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I would appericate if anyone could take a look at my attempt:
Suppose and aim for contradiction that it's a normal extension. The minimal polynomial of $\alpha = \sqrt {2+\sqrt p}$ over $\mathbb Q$ is $f(x)=x^4-4x^2+(4-p)$. Its roots are $\alpha, -\alpha, \beta, -\beta$, where $\beta = \sqrt {2-\sqrt p}$ . Since it's a normal extension, $\beta \in \mathbb Q (\alpha)$. But then $\alpha \beta =\sqrt{4-p} \in \mathbb Q (\alpha)$. Since $p>4$, $4-p<0$, thus $\sqrt{4-p}=i\sqrt{p-4}$ . But since $\mathbb Q (\alpha) \subseteq \mathbb R$, we get that $i\sqrt{p-4} \in \mathbb R \Rightarrow i \in \mathbb R$, a contradiction. So it's not a normal extension, hence not Galois.

Edit: I should probably say that since $p$ is prime, $\sqrt p \notin \mathbb Q$.

J. W. Tanner
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RatherAmusing
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  • It follows also from this duplicate, because $a^2-db^2=4-p$. This cannot be a rational square for $p>4$. So the extension is not Galois in this case. – Dietrich Burde Aug 07 '24 at 20:24
  • Please review the guidelines at https://math.stackexchange.com/tags/solution-verification/info. Please don't use "Edit:" - just revise the question to what it should have been from the start. – D.W. Aug 08 '24 at 07:15

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Looks good to me. You may want to add that for the same reason ($\sqrt p\notin\Bbb{Q}$) the polynomial $f(x)$ is, indeed, irreducible over $\Bbb{Q}$. After all, the only factor in $\Bbb{R}[x]$ would be $x^2-\alpha^2$, which has an irrational coefficient.

Jyrki Lahtonen
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