(x)$ is irreducible over $\mathbb{Q}$. If $f(x)$ has a linear factor in $\mathbb{Q}[x]$, then it has a zero in $\mathbb{C}$ which divides the constant term $1$. But, we see that $f(x)$ has no linear factors in $\mathbb{Q}[x]$. If $f(x)$ factors into two quadratic factors in $\mathbb{Q}[x]$, then it has factorisation form:
$(x^2+fx+b)(x^2+cx+d)=x^4+(f+c)x^3+(fc+b+d)x^2+(fd+bc)x+bd=x^2-2x^2+25$.
Equating coefficients we get the following:
$f+c=0$
$fc+b+d=-2$
$fd+bc=0$
$bd=1$
From the fourth one, we get $2$ cases. Either $b=d=1$ or $b=d=-1$.
$1^{st}$ case: We get the following:
$f+c=0$
$fc=-12$
For which we get that $c^2=\sqrt{12}$ which does not belong in $\mathbb{Q}$, thus it is rejected.
$2^{nd}$ Case, we have the following:
$f+c=0$
$fc=8$
from which we get:
$c^2=-8$ which is again rejected.
Thus we realized that $f(x)=x^4-2x^2+25$ is a minimal polynomial of $a$ over $\mathbb{Q}$
Then i just state that its degree is $4$.
To prove that it is $\textit{simple extension}$ I say:
$\mathbb{Q}(i\sqrt{2},\sqrt{3}):\mathbb{Q}$ is simple since $\mathbb{Q}(i\sqrt{2}+\sqrt{3})=\mathbb{Q}(i\sqrt{2},\sqrt{3})$.
Is my way of thinking correct? Do I need to explain some things in more detail? Any feedback is more than welcome.