I posted this question before, but I didn’t receive any help, which is fine, but I’m posting it again because I really need some help because my number theory exam is quite close and I’m really struggling with this question. I emailed my lecturer last week and he hasn’t responded to me either so could someone please give me some hints on how to answer it. The part I’m struggling with is part c)
If any of my working is correct, I’ve managed to show that $N(p)|28$ but I really don’t know to prove the statement in the question.
You are given that $\left( \frac{-7}{p}\right) = 1$ and since $-7$ is the discriminant of the field of fractions of $R$, this means that $p$ splits in $R$. In particular, we have prime ideals $\mathfrak{p}$ and $\bar{\mathfrak{p}}$ such that
$$pR = \mathfrak{p}\bar{\mathfrak{p}}$$
with $N\mathfrak{p} = N\bar{\mathfrak{p}} = p$. Since $R$ was assumed to be a UFD, these two ideals are principal ideals ($R$ is the ring of integers of a number field, so it is a Dedekind domain, and a Dedekind domain is a PID if and only if it is a UFD), so we have an equality of principal ideals
$$pR = (\pi_1)(\pi_2)$$
with $\pi_1, \pi_2$ prime elements of $R$. Thus, this translates to an equality of elements
$$p = u \pi_1 \pi_2$$
with $u \in R^\times$. But $R^\times = \lbrace \pm 1\rbrace$ (since $R$ is the ring of integers of an imaginary quadratic field which is not $\Bbb Q(\sqrt{-1})$ or $\Bbb Q(\sqrt{-3})$) so we have
$$p = \pm \pi_1 \pi_2.$$
Since $N(p) = p^2$ and $N(\pm 1) = 1$, we have
$$N(p) = p^2 = N(\pi_1)N(\pi_2).$$
Finally, since $\pi_1$ and $\pi_2$ are prime elements, they have prime norm, and the only prime divisors of $p^2$ are $p$, so we must have $N(\pi_1) = N(\pi_2) = p$.
Hence, there are $4$ elements with norm $p$ in $R$; namely $\lbrace \pm \pi_1, \pm \pi_2\rbrace$.