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!
 A: $Z_3[x]/(x^2-1)\cong Z_3[x]/(x-1)\times Z_3[x]/(x+1)\cong Z_3\times Z_3$.
The map is induced by $ax+b\rightarrow (a+b, -a+b)$
A: In $R:=(\mathbb Z/3\mathbb Z)[x]$, $x-1$ and $x+1$ generate two distinct maximal ideals $\mathfrak m_1, \mathfrak m_2$. We have 
$(x^2-1)R=\mathfrak m_1 \mathfrak m_2$. By Chinese Remainder Theorem, we have a canonical isomorphism 
$$ R/(x^2-1)R \to R/\mathfrak m_1\times R/\mathfrak m_1.$$ 
Now $R/\mathfrak m_i$ is isomorphic to $\mathbb Z/3\mathbb Z$. So your rings are isomorphic. 
There are exatly two isomorphisms. One is given as above, the other one is obtained by composition this isomorphism with the permutation in the second ring. (Hints: the only automorphism  of $\mathbb Z/3\mathbb Z$ is the identity). 
A: *

*A homomorphism from a quotient $A/\theta\to B$ may be given by its lift $A\to B$ such that its kernel factors through $\theta$.

*A homomorphism from a polynomial ring $R[x_1,..,x_i,..]$ to an $R$-algebra is unquely determined by the image of the indeterminants $x_i$.

*$1$ must be mapped to the unit, that is $(1,1)$, and by 2., the image of $x$ can be freely chosen, and it determines all other images.

A: Since the ideals $(x-1)$ and $(x+1)$ both contain $(x^2-1)$ (because $x-1,x+1\mid x^2-1$ in $\mathbf{F}_3[x]$), there are natural surjective homomorphisms $\pi_1:\mathbf{F}_3[x]/(x^2-1)\rightarrow\mathbf{F}_3[x]/(x-1)$ and $\pi_2:\mathbf{F}_3[x]/(x^2-1)\rightarrow\mathbf{F}_3[x]/(x+1)$. Explicitly $\pi_1(f+(x^2-1))=f+(x-1)$ and $\pi_2(f+(x^2-1))=f+(x+1)$. This gives a homomorphism
$\pi:\mathbf{F}_3[x]/(x^2-1)\rightarrow\mathbf{F}_3[x]/(x-1)\times\mathbf{F}_3[x]/(x+1)$.
Note that $f+(x^2-1)$ is in the kernel of this map if and only if both $x-1$ and $x+1$ divide $f$, and since the polynomials $x-1,x+1$ are relatively prime, their product $x^2-1$ divides $f$, so $f\in (x^2-1)$, and thus $f+(x^2-1)=0$ in $\mathbf{F}_3[x]/(x^2-1)$. Thus $\pi$ is injective. There are a couple ways to see that $\pi$ is surjective, and which one you prefer depends I guess on what you know. For example, since both the source and the target of $\pi$ are $\mathbf{F}_3$-vector spaces of dimension $2$, and the map is $\mathbf{F}_3$-linear, it has to be an isomorphism. Or, since the ideals $(x-1)$ and $(x+1)$ together generate the unit ideal $\mathbf{F}_3[x]$, since the ideal they generate contains $(x+1)-(x-1)=1+1=2$, a unit in $\mathbf{F}_3$, you can invoke the Chinese remainder theorem (which will actually tell you injectivity of the above map as well as surjectivity). Or you can try to directly find, for given $f_1,f_2\in\mathbf{F}_3[x]$, a polynomial $f$ such that $f-f_1\in(x-1)$ and $f-f_2\in(x+1)$. Then $f+(x^2-1)$ will map to $(f_1+(x-1),f_2+(x+1))$. 
In any case, the map is an isomorphism.
