Degree of a finite field extension

Question

Let $i,\sqrt{3}\in\mathbb{C}$. I know that both are algebraic over $\mathbb{Q}$.

Hence $[\mathbb{Q}(i\sqrt{3}):\mathbb{Q}]=\deg(i\sqrt{3},\mathbb{Q})$. This is equal to 2 since $\mathrm{irr}(i\sqrt{3},\mathbb{Q})=x^2+3$.

But now I am a little bit confused how to find the irreducible polynomial such that I can compute $[\mathbb{Q}(i,\sqrt{3}):\mathbb{Q}(i\sqrt{3})]$ which is equal the degree of that polynomial.

Any help is appreciated :)

Answer 1

First note that $\mathbb{Q}(i,\sqrt 3) = \mathbb{Q}(i\sqrt 3, i)$.

It remains hence to calculate the degree of the simple field extension $\mathbb{Q}(i\sqrt 3, i)/\mathbb{Q}(i\sqrt 3)$.

We find that $i$ is a zero of the polynomial $X^2+1$, which is still irreducible over $\mathbb{Q}(i\sqrt 3)$. Hence the degree is 2.