Extension field, degree of $[\mathbb Q(i,\sqrt{-3}):\mathbb Q]$

Question

I want to calculate the degree of $[\mathbb Q(i,\sqrt{-3}):\mathbb Q]$, can I do like that:

$$X=i+\sqrt{-3}\implies X=i(1+\sqrt{3})\implies X^2=-(1+\sqrt{3})^2\implies X^2=-1-2\sqrt{3}-3\implies X^2+4=\sqrt 3\implies (X^2+4)^2=3\implies X^4+8X^2-13=0$$

Since $p(X)=X^4+8X^2-13$ is irreducible, $[\mathbb Q(i,\sqrt{-3}):\mathbb Q]=4$.

If it's wrong I don't know how to do.

Answer 1

This shows that the degree of the extension is at least $4$.

Now prove the other direction of showing it is at most $4$ (this is the easier direction)

Another way of arguing the correctness would be to prove the element you considered is a primitive element of the extension

Answer 2

What you have proved is that $[\Bbb Q(i+\sqrt{-3}):\Bbb Q]=4$. It is clear that $\Bbb Q(i+\sqrt{-3})\subset \Bbb Q(i,\sqrt{-3})$, but you must prove the converse inclusion.