Is $\Bbb Z_3 [x] / \langle x^2 - 1 \rangle \cong \Bbb Z_3 \times \Bbb Z_3$? I have paired off the zero divisors in these two rings, and there are exactly 4 pairs of distinct zero divisors in both rings: 
In $\Bbb Z_3[x]/\langle x^2 - 1\rangle$ the zero divisor pairs are $$x+1, x+2 \qquad x+1, 2x+1, \qquad 2x+1, 2x+2 \qquad x+2, 2x+2$$
Similarly, in $\Bbb Z_3 \times \Bbb Z_3$ the zero divisor pairs are $$(0, 1), (1, 0) \qquad (0,1), (2,0) \qquad (0, 2), (1, 0) \qquad (0, 2), (2, 0) $$ The fact that the zero divisors match up indicates to me that these rings should indeed be isomorphic, as zero divisors are one of the structures which have failed checks in the past on rings such as $\Bbb Z_2 [x] / \langle x^2 \rangle  \ncong \Bbb Z_2 \times \Bbb Z_2$. 
My Question
Is the only homomorphism (and thus isomorphism) between these two rings an enumerative one, where I explicitly map each of the nine elements of $\Bbb Z_3[x]/ \langle x^2 - 1\rangle$ to an element in $\Bbb Z_3 \times \Bbb Z_3$? I can do that if necessary, but it seems an inelegant and brute force method. However, since these rings are so small, it may be the best way to go about it. Thanks!

$Z_3[x]/(x^2-1)\cong Z_3[x]/(x-1)\times Z_3[x]/(x+1)\cong Z_3\times Z_3$.
The map is induced by $ax+b\rightarrow (a+b, -a+b)$
