# the number of zero divisors in polynomial ring

I was looking for an answer on the question

How much zero divisors are in the ring $\dfrac{\mathbb{Z}_3[x]}{(x^4 + 2)}$?

when I came up with the brilliant/hack-isch idea that it might just be $81 - \phi(3^4) = 27$ where $\phi(n)$ denotes Euler's totient function. I thought of this idea because $\phi(n)$ denotes the number of elements in $\mathbb{Z}_n^*$, the unity group of $\mathbb{Z}_n$ and it might as well count in general.

The problem I'm dealing with now, is that I have no idea how I could justify my idea or why it would be correct. (I assume the equivalence with finding the number of elements in $\mathbb{Z}_n^*$ is not a valid explanation.)

I did already find a hint that my idea might be going in the right direction, but it's all so vague.

The structure of the ring $\mathbb{F}_p(x)/(q(x))$ obviously depends on how $q(x)$ splits over $\mathbb{F}_p$. If $q(x)$ is irreducible over $\mathbb{F}_p$, then the quotient ring is a field, hence the only zero divisor is zero. In our case, over $\mathbb{F}_3$: $$x^4+2 = x^4-1 = (x-1)(x+1)(x^2+1),$$ so the quotient ring is isomorphic to $E=\mathbb{F}_3\times\mathbb{F}_3\times\mathbb{F}_9$. Given that $\pi$ is the projection map from the quotient ring $R$ to $E$, we have that $\sigma\in R$ is a zero divisor iff one component of $\pi(\sigma)$ is zero, or, equivalently, if $\sigma$ is not an invertible element. In $E$ there are $2\times 2\times 8$ invertible elements. Can you finish from there?