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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?

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    $\begingroup$ Hint: 1) if a polynomial is not irreducible, it must be a product of lower degree factors. 2) a polynomial has a degree one factor over a field $K$ if and only if it has a zero in the field $K$. $\endgroup$ – Jyrki Lahtonen Aug 10 '18 at 4:24
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    $\begingroup$ Also, what do you mean by $\Bbb{Z}_3[i]$? $\endgroup$ – Jyrki Lahtonen Aug 10 '18 at 4:29
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    $\begingroup$ Once you figure out what exactly you are asking, then 1, 2 or 3 is likely to contain an answer. $\endgroup$ – Jyrki Lahtonen Aug 10 '18 at 4:31
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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.

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