Showing that $\pi(x)$ is not a prime in the ring $R=\Bbb Z[x]/I$ where $I$ is the ideal generated by $x^2+10$.

Question

Consider the quotient ring $R=\Bbb Z[x]/I$ where $I$ is the ideal generated by $x^2+10$. Let $\pi:\Bbb Z[x]\to R$ denote the natural projection and consider the element $\pi(x)\in R$. I am trying to show that $\pi(x)$ is irreducible but not prime in $R$. To show that $\pi(x)$ is not a prime in $R$, we have to find $f,g\in \Bbb Z[x]$ such that $\pi(x)$ divides $\pi(f)\pi(g)=\pi(fg)$ but does not divide both $\pi(f)$ and $\pi(g)$, but I can't find such $f$ and $g$. Any hints?

Answer 1

For easier notation, I'll denote $\pi(x) = \sqrt{-10}$. This is justified as $0 = \pi(x^2 + 10) = \pi(x)^2 + 10$. We therefore have $\sqrt{-10} \mid 10 = 2 * 5$. We therefore have to show that $\sqrt{-10} \nmid 2$ and $\sqrt{-10} \nmid 5$. The approach I'll use is to define a norm on $R$. Indeed, note that everything in $R$ can be uniquely written as $a + b \sqrt{-10}$ for $a, b \in \mathbb Z$. I define $N(a + b \sqrt{-10}) = (a + b \sqrt{-10})(a - b \sqrt{-10}) = a^2 + 10 b^2$. This is a map $R \longrightarrow \mathbb N$. Furthermore check that $N(\alpha \beta) = N(\alpha) N(\beta)$. Then if $\alpha \mid \beta$ in $R$, $N(\alpha) \mid N(\beta)$ in $\mathbb N$. Observe $N(\sqrt{-10}) = 10$, $N(2) = 4$, $N(5) = 25$. Hence, $\sqrt{-10}$ cannot divide $2$ nor $5$ in $R$ but it divides $10 = 2*5$ and is therefore not prime.