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.
Question: "Any help will be appreciated."
Answer: You may classify maximal ideals in $A:=\mathbb{Z}[x]/(x^2-1)$. A an ideal $I \subseteq \mathbb{Z}[x]$ is maximal iff $I=(p, f(x))$ where $p>0$ is a prime number and $\overline{f}(x)\in \mathbb{F}_p[x]$ is an irreducible polynomial (let $\mathbb{F}_p:=\mathbb{Z}/p\mathbb{Z}$, which is the field with $p$ elements). Assume $p$ is a prime, you get $A/(p) \cong \mathbb{F}_p[x]/(x-1)(x+1)$ and if $p=2$ you get
$$A/(2) \cong \mathbb{F}_2[x]/(x-1)^2$$
which is a local ring with one maximal ideal (the ideal $(x-1)$). If $p\neq 2$ you get
$$A/(p) \cong \mathbb{F}_p[x]/(x-1)(x+1) \cong \mathbb{F}_p \oplus \mathbb{F}_p$$
since $(x-1)+(x+1)=(1)$ is the unit ideal - the isomorphism is the chinese remainder lemma. $A/(p)$ has two maximal ideals $(x-1)$ and $(x+1)$. Hence the maximal ideals in $A$ are the ideals $(2,x-1)$ and $(p,x-1),(p,x+1)$ for $p\neq 2$ a prime number.
$$(*)\text{ }I=\mathfrak{m}_1^{l_1}\cdots \mathfrak{m}_d^{l_d}.$$
Taking products of maximal ideals give a class of nontrivial ideals $I$ in $A$.
Question: "Why do you believe ideals in $A$ are products of maximal ideals?"
Answer: If you instead considered $B:=\mathbb{Z}[x]/(x^2+1) \cong \mathbb{Z}[i] \subseteq \mathbb{Q}(i):=K$, this property holds for $\mathcal{O}_K \cong \mathbb{Z}[i]$. The ring $\mathcal{O}_K$ is a Dedekind domain, and this property holds in all such rings. You find a proof of this in Neukirch's book "Algebraic number theory".
Note: If you view the "decomposition" $Spec(A)=V(2)\cup D(2)$ you get $V(2)$ has ring of global sections $\mathbb{F}_2[x]/(x-1)^2$. $D(2)$ has ring of global sections
$$\mathbb{Z}[\frac{1}{2}][x]/(x-1)(x+1)\cong B:= \mathbb{Z}[\frac{1}{2}] \oplus \mathbb{Z}[\frac{1}{2}].$$
There is a canonical map
$$\phi: A\rightarrow A_2 \cong \mathbb{Z}[\frac{1}{2}] \oplus \mathbb{Z}[\frac{1}{2}]$$
and you must study ideals in $\mathbb{Z}[\frac{1}{2}] \oplus \mathbb{Z}[\frac{1}{2}]$ and the map $\phi$. It is easier to study ideals in a direct sum.
This last isomorphism follows from the chinese remainder lemma, since the ideals $\mathfrak{p}_1:=(x-1), \mathfrak{p}_2:=(x+1)$ are coprime. These two ideals are the minimal prime ideals in $A$. The ideals $\mathfrak{p}_1, \mathfrak{p}_2$ and the maximal ideals given above classify all prime ideals in $A$. An ideal $I\subseteq A$ has the property that $2\in I$ or $2\notin A$. If $2 \in A$ it follows $I$ is determined by its ideal $\overline{I} \subseteq A/(2)$. And the ideals in $A/(2)$ are $(0)$ and $(1+x)$, since $x^2=1$ is a unit. If $2 \notin I$ it follows $I\mathbb{Z}[ \frac{1}{2} ]$ gives all ideals in $\mathbb{Z}[ \frac{1}{2}]$. Hence the inverse image $\phi^{-1}(J)$ for any ideal $J \subseteq \mathbb{Z}[\frac{1}{2}]$ gives a class of ideals in $\mathbb{Z}$ not containing $2$.
Example: There is the following relation between ideals in $\mathbb{Z}$ and $\mathbb{Z}[\frac{1}{2}]:=S^{-1}(\mathbb{Z})$ with $S:=\{2^n\}$: An ideal $I\subseteq \mathbb{Z}$ satisfies $I^{ec}=\cup_{2^n \in S}(I:(2^n))$.
If $\phi: \mathbb{Z} \rightarrow \mathbb{Z}[\frac{1}{2}]$ is the canonical map it follows $\phi(3)=\phi(2^n3)$ for any $n$ since $2^n$ is a unit.
In general if $m:=2^np_1^{l_1}\cdots p_d^{l_d}$ with $p_i \neq 2$ it follows
the extended ideals $\phi((m))$ equal the ideal $(p_1^{l_1}\cdots p_d^{l_d}) \subseteq \mathbb{Z}[\frac{1}{2}]$. The inverse image of this ideal is
$$\phi^{-1}(\phi((m)))= (p_1^{l_1}\cdots p_d^{l_d}).$$
Hence the ideals $I$ in $\mathbb{Z}[\frac{1}{2}]$ are principal ideals on the form
$$I=(p_1^{l_1}\cdots p_d^{l_d})$$
with $p_i\neq 2$ primes and $l_i \geq 1$ integers. It may be easier to classify the ideals in the direct sum $B$. The prime ideals in $B$ are easy to classify. There is moreover a canonical injection
$$\psi: A \rightarrow A/\mathfrak{p}_1\oplus A/\mathfrak{p}_2 \subseteq \mathbb{Z} \oplus \mathbb{Z}$$
defined by $\psi(a+bx):=(a+b, a-b) \in \mathbb{Z}\oplus \mathbb{Z}$.
Hence $A$ is a subring of the direct sum $\mathbb{Z}\oplus \mathbb{Z}$, the subring $\mathbb{Z}\{u,v\}$ with $u:=(1,1), v:=(1,-1)$. Hence any ideal $I$ in $A$ is an ideal of the form
$$I:=(a_1u+b_1v, ..,a_lu+b_lv)$$
with $a_i,b_i\in \mathbb{Z}$. Any polynomial $f(x)\in \mathbb{Z}[x]$ may be written as $f(x)=f_1(x^2)+xf_2(x^2)$ and in $A$ you get
$$\overline{f(x)}=f_1(1)+xf_2(1)=a+bx$$
with $a,b\in \mathbb{Z}$. The ring $A$ is a free $\mathbb{Z}$-module of rank 2 on the generators $\{1,\overline{x} \}$. Under this correspondence the maximal ideals and prime ideals are the ideals
$$(2,x-1)=((2,2),(0,-2)), (p,x+1)=((p,p),(2,0)), (p,x-1)=((p,p),(0,-2))$$
and
$$(x+1)=(2,0), (x-1)=(0,-2).$$
