Primes in quadratic number fields

Question

If $p$ is a prime number such that $p≡3\;mod\;4$, prove that $\sqrt{-p}$ is prime in $\mathbb{Z}[\sqrt[ ]{-p}]$ and in $\mathbb{Z}[\displaystyle\frac{1+\sqrt[ ]{-p}}{2}]$ too.

Notes

We have seen in class that if the norm is a prime then the element is irreducible.

I tried this: If $\sqrt[]{-p}$ is not a prime, then it means it factors in two other numbers of the form $\alpha=a+\sqrt{-p}\cdot b$ and $\beta=c+\sqrt{-p}\cdot d$, such that $\alpha\beta=\sqrt{-p}$.

Answer 1

Recall some facts:

In a domain, $x$ is a prime element if and only if $(x)$ is a prime ideal.

and

$\Bbb{F}_p$, the field with $p$ elements, is the unique commutative ring with unity (up to isomorphism) with cardinality $p$.

Let's focus first on $R=\Bbb{Z}\left[ \frac{1+\sqrt{-p}}{2}\right]$. This is the ring of integers of $K=\Bbb{Q}(\sqrt{-p})$, so we can use the definition of norm of an element.

Note that $N_K(\sqrt{-p}) = \sqrt{-p}\cdot (-\sqrt{-p}) = p$, so $\sqrt{-p}$ is an irreducible element of $R$. But there's more. I hope you know what is the norm of an ideal in a ring of integers.

Consider the ideal $P=(\sqrt{-p}) \subset R$. You have that $N_K(P)=N_K(\sqrt{-p}) = p$, so $R/P$ is a ring with $p$ elements. But there is only one ring with $p$ elements, i.e. the finite field of order $p$, so $P$ is actually a maximal (hence prime) ideal. Using the fact above, this proves that $\sqrt{-p}$ is prime in $R$.

As for the other ring, call $Q = (\sqrt{-p}) \subset \Bbb{Z}[\sqrt{-p}]$. Clearly $\Bbb{Z}[\sqrt{-p}] \subset R$ is an extension of rings, and $Q=\Bbb{Z}[\sqrt{-p}] \cap P$. Now one has that $Q$ is a prime ideal of $\Bbb{Z}[\sqrt{-p}]$ applying the following proposition:

Let $A\subset R$ be an extension of rings. If $P\subset R$ is a prime ideal of $R$, then $P \cap A \subset A$ is a prime ideal of $A$.

This proves that $\sqrt{-p}$ is a prime element in $\Bbb{Z}[\sqrt{-p}]$ as well.