Let $K = \mathbb{Q}( \sqrt{2}, i)$ be the field generated over $\mathbb{Q}$ by $\sqrt 2$ and $i$ [Delhi-University PhD Screening test]

Let $K = \mathbb{Q}( \sqrt{2}, i)$ be the field generated over $\mathbb{Q}$ by $\sqrt 2$ and $i$. Then the dimension of $\mathbb{Q}( \sqrt2, i)$, as a $\mathbb{Q}$-vector space is equal to
(a)1
(b)2
(3)3
(4)4

Basis for $\mathbb{Q}(\sqrt2)$ and $\mathbb Q( i)$ are $\{1,\sqrt 2\}$ and $\{1,i\}$ respectively. So Basis for $\mathbb Q( \sqrt 2, i)$ is $\{1,\sqrt 2\} \times \{1,i\}  =\{1,\sqrt 2,i,i\sqrt2\}$. Am I Right?
 A: For a given Extension of fields $L/K$ i simply write $[L:K]$ for the degree of $L$ over $K$ and $[a:K]$ for the degree of $a$ over $K$, that is defined as the degree of the minimum polynomial of $a$ over $K$. As we know it holds $[K(a):K] = [a:K]$ for algebraic(over $K$) $a$. Furthermore we need $K(a,b) = K(a)(b) = K(b)(a)$, it doesn't make any difference how you adjungate elements to a field (in which order). Last but not least we know $[K(a):K] = [a:K] = 1 \Longleftrightarrow a \in K$. Let's start:
Well there is a theorem that says:
Let $L/Z/K$ be a given chain of field-extensions. Then
$$[L:K] = [L:Z] \cdot [Z:K].$$ 
In your case we have $ L =\mathbf{Q}(\sqrt{2},i) = \mathbf{Q}(\sqrt{2})(i), Z = \mathbf{Q}(\sqrt{2})$ and of course $K = \mathbf{Q}$, thus it holds
$$[\mathbf{Q}(\sqrt{2},i):\mathbf{Q}] = [\mathbf{Q}(\sqrt{2},i):\mathbf{Q}(\sqrt{2})] \cdot [\mathbf{Q}(\sqrt{2}): \mathbf{Q}] = [\mathbf{Q}(\sqrt{2},i):\mathbf{Q}(\sqrt{2})] \cdot 2.$$ So lets compute $[\mathbf{Q}(\sqrt{2},i):\mathbf{Q}(\sqrt{2})]$. As one can check, it holds
$$[\mathbf{Q}(\sqrt{2},i):\mathbf{Q}(\sqrt{2})] = [\mathbf{Q}(\sqrt{2})(i):\mathbf{Q}(\sqrt{2})] =  [i:\mathbf{Q}(\sqrt{2})] \leq_* [i:\mathbf{Q}] = 2.$$
The inequality * follows from the following idea:
$[a:K]$ is defined as the degree of the minimum polynomial of $a$ over $K$.
Since the minimum polynomial $m_{i,\mathbf{Q}}$ of $i$ over $Q$ is also a polynomial in $\mathbf{Q}(\sqrt{2})$, that has $i$ as root, the degree of the minimum polynomial $m_{i,\mathbf{Q}(\sqrt{2})}$ of $i$ over $\mathbf{Q}(\sqrt{2})$ is bounded upwards to the degree of $m_{i,\mathbf{Q}}$, this shows the inequality.
Thus $[\mathbf{Q}(\sqrt{2},i):\mathbf{Q}(\sqrt{2})]\leq  [i:\mathbf{Q}] = 2$. But $[\mathbf{Q}(\sqrt{2},i):\mathbf{Q}(\sqrt{2})] = 1$ cannot hold since $i \notin \mathbf{Q}(\sqrt{2})$ as $\mathbf{Q}(\sqrt{2}) \subset \mathbf{R}$ holds.
Thus $[\mathbf{Q}(\sqrt{2},i):\mathbf{Q}(\sqrt{2})] = 2$ and you get
$$[\mathbf{Q}(\sqrt{2},i):\mathbf{Q}] = [\mathbf{Q}(\sqrt{2},i):\mathbf{Q}(\sqrt{2})] \cdot [\mathbf{Q}(\sqrt{2}): \mathbf{Q}] = [\mathbf{Q}(\sqrt{2},i):\mathbf{Q}(\sqrt{2})] \cdot 2 = 2 \cdot 2 = 4$$
as desired.
