Find the minimal polynomial of $\sqrt{3} +i$ over $Q(i)$ and over $Q(\sqrt{3})$ 
Find the minimum polynomial of $\sqrt{3} +i$ over $Q(i)$ and $Q(\sqrt{3})$.

My solution:


*

*Over $Q(i)$ : Suppose $a = \sqrt{3} +i$, $a-\sqrt{3} =i$;
so, the minimal polynomial is $x-\sqrt{3} -i=0$.

*Over $Q(\sqrt{3})$:  Suppose a=$\sqrt{3} +i$, $a-\sqrt{3} =i$; by squaring both side of this equation, $a^2-2\sqrt{3}a+3=-1$. So, we get $x^2-2\sqrt{3}x+4=0$.
Is this correct?

Hint Consider the product of the monomial $$x - (\sqrt{3} + i)$$ with its conjugate, in two different senses of the word conjugate. (One of the products recovers the correct partial solution from the question: The minimal polynomial of $\sqrt{3} + i$ over $\Bbb Q (\sqrt{3})$ is $x^2 - 2\sqrt{3} x + 4 = 0$.)
