Constructing an isomorphism between $\mathbb{Q}(i,\sqrt{2})$ and $\mathbb{Q}(i + \sqrt{2})$

Question

So, we know that $\mathbb{Q}(i, \sqrt{2}) = \mathbb{Q}(i + \sqrt{2})$ by constructing the two extensions separately and comparing: building first $\mathbb{Q}(i)$ and then $\mathbb{Q}(i)(\sqrt{2})$ as a tower of fields, by one side, for example, and for the other side, construct $\mathbb{Q}(i + \sqrt{2})$ "by hand", finding its minimal polynomial. This process is sincerely boring and demands a lot of work, specially when we have extensions of higher order, so I was wondering if there is a way to construct an explicit $\mathbb{Q}$-automorphism that gives us more elegantly the desired relation.

Answer 1

Yes, there is a way to construct an explicit $\Bbb Q-$isomorphism by hand.

$\Bbb Q(i,\sqrt{2})$ is a $\Bbb Q-$vector space of dimension four, and a basis for such space is $\{1,i,\sqrt{2},i\sqrt{2}\}$. On the other hand $\Bbb Q(i+\sqrt{2})$ is also a $\Bbb Q-$vector space of dimension four. Set $\xi=i+\sqrt{2}$, a basis for $\Bbb Q(i+\sqrt{2})$ is $\{1,\xi,\xi^2,\xi^3\}$,that is $\{1,i+\sqrt{2},1+2i\sqrt{2},5i-\sqrt{2}\}$.

Now, take $\phi:\Bbb Q(i,\sqrt{2})\longrightarrow \Bbb Q(i+\sqrt{2})$ such that $$\phi(1)=1$$ $$\phi(i)=\frac{1}{6}(\xi+\xi^3)=i$$ $$\phi(\sqrt{2})=\frac1 2(5\xi-\xi^3)=\sqrt{2}$$ $$\phi(i\sqrt{2})=\frac12(-1+\xi^2)=i\sqrt{2}$$

The matrix associated to $\phi$ with respect to those basis is

$$ M=\begin{pmatrix} 1 & 0 & 0 & -\frac12\\ 0 & \frac16 & \frac 52 & 0\\ 0 & 0 & 0 & \frac12\\ 0 & \frac16 & -\frac12 & 0 \end{pmatrix} $$

Now, you may easily compute the determinant of $M$ which is different to zero, hence $M$ is invertible. Thus, $\phi$ defines the desired isomorphism.

Answer 2

Observe that

$$(\sqrt2+i)^3=2\sqrt2+6i-3\sqrt2-i=-\sqrt2+5i\implies\left(\sqrt2+i\right)^3+\sqrt2+i=6i\in\Bbb Q(\sqrt2+i)$$

$$\implies\frac16\cdot6i=i\in\Bbb Q(\sqrt2+i)\;$$

Likewise

$$(\sqrt2+i)^3-5(\sqrt2+i)=-6\sqrt2\implies \sqrt2\in\Bbb Q(\sqrt2+i)$$

and thus $\;\Bbb Q(\sqrt2,\,i)\subset\Bbb(\sqrt2+i)\ldots\;$

Answer 3

It does depend on how you constructed the fields themselves.

For example, if $\mathbb{Q}(i, \sqrt{2})$ means the smallest subfield of $\mathbb{C}$ containing $\mathbb{Q}$, $i$, and $\sqrt{2}$, and similarly for $\mathbb{Q}( i + \sqrt{2})$, then they are actually equal fields, so the identity function is such an isomorphism.