Find the dimension and a basis of the extension field $\mathbb Q(\sqrt{5},i)$ of $\mathbb Q$.
Consider the following chain of fields $$\mathbb Q \subseteq \mathbb Q (\sqrt{5}) \subseteq \mathbb Q(\sqrt{5})(i)=\mathbb Q(\sqrt{5},i).$$ Now $$[\mathbb Q (\sqrt{5},i):\mathbb Q]= [\mathbb Q(\sqrt{5})(i):\mathbb Q(\sqrt{5})] [\mathbb Q(\sqrt{5}):\mathbb Q].$$
- $u= \sqrt{5} \iff u^2 = 5 \iff u^2 -5 = 0$, so the minimal polynomial of $\sqrt{5}$ over $\mathbb Q$ is $x^2 - 5,$ i.e. $ [\mathbb Q(\sqrt{5}):\mathbb Q] =2$.
- Consider $x^2 + 1$ in $\mathbb Q[x]$. Clearly this is monic and irreducible in $\mathbb Q$, but we must confirm that $x^2 +1$ is irreducible in $\mathbb Q(\sqrt{5})$:
If $\pm i \in \mathbb Q(\sqrt{5})$, then $\pm i= a + b \sqrt{5}$ for some $a,b \in \mathbb Q$. Then $$i^2 = (a+b\sqrt{5})^2 \iff \sqrt{5}= \frac{-1-a^2-5b^2}{2ab} \in \mathbb Q,$$ which is absurd. Similarly we can show that if $a=0$ or $b=0$, then $\pm i \not \in \mathbb Q (\sqrt{5})$. Hence we clearly have that $x^2 + 1$ is irreducible over $\mathbb Q(\sqrt{5})$, so $x^2 + 1$ is the minimal polynomial of $i$ over $\mathbb Q(\sqrt{5})$, i.e. $[\mathbb Q(\sqrt{5})(i):\mathbb Q(\sqrt{5})]=2.$
So we have that $$[\mathbb Q (\sqrt{5},i):\mathbb Q]= 4.$$
Now how do I find a basis, is it simply $$\{1, \sqrt{5},i,\sqrt{5}i\}?$$
For fields $K \subset L \subset F$ we have $[F:K] = [F:L][L:K]$ we have that the basis of $F$ is $\{ab | \text{a is an element of a basis of L, b is an element of a basis of K}\}$. Actually this follows from the proof of the fact:
$$[F:K] = [F:L][L:K]$$