Maximal ideals in $\mathbb{Z}[\sqrt{-5}]$

Question

This is problem 8.3.8 in D&F

Let R be the quadratic integer ring $Z[\sqrt{-5}]$ and define the ideals $I_2 = (2, 1 + \sqrt{-5})$, $I_3 = (3, 2 + \sqrt{-5})$, and $I_3' = (3, 2 - \sqrt{-5})$.

Prove that $I_2$, $I_3$ and $I_3'$ are prime ideals in $R$. [One approach: for $I_3$ observe that $R/I_3 \cong (R/(3))/(I_3/(3))$ by the Third Isomorphism Theorem for Rings. Show that $R/(3)$ has $9$ elements, $(I_3/(3))$ has $3$ elements, and that $R/I_3 \cong \mathbb{Z}/3\mathbb{Z}$ as an additive abelian group. Conclude that $I_3$ is a maximal (hence prime) ideal and that $R/I_3 \cong \mathbb{Z}/3\mathbb{Z}$ as rings.]

The last line is giving me a bit of trouble. I've seen these posts (1, 2, there are several more) but the method of writing $R/I_3$ as a polynomial ring over an irreducible is unfamiliar. I assume I'm being asked to show $R/I_3 \cong \mathbb{Z}/3\mathbb{Z}$ first and then that $I_3$ is prime. I tried using the first isomorphism theorem and calculating by hand but I'm not sure how to get $3$ representatives from $9$. For instance, I believe $2 \equiv2\sqrt{-5}$ and $2 + \sqrt{-5} \equiv 1 + 2\sqrt{-5}$ but I'm also getting $1 \equiv 2$ from $\sqrt{-5} \equiv -2$, and I'm not really sure what I should be doing.

Update: I figured out that I was making calculation mistakes. You can find $1 \equiv \sqrt{-5} \pmod I$ so that, as per the hint, \begin{gather*} 0 \equiv 1 + 2\sqrt{-5} \equiv 2 + \sqrt{-5} \pmod{I} \\ 1 \equiv \sqrt{-5} \equiv 2 + 2\sqrt{-5} \pmod{I} \\ 2 \equiv 1 + \sqrt{-5} \equiv 2\sqrt{-5} \pmod{I} \end{gather*} gives all three elements in $R/I_3$ which has the same multiplicative structure as $\mathbb{Z}/3\mathbb{Z}$.

Answer 1

Let $I=(3,2+\sqrt{-5})$ and $R=\Bbb Z[\sqrt{-5}]$. Note that every element of $R$ is of the form $m+n\sqrt{-5}$, where $m,n\in\Bbb Z$.

Also notice that $I$ contains $-3$ and therefore $-3+2+\sqrt{-5}=\sqrt{-5}-1$. When we quotient by an ideal containing $\sqrt{-5}-1$, we are really applying the relation $\sqrt{-5}-1=0$, or $\sqrt{-5}=1$. Following this, in the quotient $R/I$, we can write $m+n\sqrt{-5}=m+n\pmod{I}$. If you want to say this more precisely, we have \begin{align} m+n\sqrt{-5}&\equiv m+n\sqrt{-5}-n(\sqrt{-5}-1)\pmod{I}\\ &\equiv m+n\pmod{I} \end{align} since $\sqrt{-5}-1\in I$.

This suggests we can define a map $R/I\to \Bbb Z/3$ by taking $m+n\sqrt{-5}$ to $m+n\pmod 3$. Let $f\colon R\to\Bbb Z/3$ be the map $f(m+n\sqrt{-5})=m+n\pmod 3$. Verify that $f$ is a homomorphism satisfying $f(3)=0$ and $f(2+\sqrt{-5})=0$. It follows that $f$ descends to a map $\overline f\colon R/I\to\Bbb Z/3$. It is easy to see that this map is surjective. To show $\overline f$ is injective, suppose $m+n\sqrt{-5}\in\ker \overline f$. Then $\overline f(m+n\sqrt{-5})=m+n\equiv 0\pmod 3$, so $3|(m+n)$. That means $m+n=3k$ for some integer $k$, whence \begin{align} m+n\sqrt{-5}&= m+(3k-m)\sqrt{-5}\\ &\equiv m+(3k-m)\sqrt{-5}-(3k-m)(\sqrt{-5}-1)\pmod{I}\\ &\equiv m+3k-m\pmod{I}\\ &\equiv 3k\pmod{I}\\ &\equiv 0\pmod{I}. \end{align}

Make sure you understand why each of these statements is true, and conclude that $\overline f$ is injective.

Returning to the original problem, notice that, since $\Bbb Z/3$ is a field, $I$ is a maximal ideal, and therefore prime.