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Given $\alpha = \sqrt{7 + \sqrt{7}}$, I want to find 2 entities:

  1. Determine, with justification, the minimal polynomial, $m_\mathbb{Q}(\alpha)$ of $\alpha$ over $\mathbb{Q}$.
  2. 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.

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  • $\begingroup$ Do you know any galois theory? $\endgroup$ – Dionel Jaime Aug 2 '17 at 2:35
  • $\begingroup$ I've done a little galois theory. In general, it's been a while since I've done any abstract algebra, and at the moment I'm going back through some texts, and working out problems. $\endgroup$ – Pistol Pete Aug 2 '17 at 3:19
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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.

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  • $\begingroup$ Hopefully I didn't mess up any negative signs. $\endgroup$ – Dionel Jaime Aug 2 '17 at 4:02
  • $\begingroup$ Dionel Jaime after looking at this again, I think that immediately after "proof" one of the $\sqrt{7}$ terms should have a "-" instead of "+". Also, $2cd-1$ should be $2cd+1$. $\endgroup$ – Pistol Pete Aug 7 '17 at 3:23
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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 $\mathbf{Q}$.

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