# Prove that $[\mathbb{Q}(\sqrt{3},i):\mathbb{Q}=4$. Find a number $a$, such that $\mathbb{Q}(a)=\mathbb{Q}(\sqrt{3},i)$

The first part follows from the fact that for any algebraic elements $$a_i$$ the following holds:

$$F(a_1,...,a_n)=(F(a_1,...,a_k))(a_{k+1},...,a_n)$$

where $$F$$ is the field generated by the $$a_i$$. The second part is giving me trouble. The $$a$$ has to be such that for every $$\lambda_i$$ there exist $$\mu_j$$ such that the following holds (and vice versa since the fields must be the same):

$$\lambda_0+\lambda_1a+...+\lambda_3a^{3}=\mu_0+\mu_1\sqrt{3}+\mu_2i+\mu_3\sqrt{3}i$$

I don't know where to go from here. Any help would be greatly appreciated.

In many cases $$a=\sqrt{3}+i$$ is a good guess. Let's check that this $$a$$ works. Obviously $$a\in\mathbb{Q}(\sqrt{3},i)$$ and hence $$\mathbb{Q}(a)\subseteq\mathbb{Q}(\sqrt{3},i)$$. For the other direction note that:
$$\frac{1}{a}=\frac{1}{\sqrt{3}+i}=\frac{\sqrt{3}-i}{4}$$
This is an element in $$\mathbb{Q}(a)$$. if we multiply by $$4$$ we get $$\sqrt{3}-i\in\mathbb{Q}(a)$$. Hence $$\sqrt{3}=\frac{(\sqrt{3}+i)+(\sqrt{3}-i)}{2}\in\mathbb{Q}(a)$$, and then $$i=\sqrt{3}-(\sqrt{3}-i)\in\mathbb{Q}(a)$$. So $$\mathbb{Q}(\sqrt{3},i)\subseteq\mathbb{Q}(a)$$.