Prove that $\sqrt{-5}$ is a prime in the ring $R=ℤ[\sqrt{-5}]$.

Question

If $R=ℤ[\sqrt{-5}]$ is a ring but not a UFD, prove that the irreducible element $\sqrt{-5}$ is a prime.

This is what I have so far.

Proof: Let $R=ℤ[\sqrt{-5}]$ be a ring but not a UFD. Since $\sqrt{-5}$ is a irreducible element then we can write $\sqrt{-5}=bc$ for some $b$, $c$ in $R$ where $b$ or $c$ is a unit that does not equal $0$.

I'm not sure where to go from here, any help would greatly be appreciated.

Answer 1

This is what you need to show in order to prove that $\sqrt{-5}$ is a prime in $\mathbb Z[\sqrt{-5}]$:

For any $a, b \in \mathbb Z[\sqrt{-5}]$, if $\sqrt{-5}$ divides $ab$, then $\sqrt{-5}$ divides $a$ or $b$.

First of all, I suggest we work out which elements in $\mathbb Z[\sqrt{-5}]$ are divisible by $\sqrt{-5}$. So see if you can prove this:

$\sqrt{-5}$ divides the element $m + n\sqrt{-5} \in \mathbb Z[\sqrt{-5}]$ if and only if $m$ is a multiple of $5$.

This gets you most of the way there.

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Alternatively, you could aim to prove the following:

$(\sqrt{-5})$ is a prime ideal; or equivalently, $\mathbb Z[\sqrt{-5}] / (\sqrt{-5}) $ is an integral domain.

This is what @reuns is hinting at. If you want to go for this approach, then prove the following:

The ideal $(\sqrt{-5})$ consists precisely of the elements $m + n\sqrt{-5} \in \mathbb Z[\sqrt{-5}]$ where $m$ is a multiple of $5$. Therefore, $\mathbb Z[\sqrt{-5}] / (\sqrt{-5})$ is isomorphic to $\mathbb Z_5$.

Good luck.