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Let $R = \mathbb{F}_3[x]$ and let $I$ be the set $\{ (x^2+1)p(x)|p(x) \in R \}$ which is an ideal of R. By showing that $R/I$ is an integral domain, deduce that $R/I$ is a field.

I know that $R/I$ is commutative and that it's finite. I am having trouble proving that it has no zero-divisors.

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If $q(x),r(x)\in\mathbb F_3[x]$ are such that $[q(x]\times[r(x)]=0$, then $q(x)r(x)=(x^2+1)p(x)$ for some $p(x)\in\mathbb F_3[x]$. So, $x^2+1\mid q(x)r(x)$. But $x^2+1$ is irreducible in $\mathbb F_3[x]$, and therefore, since it divides the producte $q(x)r(x)$, it divides one of its factors. But this means the $[q(x)]$ or $[r(x)]$ is equal to $0$ in your ring.

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