Find a suitable number $a$ such that $\mathbb Q(\sqrt 3, i)=\mathbb Q(a)$

Question

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

Answer 1

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 $\frac12\in\Bbb Q$.

Answer 2

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.

Answer 3

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 $Q(a)=K$.