Finding minimal polynomial and splitting field Given $\alpha = \sqrt{7 + \sqrt{7}}$, I want to find 2 entities:


*

*Determine, with justification, the minimal polynomial, $m_\mathbb{Q}(\alpha)$ of $\alpha$ over $\mathbb{Q}$.

*Recall that the splitting field $\mathbb{E}$ of $m_\mathbb{Q}(\alpha)$ is the smallest extension of $\mathbb{Q}$ that contains all roots of $m_\mathbb{Q}(\alpha)$. Determine, with justification $|\mathbb{E}: \mathbb{Q}|$.


In regards to the first, it's straightforward to show that $m_\mathbb{Q}(\alpha) = \alpha^4 - 14\alpha^2 + 42$, and that $m_\mathbb{Q}(\alpha)$ is irreducible over $\mathbb{Q}$. However, it is the second part that I am stuck on. The roots of $m_\mathbb{Q}(\alpha)$ are easily shown to be $\pm\sqrt{7\pm\sqrt{7}}$, but I am unsure how to proceed.
 A: These problems are usually quite annoying without galois theory, and usually trivial with galois theory, so I'll assume you're unfamiliar with it.  
We know that $|\mathbb{Q}(\sqrt7) : \mathbb{Q}| = 2$ 
Now we want to show  $|\mathbb{Q}(\sqrt{7+\sqrt7}) :  \mathbb{Q}(\sqrt7)| = 2.$
And similarly  $|\mathbb{Q}(\sqrt{7-\sqrt7}) :  \mathbb{Q}(\sqrt7)| = 2.$
We then can conclude  $|\mathbb{Q}(\sqrt{7+\sqrt7}) :  \mathbb{Q}| =  |\mathbb{Q}(\sqrt{7-\sqrt7}) :  \mathbb{Q}| = 4$
Proof: 
By the work you've done you should easily see that $p(x) = x^2 +  (-7 +\sqrt7)$ has $\pm \sqrt{7+\sqrt7}$ as roots.  So we let's hope $p(x)$ is irreducible over  $\mathbb{Q}(\sqrt7) $.
Suppose it were reducible, meaning $p(x) = (x-a)(x-b) = x^2 -x(a+b) +ab$ with $a,b \in \mathbb{Q}(\sqrt7)$.
We find that $a+b = 0$ since there is no $x$ coefficient in $x^2 +  (-7 +\sqrt7)$, so $a = -b.$
Now we have $ab = -a^2 =  -7 +\sqrt7$ and since $a \in \mathbb{Q(\sqrt7)}, \ \ a = c +d\sqrt7$ with $ \ c,d \in \mathbb{Q} $ 
So $(c+d\sqrt7)^2 = c^2 +7d^2 + 2cd\sqrt7 = 7 - \sqrt7$ which implies
$c^2 + 7d^2 - 7 = -2cd\sqrt7 - \sqrt7 = -\sqrt7(2cd -1)$ and we have that 
$$\frac{c^2 +7d^2 -7}{2cd -1} = -\sqrt7$$ 
Now if $2cd - 1 \neq 0$ then the left hand side is in $\mathbb{Q}$ and the right hand side isn't, so that's a contradiction and therefore $x^2 + (\sqrt7 - 7)$ is irreducible over $\mathbb{Q(\sqrt7)}$
Suppose $2cd - 1 = 0$.  Then $c = \frac{1}{2d}$ and our earlier equation reads as $$(\frac{1}{2d} + d\sqrt7)^2 = \frac{1}{4d^2} + \sqrt7 +7d^2 = 7 - \sqrt7$$ so $$\frac{1}{4d^2} +7d^2 - 7 = -2\sqrt7 $$
Again the L.H.S is in $\mathbb{Q}$ and the right hand side isn't, so we've shown that   $x^2 + (\sqrt7 - 7)$ is irreducible over $\mathbb{Q(\sqrt7)}$.
Very similar steps can be taken to show that $(x - (\sqrt{7 - \sqrt7})(x -
 (-\sqrt{7-\sqrt7})$ is irreducible $\mathbb{Q(\sqrt7)}$
So now we have that 
 $|\mathbb{Q}(\sqrt{7+\sqrt7}) :  \mathbb{Q}(\sqrt7)| = |\mathbb{Q}(\sqrt{7-\sqrt7}) :  \mathbb{Q}(\sqrt7)| = 2$ 
Now if we show that $\mathbb{Q}(\sqrt{7+\sqrt7}) / \mathbb{Q(\sqrt7)} \neq  \mathbb{Q}(\sqrt{7-\sqrt7}) / \mathbb{Q(\sqrt7)}$ we can conclude that $|\mathbb{E}:\mathbb{Q}| = 8$
This can be done without too much trouble by showing that $\sqrt{7 - \sqrt7} \notin \mathbb{Q}(\sqrt{7+\sqrt7}) / \mathbb{Q\sqrt7)} $
The rest of the details should be filled in without too much trouble.
A: I claim that the minimal polynomial of $\sqrt{7 - \sqrt{7}}$ over $\mathbf{Q}[\alpha]$ is $$x^2 - (7 - \sqrt{7}) = x^2 + \alpha^2 - 14.$$
We will make use of Eisenstein's criterion.
First, let us see how $\alpha^2 - 14$ factors into prime ideals:
\begin{align}
\frac{\mathbf{Z[\alpha]}}{\alpha^2 - 14} &\cong \frac{\mathbf{Z[x]}}{(x^4 - 14x^2 + 42, x^2 - 14)} \\
&\cong \frac{\mathbf{Z}[\sqrt{14}]}{(\sqrt{14})^4 - 14(\sqrt{14})^2 + 42} \\
&= \frac{\mathbf{Z}[\sqrt{14}]}{42} \\
&\cong \frac{\mathbf{Z}[\sqrt{14}]}{2} \times \frac{\mathbf{Z}[\sqrt{14}]}{3} \times \frac{\mathbf{Z}[\sqrt{14}]}{7} \\
&\cong \mathbf{F}_2[x]/(x^2) \times \mathbf{F}_9 \times \mathbf{F}_7[x]/(x^2).
\end{align}
So there is a prime ideal $\mathfrak p$ with $3 \in \mathfrak p$, $\alpha^2 -14 \in \mathfrak p$ and $\mathfrak p^2 \not\mid \alpha^2 - 14$. This ideal is
$$ \mathfrak p = (3, \alpha^2 - 14). $$
We can see that it is prime because
\begin{align}
\frac{\mathbf{Z}[\alpha]}{\mathfrak p} &\cong \frac{\mathbf{Z}[x]}{(3, x^2 -14,x^4-14x^2+42)} \\
&\cong \frac{\mathbf{Z}/3}{(x^2-14, x^2(x^2 - 14))} \\
&\cong \frac{\mathbf{Z}/3}{(x^2-14)} \\
&=\mathbf{F}_9.
\end{align}
Since $\mathbf{Z}[\alpha]/\mathfrak p$ is a field, $\mathfrak p$ is a maximal ideal. By definition, $\mathfrak p \ni \alpha^2 - 14$.
To see that $\alpha^2 - 14 \notin \mathfrak{p}^2$ we first compute
$$ \mathfrak{p}^2 = (9, 3\alpha^2 - 42, (\alpha^2 - 14)^2) = (9,3\alpha^2 - 42, -14\alpha^2 + 154). $$
If $\alpha^2  - 14 \in \mathfrak{p}^2$ then
$$ 14(\alpha^2  - 14) + (-14\alpha^2 + 154) + 5 \cdot 9 = 3 \in \mathfrak{p}^2 $$
Thus $\mathfrak{p} = (3, \alpha^2 - 14) \subseteq \mathfrak{p}^2$, a contradiction.
Therefore $\alpha^2 - 14 \in \mathfrak{p}\setminus\mathfrak{p}^2$ and the polynomial $x^2 + \alpha^2 - 14$ is irreducible by Eisenstein's criterion. Therefore $\sqrt{7 - \sqrt{7}}$ has degree $2$ over $\mathbf{Q}[\alpha]$ whence we conclude that the splitting field, $\mathbf{Q}[\alpha, \sqrt{7 - \sqrt{7}}]$, is a degree $8$ extension of $\mathbf{Q}$.
