Find a suitable number $a$ such that $\mathbb Q(\sqrt 3, i)=\mathbb Q(a)$ Find a suitable number $a$ such that $\mathbb Q(\sqrt 3, i)=\mathbb Q(a)$
I'm thinking about $a=\sqrt 3 + i$, but I don't know how to prove it.
I need help
Thanks
 A: Yes, your idea seems correct.
All you have to prove is, that $a\in\Bbb Q(\sqrt3,i)$ -which is obvious,- and that $\sqrt3,i\in\Bbb Q(a)$. For this,
Hint: $(\sqrt3-i)a=?$, and use $\frac12\in\Bbb Q$.
A: This is probably a bit advanced but I would like to propose a solution
that doesn't really require calculations:
Note that a basis for $K=\mathbb{Q}(\sqrt{3},i)$ over $F=\mathbb{Q}$is
$\{\alpha_{i}\beta_{j}\}_{1\leq i,j,\leq2}$ where $\alpha_{1}=1,\alpha_{2}=\sqrt{3},\beta_{1}=1,\beta_{2}=i$.
We have $4$
maps defined by $\varphi:K\to K$ by $1\to1$ and $\sqrt{3}\to\pm\sqrt{3},i\to\pm i$.
Verify that all $4$ $\varphi$ are automorphisms of $K$ that fix
$F$. Since this is the splittinf field of $(x^2-3)(x^2+1)$ over a perfect field we have it that $K/F$
is Galois since the degre of the extension is $4$ we have it that $Gal(K/F)=\{\varphi_{i}\}_{i=1}^{i=4}$ where the
$\varphi_{i}'s$ are the ones I defined above.
Note that the only automorphism of $K$ that fix $\sqrt{3}+i$ is
$Id_{K}$ hence it is a primitive element of the extension, since
otherwise $\mathbb{Q}(\sqrt{3}+i)$ is a proper subfield of $K/F$
hence correspond to a proper subgroup of $K/F$, in contradiction.
A: You are correct. $K=Q(\sqrt{3},i)$ has degree 4 and $Q(a)\subset K$.  The minimal polynomial for $a$ has degree 4, it is $x^4-4x^2+16$; so $Q(a)=K$.
