Prove that $\mathbb{Q}(\sqrt{2+\sqrt{2}}) / \mathbb{Q}$ is a normal extension

Question

I need to prove that $\mathbb{Q}(\sqrt{2+\sqrt{2}}) / \mathbb{Q}$ is a normal extension and $\mathbb{Q}(\sqrt{2+\sqrt{2}}, i)=\mathbb{Q}(\phi)$, where $\phi^{4}=i$.

For the first part i use the theorem that says: every extension of degree 2 is normal.

Note that the polynomial \begin{equation*} q(x)=x^2 -(2+\sqrt{2}) \in \mathbb Q[x] \end{equation*} is minimal of $\gamma = \sqrt{2+\sqrt{2}}$ over $\mathbb Q$, where $\gamma$ is a root of $q(x)$. Hence $[\mathbb{Q}(\gamma):\mathbb Q]=2$ and $\gamma$ is a root of $q(x)$ then $\mathbb{Q}(\gamma)$ is a splitting field of the polynomial $q(x)$ over $\mathbb Q$, then $\mathbb{Q}(\sqrt{2+\sqrt{2}}) / \mathbb{Q}$ is normal because every extension of degree 2 is normal.

my question is if the prove is right? and any idea to prove the second part? $\mathbb{Q}(\sqrt{2+\sqrt{2}}, i)=\mathbb{Q}(\phi)$, where $\phi^{4}=i$.

Thanks.

Answer 1

Let $a=\sqrt {2+\sqrt{2}}$ and then $a$ is a root of $$f(x) =(x^2-2)^2-2=x^4-4x^2+2\in\mathbb {Q} [x] $$ The polynomial above is irreducible over $\mathbb {Q} $ (via Eisenstein criterion with $p=2$) and hence $[\mathbb {Q} (a):\mathbb{Q }] =4$.

We now show that $f(x) $ splits over $\mathbb {Q}(a)$ and hence $\mathbb {Q} (a) $ is normal over $\mathbb {Q} $. We can observe that $-a$ is also a root of $f(x) $ and hence $$f(x) =(x^2-a^2)(x^2-c)$$ for some $c\in\mathbb {Q} (a) $ (this involves that $f(x) $ has no term of degree $3$). Comparing coefficients we get $c=2/a^2$ and then $$c=\frac{(a^2-2)^2}{a^2}$$ so that the $$\pm\frac{a^2-2}{a}$$ are the roots of $x^2-c$. It follows that the roots of $f(x) $ are $$\pm a, \pm\frac{a^2-2}{a}$$ (check that they are distinct) and $\mathbb{Q} (a) $ is the splitting field of $f(x) \in\mathbb {Q} [x] $.

Next we show $\mathbb {Q} (\phi) =\mathbb{Q} (a,i)$. Since $a$ is real $i\notin \mathbb {Q} (a) $ and therefore $[\mathbb {Q} (a, i) :\mathbb {Q} (a)] =2$ and then $[\mathbb {Q} (a, i) :\mathbb{Q}] =8$.

Next we note that $\phi$ is a root of $x^8+1\in\mathbb {Q} [x] $ which is again irreducible (why? Because any polynomial of type $x^n-k\in F[x] $ is irreducible if it does not have any roots in $F$). Hence $[\mathbb {Q} (\phi) :\mathbb{Q}] =8$.

Finally let us show that both $a, i$ lie in $\mathbb {Q} (\phi) $. Now $i=\phi^4\in\mathbb {Q} (\phi) $. Next consider $b=\phi+\phi^{-1}$. We have $\phi^8+1=0$ or $$\phi^4+\phi^{-4}=0$$ or $$(\phi^2+\phi^{-2})^2-2=0$$ or $$((\phi+\phi^{-1})^2-2)^2-2=0$$ so that $b$ is a root of $$(x^2-2)^2-2=x^4-4x^2+2\in\mathbb{Q}[x] $$ which also turns out to be the minimal polynomial $f(x) $ for $a$. As we have noted earlier this polynomial is such that all its roots can be expressed as rational functions of any of its roots and thus $a\in\mathbb {Q} (b) \subseteq \mathbb {Q} (\phi) $ (we also have $b\in\mathbb {Q} (a) $ as well so that $\mathbb {Q} (a) =\mathbb{Q} (b) $).

We thus have $\mathbb {Q} \subseteq \mathbb {Q} (a, i) \subseteq \mathbb {Q} (\phi) $ and since both fields $\mathbb {Q} (\phi), \mathbb {Q} (a, i) $ are of degree $8$ over $\mathbb {Q} $ it follows that they are same field.

Answer 2

To show that $$\mathbb{Q}(\sqrt{2+\sqrt{2}}) / \mathbb{Q}$$ is a normal extension we can show that all conjugates lie in the field.

The conjugates are

• $\alpha_1 = \sqrt{2+\sqrt{2}}$
• $\alpha_2 = \sqrt{2-\sqrt{2}}$
• $\alpha_3 = -\sqrt{2+\sqrt{2}}$
• $\alpha_4 = -\sqrt{2-\sqrt{2}}$

We are done if we show that we can express $\alpha_2$ in terms of $\alpha_1$.

Note that $\alpha_1^2 - 2 = \sqrt{2}$ and $\alpha_1 \alpha_2 = \sqrt{2}$.

---

For the second part, try to write $\phi$ in component form.