I am trying to solve the following exercise. But I have trouble understanding the theory and I am probably not doing the right thing. Can you help me, especially with part b) and c) ?
For $K := \mathbb{Q}(i, \sqrt{3})$ and $\alpha: = i + \sqrt{3}$. I have to
(a) Find the minimal polynomials of $\sqrt{3}$
over $\mathbb{Q}$ and $i$ over $\mathbb{Q}(\sqrt{3})$
(b) Show that $\lbrace 1, i, \sqrt{3}, i \sqrt{3} \rbrace$
is a basis of K over $\mathbb{Q}$.
(c) Prove that the minimal polynomial of $\alpha$ over $\mathbb{Q}(\sqrt{3})$
is $$(x − \alpha)(x − \overline{\alpha}),$$ where $\overline{\alpha}$
denotes the complex conjugate of $\alpha$.
For (a) I thought: Because of $[\mathbb{Q}[\sqrt{3}]:\mathbb{Q}] = 2$, the degree of the minimal polynomial must be 2, so I would say $x^2 - 3$ is my minimal polynomial for $\sqrt{3}$ in $\mathbb{Q}$. Since $[\mathbb{Q}[i]:\mathbb{Q}[\sqrt{3}]] = 2$, my minimal polynomial of $i$ over $\mathbb{Q}[\sqrt{3}]$ has to have degree two, so it must be $x^2 + 1$.
For b) I know that a basis must contain 4 elements, because $[\mathbb{Q}(i,\sqrt{3}):\mathbb{Q}] = 4$. Do I have to show here that $ \mathbb{Q}(i,\sqrt{3})= \mathbb{Q}[1, i, \sqrt{3}, i \sqrt{3} ] ? $
Then, this direction $\mathbb{Q}(i,\sqrt{3}) \subseteq \mathbb{Q}[1, i, \sqrt{3}, i \sqrt{3} ] $ seems obvious, and for the other direction I use that $\mathbb{Q}(i,\sqrt{3})$ is a field, hence $i, \sqrt{3} \in \mathbb{Q}(i,\sqrt{3}) \Longrightarrow i\sqrt{3} \in \mathbb{Q}(i,\sqrt{3})$ ?
I do not know how to start on c).