Prove that $K = \mathbb{Q}[i, \sqrt{2}] = \mathbb{Q}[i + \sqrt{2}]$

Question

Prove that the field extension $K = \mathbb{Q}[i, \sqrt{2}] = \mathbb{Q}[i + \sqrt{2}]$

I have that $\mathbb{Q}[i, \sqrt{2}]$ = {$a + bi +c\sqrt{2} | a, b, c \in \mathbb{Q}$}

and $\mathbb{Q}[i +\sqrt{2}]$ = {$a + b(i +\sqrt{2}) | a, b, c \in \mathbb{Q}$}

If $x \in \mathbb{Q}[i + \sqrt{2}], x = a_x + b_x(i + \sqrt{2}) = a_x + b_x i + b_x \sqrt{2} \in \mathbb{Q}[i, \sqrt 2]$

I am having trouble proving that an element of $\mathbb{Q}[i, \sqrt{2}]$ would be in $\mathbb{Q}[i + \sqrt{2}]$ unless the constants on $i$ and $\sqrt{2}$ are the same.

Answer 1

$\mathbb{Q}(i+\sqrt2)\subset\mathbb{Q}(i,\sqrt2)$.

On the other hand, ${{1\over{i+\sqrt2}}}=$

${{1\over{i+\sqrt2}}}{{i-\sqrt2}\over{i-\sqrt2}}$

$=-{1\over 3}(i-\sqrt2)$ implies that $i-\sqrt2\in\mathbb{Q}(i+\sqrt2)$

$i-\sqrt2+i+\sqrt2=2i\in\mathbb{Q}(i+\sqrt2)$ implies $i\in \mathbb{Q}(i+\sqrt2)$ and $\sqrt2\in\mathbb{Q}(i+\sqrt2)$.

Answer 2

If $x=i+\sqrt2$, can you find what are $$a,b=x\pm\frac3x$$

Answer 3

We know that

\begin{align*}\operatorname{Gal}(\mathbf{Q}(i,\sqrt{2}) /\mathbf{Q})&\xrightarrow{\sim} \{\pm 1\}^2 \\ \sigma&\mapsto \left(\frac{\sigma(i)}{i},\frac{\sigma(\sqrt{2})}{\sqrt{2}} \right). \end{align*}

Consider $\mathbf{Q}(\alpha)$ with $\alpha=i+\sqrt{2}$. Only the identity element in the Galois group fixes $\alpha$, hence by the fundamental theorem of Galois theory, $\mathbf{Q}(i,\sqrt{2})=\mathbf{Q}(\alpha)$.