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

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 $$\mathbb{Q}$$?

• $p$ certainly does not need to be prime: e.g., the extension is Galois whenever $p$ is a square. Commented Oct 11, 2022 at 6:07
• @EmilJeřábek Oh yes! Can’t believe that I missed something so obvious. Was thinking of the case when $p=a^2b$, where $b$ is non-square. For example, $p=2^2\times 3$ Commented Oct 11, 2022 at 6:11
• Moreover, for the Eisenstein irreducibility criterion you only need $p$ to not be a powerful number. Commented Oct 11, 2022 at 6:34

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

• But $p^2-p$ cannot be square, as it is divisible by $p$ but not $p^2$. Commented Oct 11, 2022 at 9:45
• Take $p = 4/3$... (the criterion is valid for rational $p$ such that the polynomial is irreducible). Commented Oct 11, 2022 at 10:24
• With $p = 4/3$, the polynomial is reducible, but $p = 9/5$ gives a Klein group example. Commented Oct 11, 2022 at 10:49