When is $\mathbb{Q}(\sqrt{p+\sqrt{p}})$ a Galois extension of $\mathbb{Q}$?

Question

When is $\mathbb{Q}(\sqrt{p+\sqrt{p}})$ a Galois extension of $\mathbb{Q}$?

I was motivated by the question that $\mathbb{Q}(\sqrt{5+\sqrt{5}})$ is a Galois extension of $\mathbb{Q}$. Here is a rough sketch of the proof.

First, note that the polynomial $$f(x)=(x-\sqrt{5+\sqrt{5}}) (x+\sqrt{5+\sqrt{5}})(x-\sqrt{5-\sqrt{5}})(x+\sqrt{5-\sqrt{5}})$$ is irreducible by the Eisenstein irreducibility criterion. So we claim that $\mathbb{Q}(\sqrt{5+\sqrt{5}})$ is the splitting field of the separable polynomial $f(x)$. It suffices to show that $\sqrt{5-\sqrt{5}}\in\mathbb{Q}(\sqrt{5+\sqrt{5}})$. To do this, we first note that $5+\sqrt{5}$ and hence $\sqrt{5}$ is in $\mathbb{Q}(\sqrt{5+\sqrt{5}})$.

Then, by noticing that $$\sqrt{5-\sqrt{5}}=\frac{\color{red}{\sqrt{5-1}}\sqrt{5}}{\sqrt{5+\sqrt{5}}}= \frac{\color{red}2\sqrt{5}}{\sqrt{5+\sqrt{5}}},\tag{1}$$

we see that $\sqrt{5-\sqrt{5}}$ is in $\mathbb{Q}(\sqrt{5+\sqrt{5}})$ as desired.

By the above argument, we see that $\mathbb{Q}(\sqrt{p+\sqrt{p}})$ is a Galois extension of $\mathbb{Q}$ when $p-1$ is a square number. (The importance of $p-1$ being a square number can be seen from equation (1)-See the part highlighted in red)

I am wondering if this is both a necessary and sufficient criterion for $\mathbb{Q}(\sqrt{p+\sqrt{p}})$ to be a Galois extension of $\mathbb{Q}$. In other words, $\mathbb{Q}(\sqrt{p+\sqrt{p}})$ is a Galois extension of $\mathbb{Q}$ if and only if $p$ is prime and $p-1$ is a square number.

Some things that I have noticed regarding this problem:

1. $p=3$ doesn’t satisfy the suggested criterion and isn’t a Galois extension. See (https://math.stackexchange.com/questions/4052247/showing-that-mathbbq-sqrt3-sqrt3-is-not-galois)

2. For general a $p$, $f(x)=x^4-2px^2+(p^2-p)$.

3. I am not sure if we need $p$ to be prime. The only place where we used $p$ is prime is when we used the Eisenstein irreducibility criterion. But do we really need $f(x)$ to be irreducible?

4. This is where I am probably getting a bit ahead of myself. But when we say that $p-1$ is a square number, most of us naturally assume that $p$ is in $\mathbb{Z}^+$ and $p-1$ is square in $\mathbb{Z}^+$. But can we further generalise the result by allowing $p$ to be in $\mathbb{Q}$ and $p-1$ be square in $\mathbb{Q}$?

Answer 1

Note that $f$ cannot have an irreducible factor of degree $3$ since it is even, so if $f$ is reducible, it means that the minimal polynomial of $\sqrt{p+\sqrt{p}}$ has degree at most $2$, so your extension is automatically Galois in this case.

Hence we may assume without loss of generality that $f$ is irreducible.

Your question then may be rephrased as follows: when is the Galois group of $f=X^4-2pX^2+p^2-p$ is the Klein group or the cyclic group of order $4$ ?

The answer (and much more!) is given in Corollary 4.5. of this paper by Keith Conrad

Applied to your specific polynomial, the answer is : exactly when $p^2-p$ is a square (Klein group case) or when $p-1$ is a square (cyclic case).

Another link with a direct proof may be found here