I want to find/describe all the ideals in the quotient ring $A:=\mathbb{Z}[x]/(x^2-1)$. I know from the correspondence theorem that the ideals are the images under quotient map of ideals of $\mathbb{Z}[x]$ that contain $(x^2-1)$. Since $x^2-1\in (x\pm 1)$ we can say that these are two ideals which contain this. There are of course more ideals like $(2,x\pm 1)$ ... . I have little idea on how to proceed with such computations. Any help will be appreciated.
Thanks.
By the correspondence theorem, it suffices to analyze ideals $I$ of $\Bbb Z[x]$ containing $(x^2-1)$. Write such an ideal as $I=(x^2-1,f_1,\cdots,f_m)$ (if there are no $f_i$, then $I=(x^2-1)$). Then up to subtracting multiples of $x^2-1$ from the $f_i$ and flipping signs if necessary, we may assume that all the $f_i$ are either constant or linear with positive leading coefficient. Now we can break in to two cases: write $I\cap\Bbb Z=(n)$ for $n\geq 0$, and either $n=0$ or $n>0$.
When $n=0$, then it must be the case that all the $f_i$ are either all multiples of $x-1$ or all multiples of $x+1$: if there is some linear form $ax+b$ with $|a|\neq |b|$, then $(ax+b)(ax-b)=a^2(x^2-1)+(a^2-b^2)$ gives rise to a nonzero element of $I\cap\Bbb Z$, in contradiction to our assumption. By taking the smallest positive multiple of $x\pm1$, we can write $I=(x^2-1,ax\pm a)$ for some $a\geq 0$. When $a=1$, we get $(x-1)$ and $(x+1)$.
When $n\neq 0$, things get a little more interesting. If all the coefficients of the linear $f_i$ are multiples of $n$, then $I=(x^2-1,n)$. When this doesn't occur, then we may write $I=(x^2-1,ax+b,n)$ for $a,n>0$ and $a\mid n$. But not every choice of $a,b,n$ satisfies $(x^2-1,ax+b,n)\cap\Bbb Z=(n)$: given any $p(ax+b)+qn\in I$ and any $cx+d\in \Bbb Z[x]$, we need to make sure that if $(p(ax+b)+qn)(cx+d)$ has no $x$ term, then it has remainder divisible by $n$ after division by $x^2-1$. In order for this product to have no $x$ term, we must have $(c,d)=\frac1r(pa,-pb-qn)$ for some $r$ dividing both $pa$ and $pb+qn$, which gives us $$ \frac{p^2a^2}{r}(x^2-1) + \frac{p^2a^2-(pb+qn)^2}{r}. $$ So $n$ must divide $\frac{p^2a^2-(pb+qn)^2}{r}$ for any choice of $p,q,r$. It's possible that there's a nice simplification of this condition based on just $a$ and $b$, but I'm not seeing it yet and I figured it would be better to post this and get suggestions on how to finish the last bit rather than sit on it.
In summary: the ideals of $\Bbb Z[x]/(x^2-1)$ are of the form $0$; $(n)$ for $n\in\Bbb Z_{>0}$; $(ax\pm a)$ for $a>0$; and $(ax+b,n)$ with $a,n>0$, $a\mid n$, and subject to the above condition.
