As it says in the title, but there is a caveat: I cannot use or prove the fact that $\mathbb{Q}[\sqrt{p}]$ is a field beforehand, since the whole point of the exercise is to prove this from the title statement.
I have been attempting this for a few hours now and have no idea how to proceed. I have proven this sort of thing before, by contradiction, but, in this case, I only managed to get $1 \in \mathbb{Q}[\sqrt{p}]$...
I'd love a suggestion or an indication of a way to proceed more than an actual proof.
Thanks in advance!
EDIT: I do know that $\mathbb{Q}[\sqrt{p}] = \{x + y\sqrt{p} \,|\, x, y, \in \mathbb{Q}\}$
Considering that $\mathbb Q$ is a field, it follows that $\mathbb Q[x]$ is a PID (principal ideal domain). We may assume therefore that $J = (q(x))$ for some polynomial $q(x) \in \mathbb Q[x]$ such that $q(\sqrt p) = 0.$ Considering that $\sqrt p$ is a root of $q(x),$ it follows that $x - \sqrt p$ is a factor of $q(x)$ in $\mathbb R[x].$ But by hypothesis that $q(x)$ is in $\mathbb Q[x],$ we must have that $x + \sqrt p$ is also a factor of $q(x)$ in $\mathbb R[x].$ (Why?) We have therefore that $x^2 - p$ is a factor of $q(x)$ in $\mathbb Q[x]$ so that $J = (q(x)) = (x^2 - p).$
Every prime ideal of a PID is maximal, hence it suffices to show that $J$ is prime. Of course, a principal prime ideal is generated by a single prime element, so it suffices to prove that $x^2 - p$ is prime. Further, the prime elements of a PID (or more generally a UFD) are the same as the irreducibles, hence it suffices to prove that $x^2 - p$ is irreducible. Can you finish the proof?