Find all units, zero divisors, nilpotent elements and idempotent elements in $\Bbb Z[x]/(x^2-1)$ I have no clue on how to do unit elements (inversible), zero divisors (might not exist),  I know only for stuff like $\Bbb Z_n$, and I guessed that nilpotents you get from $(ax+b)^n=0$ and we have $a=0, b=0$ so only $0$ is nilpotent (like $b^n=0$ so b=0 and $a^nx^n=0$ so $a=0$ as well) and for idempotents, you just solve $(ax+b)^2=ax+b$ and you find $a,b$ which have to be something like $0$ and $1$?
Is this true for any case, like what has my function, $x^2-1$ have to do with any of the nilpotent/idempotent stuff?
Maybe I need like $x=\pm 1$, from the function and I put in the equation for idempotents that $x^2=1$? So this is literally the only information I have in all seminars, tutorials etc. I can't find anything related to this stuff, also I'm very bad at math, simple stuff is good.
 A: You just need to do the same, but instead of $(ax+b)^n$ is zero, you have $(ax+b)^n$ is a multiple of $x^2-1$ (that's an example for nilpotent elements). A polynomial $f$ is a multiple of $x^2-1$ iff $f(1)=f(-1)=0$. Using that, you can easily observe that we have no non-zero nilpotents: $(a+b)^n=0$ and $(-a+b)^n=0$ means $a=b=0$.
For other cases, you'll get similar equations, but you must remember that $x^2=1$ in your ring, so all elements can be reduced to linear functions. For example, in this ring
$$ (ax+b)^2 = a^2x^2 + 2abx + b^2 = 2abx + a^2+b^2, $$
so your idempotents have to satisfy $a=2ab$ and $b=a^2+b^2$. I leave details of the rest of the problem for you.
A: When you make a quotient ring out of a ring $R$ and one of its ideals $I$, the elements present in that quotient are subsets of $R$ of the form $a + I := \{ a + b \colon b \in I \}$, where $a \in R$. It forms a ring under the operations $(a + I) + (b + I) := ((a+b) + I)$ and $(a + I)(b + I):= (ab +I)$.
In this ring, it is easy to check that its zero (the addition identity) is $0 + I$. So, what you're looking for in that ring are (in the nilpotent case) elements $(a +I) \in R/I$ with $(a + I)^n = (a^n +I) = (0 + I)$.
This in turn is equivalent to stating that $a^n \in I$. Now you can check from the definition of an ideal that in this case $I := (x^2 -1) = \{ g(x)(x^2-1) \colon g(x)\in \mathbb{Z}[x] \}$.
So for an element of $R/I$ to be nilpotent, it must be of the form $q(x) + I$ with $q(x)^n = g(x)(x^2 -1)$ for some $n \in \mathbb{N}$ and some $g(x) \in \mathbb{Z}$.
To give you an example, $((x-1) + I)((x+1) + I) = ((x^2 - 1) + I) = (0 + I)$ since $x^2 -1 \in I$. Note also that $(x-1), (x+1) \notin I$, so $((x-1) + I), ((x+1) + I) \neq (0 + I)$, that is, they are not zero in the ring $\mathbb{Z}[x]/(x^2-1)$
Hope this helps :)
