# Is $(x^3-x^2+2x-1)$ prime in $\mathbb{Z}/(3)[x]$?

This is somewhat of a follow up on this question: Why is $(3,x^3-x^2+2x-1)$ not principal in $\mathbb{Z}[x]$?

I'm curious, is $\mathbb{Z}[x]/I$ a domain, with $I=(3,x^3-x^2+2x-1)$? I know $I$ is not principal. Also, I took the sequence of epimorphisms $$\mathbb{Z}[x]\stackrel{\varphi}{\longrightarrow}\bar{\mathbb{Z}}[x]\stackrel{\pi}{\longrightarrow}\bar{\mathbb{Z}}[x]/\bar{I}$$ where $\bar{I}=(\bar{x}^3-\bar{x}^2+\bar{2}\bar{x}-\bar{1})$ is the image of $I$ in $\bar{\mathbb{Z}}:=\mathbb{Z}/(3)$. Since the kernel of $\pi\circ\varphi=I$, I know $\mathbb{Z}[x]/I\simeq\bar{\mathbb{Z}}[x]/\bar{I}$. The latter is a ring of $27$ elements, but I don't want to go through and verify by hand that it is a domain. I know it's a domain iff $\bar{I}$ is prime, but I can't tell if it is or isn't. How can I efficiently tell if $\bar{\mathbb{Z}}[x]/\bar{I}$ is a domain? Thank you.

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Let $F$ be the field with $3$ elements. Since $F[x]$ is a unique factorization domain, a principal ideal $(f)$ being prime is equivalent to $f$ being irreducible. In your case $f$ is a cubic polynomial, so being irreducible is equivalent to not having any roots in $F$. Just check the three values: $f(0)=-1$, $f(1)=1$, and $f(2)=1$ (all modulo $3$). Thus $f$ is irreducible and the ideal it generates is prime.
Thanks Steve ${}$. – Son Bi Jul 14 '12 at 2:25