Find dimension of $\mathbb{Q}(\sqrt{2}, i)$ over $\mathbb{Q}$

Question

One of the tasks that I have is to first (1) find the minimal polynomial of $\sqrt{2} + i$ over $\mathbb{Q}$, and (2) find the degree of $\mathbb{Q}(\sqrt{2}, i)$ over $\mathbb{Q}$, i.e. $[\mathbb{Q}(\sqrt{2}, i) : \mathbb{Q}]$.

For (1), I first find the minimal polynomial as follows: If $x = \sqrt{2} + i$, then \begin{align*} x^2 &= (\sqrt{2} + i)^2 \\ x^2 &= 2 + 2\sqrt{2}i - 1 = 1 + 2\sqrt{2}i \\ x^2 - 1 &= 2\sqrt{2}i \\ (x^2 - 1)^2 &= (2\sqrt{2}i))^2 \\ x^4 - 2x^2 + 1 &= -4 \\ x^4 - 2x^2 + 5 &= 0 \end{align*} So the minimal polynomial of $\sqrt{2} + i$ is $x^4 - 2x^2 + 5 = 0$.

Now for (2), I know I can use the fact that \begin{equation*} [\mathbb{Q}(\sqrt{2}, i) : \mathbb{Q}] = [\mathbb{Q}(\sqrt{2}, i) : \mathbb{Q}(\sqrt{2} + i)][\mathbb{Q}(\sqrt{2} + i) : \mathbb{Q}] \end{equation*} in which because $x^4 - 2x^2 + 5$ is the minimal polynomial of $\sqrt{2} + i$ over $\mathbb{Q}(\sqrt{2} + i)$, then $[\mathbb{Q}(\sqrt{2} + i) : \mathbb{Q}] = 4$. Now the only issue that I am running into is to show that $[\mathbb{Q}(\sqrt{2}, i) : \mathbb{Q}(\sqrt{2} + i)] = 1$. Some help would be much appreciated!

Answer 1

It suffices to show that $[\mathbb{Q}(\sqrt{2},i):\mathbb{Q}]=4$. We have $$[\mathbb{Q}(\sqrt{2},i):\mathbb{Q}]=[\mathbb{Q}(i,\sqrt{2}):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2.2=4,$$ in which $[\mathbb{Q}(i,\sqrt{2}):\mathbb{Q}(\sqrt{2})]=2$ since $[\mathbb{Q}(i,\sqrt{2}):\mathbb{Q}(\sqrt{2})]\leq 2$ and it can not be $1$ since $i\not \in \mathbb{Q}(\sqrt{2}).$

Answer 2

It is trivial that $\mathbb Q\left(\sqrt2+i\right)\subset \mathbb Q\left(\sqrt2,i\right)$. Convsersely, since $$\frac 3{\sqrt 2+i}=\sqrt2-i,$$ we know that $\sqrt2-i$ belongs to $\mathbb Q\left(\sqrt2+i\right)$ and hence $\sqrt2$ and $i$ are in $\mathbb Q\left(\sqrt2+i\right)$.

Answer 3

It's obvious that $\sqrt 2 + i \in \Bbb Q(\sqrt 2,i)$, so $\Bbb Q(\sqrt 2+i)\subseteq \Bbb Q(\sqrt 2,i)$.

On the other hand, $i = \frac{1}{6}\left((i+\sqrt{2})+\left(\sqrt{2}+i\right)^3\right)\in\Bbb Q(\sqrt 2+i)$ and $\sqrt 2 = \frac{1}{6} \left(5 \left(\sqrt{2}+i\right)-\left(\sqrt{2}+i\right)^3\right)\in\Bbb Q(\sqrt 2+i)$, so $\Bbb Q(\sqrt 2,i)\subseteq \Bbb Q(\sqrt 2+i)$

Then $\Bbb Q(\sqrt 2,i) = \Bbb Q(\sqrt 2+i)$.

Answer 4

Your original question seems to be about just the dimension of $K=\mathbb{Q}(\sqrt{2},i)$ over $\mathbb{Q}$, not about finding a primitive element for $K$.

If so, to conclude that $[K:\mathbb{Q}]=4$ is enough to note that we have a tower $$ \mathbb{Q}\subset\mathbb{Q}(\sqrt{2})\subset\mathbb{Q}(\sqrt{2},i) $$ since each step is quadratic and the degree is multiplicative on towers.