$\mathbb{Q}(\sqrt{p+\sqrt{p}})$ is Galois over $\mathbb{Q} \iff p=n^2+1$ for a prime $p$.

Question

Recently, I have encountered a question saying that $\mathbb{Q}(\sqrt{p+\sqrt{p}})$ is not always Galois over $\mathbb{Q}$, with the counterexample $p=3$.

My professor then followed up with that question and said that $\mathbb{Q}(\sqrt{p+\sqrt{p}})$ is Galois over $\mathbb{Q}$ if and only if $p$ is a prime of the form $n^2+1$.

So far, I have managed to prove the $\Leftarrow$ direction, but got stuck trying to prove the other.

I considered the polynomial $h(x)=x^4-2px^2+p^2-p$, and noted that it is both a separable and irreducible polynomial. Since $\mathbb{Q}(\sqrt{p+\sqrt{p}})$ is a Galois extension, the roots of $h(x)$, that is, $\pm\sqrt{p+\sqrt{p}}, \pm\sqrt{p-\sqrt{p}}$ all lie in $\mathbb{Q}(\sqrt{p+\sqrt{p}})$.

I have also shown that $\sqrt{p-1}\in\mathbb{Q}(\sqrt{p+\sqrt{p}})$, and then I'm stuck.

So my question is, how do I proceed from here?

Answer 1

Suppose $p$ is prime. The Galois closure of $L=\Bbb Q(\sqrt{p+\sqrt p})$ is $\Bbb Q(\sqrt{p+\sqrt p},\sqrt{p-\sqrt p})$. So $L$ is Galois over $\Bbb Q$ iff $p-\sqrt p$ is a square in $L$.

By Kummer theory, $p-\sqrt p$ is a square in $L$ iff either it is a square in $K=\Bbb Q(\sqrt p)$ or if $(p+\sqrt p)(p-\sqrt p)$ is a square in $K$. In the former case $N(p-\sqrt p)=p^2-p$ must be a square in $\Bbb Q$, which it isn't as it's a multiple of $p$ but not of $p^2$. In the latter case $(p+\sqrt p)(p-\sqrt p)=p(p-1)$ is a square in $K$. The rationals that are squares in $K$ are the squares of rationals or $p$ times the squares of rationals. $p(p-1)$ is not a square of a rational, and is $p$ times a square of a rational iff $p-1=n^2$ where $n$ is an integer.

Answer 2

One can directly apply the criterion which appears in the question of @user289143, with $K=\mathbf Q$ and $L=\mathbf Q(p+\sqrt p)=\mathbf Q(\sqrt p)$. Then $L(\sqrt {p+\sqrt p})$ is galois over $\mathbf Q$ iff $(p-\sqrt p)/(p+\sqrt p)$, or equivalently $(p-\sqrt p)(p+\sqrt p)=p^2-p=p(p-1)$, is a square in $L$. But $p-1$ is a square in $L$ iff it is of the form $a^2+pb^2+2ab\sqrt 2$, with $a,b \in \mathbf Q$. The only possibility is $b=0$, from which we derive easily that $p-1$ is the square of an integer.