Let $a = \sqrt{2+i}$ and $K$ is the splitting field of minimal polynomial of $a$ over $\mathbb{Q}$. Prove that $Gal(K/\mathbb{Q})$ is $D_4$.
I find the minimal polynomial of $a$ is $p(x)=x^4-4x^2+5$ and its roots are $\sqrt{2+i},-\sqrt{2+i},\sqrt{2-i},-\sqrt{2-i}$. Let $b=\sqrt{2-i}$. So the splitting field of $p$ is $K=\mathbb{Q}(a,b)$. Also by Eisenstein's criterion to $p(x-1)=x^4 - 4 x^3 + 2 x^2 + 4 x + 2$ with the prime 2 we can conclude, $p$ is irreducible over $\mathbb{Q}$. Thus $[\mathbb{Q}(a):\mathbb{Q}]=4$. Also $b\not\in\mathbb{Q}(a)$ (since if it were then $\sqrt{2+i}\cdot\sqrt{2-i}=\sqrt{5}$ would also be in $\mathbb{Q}(a)$), and minimal polynomial of $b$ over $\mathbb{Q}(a)$ is $x^2-2+i$. Thus $[\mathbb{Q}(a,b):\mathbb{Q}]=[\mathbb{Q}(a,b):\mathbb{Q}(a)]\cdot[\mathbb{Q}(a):\mathbb{Q}]=2\cdot 4=8.$ Now since $K$ is the splitting field over $\mathbb{Q}$ of a separable polynomial, $K/\mathbb{Q}$ is Galois. Hence $|Gal(K/\mathbb{Q})|=8$. Hence the 8 automorphisms are $$a\to \begin{cases}a\\-a\\b\\-b\end{cases}\quad\text{and}\quad b \to \begin{cases}b\\-b\end{cases}.$$ But there is no automorphism of order 4. Then how can the Galois group be $D_4$?
Can somebody correct me what I am missing?
We put as before $\;a=\sqrt{2+i}\;,\;\;b=\sqrt{2-i}$ . I think the basic automorphisms are ("copying" the embedding $\;D_4\hookrightarrow S_4$ ):
$$\begin{cases}\sigma:\;\;a\mapsto -b\;,\;\;b\mapsto a\;\\{}\\\tau:\;\;a\mapsto a\;,\;\;b\mapsto -b\;\end{cases}$$
Observe then
$$\begin{cases}\sigma^2(a)=\sigma(-b)=-a\;,\;\;\sigma^2(b)=\sigma(a)=-b\;,\;\\{}\\ \sigma^3(a)=\sigma(-a)=b\;,\;\;\sigma^3(b)=\sigma(-b)=-a\;,\\{}\\ \sigma^4(a)=\sigma(b)=a\;,\;\;\sigma^4(b)=\sigma(-a)=b\;.\end{cases}\;\;\;\;\;\;\;\;\;\;\implies \text{ord}\,\sigma=4$$
$${}$$
$$\tau^2(a)=a\;,\;\;\tau^2(b)=b\;\;\;\;\implies \text{ord}\,\tau=2$$
And also
$$\begin{cases}\tau\sigma\tau(a)=\tau\sigma(a)=\tau(-b)=b=\sigma^3(a)\\{}\\ \tau\sigma\tau(b)=\tau\sigma(-b)=\tau(-a)=-a=\sigma^3(b)\end{cases}\;\;\;\;\;\;\;\;\;\;\implies \tau\sigma\tau=\sigma^3$$
and we thus got
$$\text{Gal}\,\left(K/\Bbb Q\right)=\left\{\;\sigma,\,\tau\;/\;\sigma^4=\tau^2=1\;,\;\;\tau\sigma\tau=\sigma^3\;\right\}\cong D_4$$