Proving whether ideals are prime in $\mathbb{Z}[\sqrt{-5}]$

Question

I am working with the ideals $\mathfrak{p}=\left<2,1+\sqrt{-5}\right>, \mathfrak{q}=\left<3,1+\sqrt{-5}\right>, \mathfrak{t}=\left<3,1-\sqrt{-5}\right>$ and I am trying to prove that they are prime in $\mathbb{Z}[\sqrt{-5}]$.

I understand a good method to do this involves taking norms of each element that generates the ideals. This gives, for instance, the norms of the generators of $\mathfrak{p}=4,6$. These are both divisible by $2$. I have also calculated $\mathfrak{p}^2=\left<2\right>$. I get the feeling that since both norms are divisible by the generator of the ideal squared this proves it is prime, but I am not quite sure why.

In a similar vein, the norms of the generators of $\mathfrak{q,t}=9,6$ which are both divisible by $3$. I know $\mathfrak{qt}=\left<3\right>$, but I am not sure if this is relevant, since it is not either ideal squared. I think the norm of the ideal itself being prime implies the ideal is prime, but I am not sure how to find the norm of the ideal from the norm of its generating elements.

This question is similar to: Prove that ideals are prime, but I don't quite understand the reasoning behind the chosen answer. I am not sure how the answerer deduces that the example there is prime either. Following their reasoning as far as I can, the fact that $2$ divides both norms means that $\left<2\right> \subset \mathfrak{p} \subset \mathbb{Z}[\sqrt{-5}]$ but I'm not sure how, and then I don't know how this proves it is prime.

Answer 1

There are several ways to deduce that these are ideals are prime. The easiest might be to just compute the quotient:

$$\mathbb Z[\sqrt{-5}]/(2,1+\sqrt{-5}) \cong \mathbb Z[X]/(X^2+5,2,1+X) = \mathbb Z[X]/(2,1+X) \cong \mathbb Z/2\mathbb Z$$

But you can also use some theory (And this somehow fits to your norm approach). Whenever we have an quadratic integer ring and an integer prime number $p \in \mathbb Z$, then the ideal $(p) \subset \mathcal O_K$ behaves in three ways:

• $(p)$ is prime.
• $(p) = \mathfrak p^2$ for some prime ideal.
• $(p) = \mathfrak p_1 \mathfrak p_2$ for two different prime ideals.

Together with the fact, that there exists a unique prime ideal factorization, we get the following corollary: Whenever we have $(p)=IJ$ for some ideals $I,J$, then $I$ and $J$ are necessarily prime (If one of them would factor into primes, $(p)$ would factor into at least $3$ primes).

Answer 2

In that answer, the answerer speaks about norm of ideals, not norm of numbers. You can think geometrically, if $\langle 2, 1 + \sqrt{-5} \rangle \subset A$ then its lattice is a sub lattice of $A$, so the area of the fundamental area (the area of the smallest parallelogram in the lattice ) of $A$ divide the area of the fundamental area of $\langle 2, 1 + \sqrt{-5}\rangle$ but this area is two so the fundamental area of $A$ is one and $A = \mathbb{Z}[\sqrt{-5}]$. So $\langle 2, 1 + \sqrt{-5}\rangle$ is a maximal ideal, and so it is a prime ideal.

Answer 3

The norm of $\mathfrak{p}$ is \begin{equation*} \big\lvert \mathbb{Z}[\sqrt{-5}]\big/(2,1+\sqrt{-5})\big\rvert = 2, \end{equation*} as was observed in another answer. So if $\mathfrak{p}$ is a product of two ideals, their norms must have product $2$. Thus one of the ideals has norm $1$ so is the whole ring. Thus $\mathfrak{p}$ is prime.

Answer 4

Here is how I see it. $\mathbb{Z}[\sqrt{-5}] = \mathbb{Z}[1+\sqrt{-5}]$. Let $1+\sqrt{-5} = \alpha$. Then $(\mathbb{Z}.1 \oplus \mathbb{Z}.\alpha)/(2\mathbb{Z} \oplus \mathbb{Z}.\alpha) \equiv \mathbb{Z}/\mathbb{2Z}$, since $\mathbb{Z}/\mathbb{2Z}$ is a field, our ideal is maximal and hence prime.