Are these two fields the same? I wanted to know if the field $\mathbb{Q}(i\sqrt{7}) = \mathbb{Q}(\sqrt{7}, i)$ are the same. I don't  think they are because $i \notin \mathbb{Q}(i\sqrt{7})$?
 A: Since $i\sqrt7$ is a root of the polynomial $X^2+7$, the extension $[\Bbb Q(i\sqrt 7):\Bbb Q]$ has degree $2$. So if $i$ belongs to it, it must be
$$i=a+bi\sqrt7$$
for some rationals $a,b$. Note that $b$ can't be $0$.
Isolating $a$ and squaring,
$$-1-7b^2+2b\sqrt 7=a^2$$
or
$$\sqrt7=\frac{1+7b^2+a^2}{2b}$$
which is impossible.
A: $$r:=i\sqrt7\implies r^2=-7\implies\;\;\text{the rational polynomial}\;\; x^2+7$$
is the minimal polynomial of $\;i\sqrt7\;$ over $\;\Bbb Q\;$ (use Eisenstein to prove it is irreducible or directly with the discriminant), and thus
$$[\Bbb Q(i\sqrt7):\Bbb Q]=2\;.$$
On the other hand, and since $\;i\notin\Bbb Q(\sqrt7)\;$ , we have that $\;x^2+1\;$ is the minimal polynomial of $\;i\;$ over $\;\Bbb Q(\sqrt7)\;$ , and thus
$$[\Bbb Q(i,\sqrt7):\Bbb Q]=[\Bbb Q(\sqrt7)(i):\Bbb Q(\sqrt7)]\cdot[\Bbb Q(\sqrt7):\Bbb Q]=2\cdot2=4\;$$
A: Writing  $\alpha=i\sqrt7$, in the field Q[$\alpha$] the elelments are of the form
$a+b\alpha$, with $a,b$ rationals, and hence the only real numbers in that field are rationals, so it does not  contain the number $\sqrt7$ from the other field.
A: Hint $\ [\Bbb Q(\sqrt a,\sqrt b):\Bbb Q] = 4\,$ if $\,\sqrt a,\sqrt b,\sqrt{ab}\,$ all $\,\not\in \Bbb Q,\,$ by the simple Lemma here.
