Q:

Show that $\mathbb{Z}_3[i]$ is a field

I know that $\dfrac{\mathbb{Z}_3[x]}{\langle x^2+1\rangle}$ is isomorphic to $\mathbb{Z}_3[i]$, does this help me prove that $\mathbb{Z}_3[i]$ is a field?
$\langle x^2+1\rangle$ is the ideal generated by $x^2+1$.
I could also use that $x^2+1$ is irreducible on $\mathbb{Z}_3$, but how?

A:

The fact that $p(x)=x^2+1\in\mathbb{Z}_3[x]$ is irreducible immediately tells us that this quotient ring is a field. So you're pretty much done!
More specifically, in a commutative ring $R$, an element $p\in R$ is irreducible iff the ideal $\langle p\rangle$ is prime. But then recall that a polynomial ring $K[x]$ over a field $K$ is a PID, and in a PID every nonzero prime ideal is maximal. And the quotient of a commutative ring by a maximal ideal is a field.

