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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.

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    $\begingroup$ $\Bbb{Z}[a]/a\Bbb{Z}[a] = \Bbb{Z}/(a\Bbb{Z}[a] \cap \Bbb{Z})$ $\endgroup$
    – reuns
    Commented Mar 20, 2023 at 1:52

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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.

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.

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    $\begingroup$ +1 Nice answer. Alternatively, one can first prove that $\mathbb{Z}[\sqrt{-5}]$ is isomorphic to $\mathbb{Z}[X]/\langle X^{2}+5 \rangle$ (assuming that this is not already your definition of $\mathbb{Z}[\sqrt{-5}]$) by showing that the ring morphism $\mathbb{Z}[X] \to \mathbb{Z}[\sqrt{-5}]$ sending $X$ to $\sqrt{-5}$ has kernel $\langle X^{2}+5 \rangle$. Then observe that this isomorphism descends to an isomorphism $\mathbb{Z}[\sqrt{-5}]/\langle \sqrt{-5} \rangle \cong \mathbb{Z}[X]/\langle X, X^{2}+5 \rangle$, and prove that the latter is isomorphic to $\mathbb{Z}/5\mathbb{Z}$. $\endgroup$
    – Alex Wertheim
    Commented Mar 20, 2023 at 5:56

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