I want to find all ideal of $J=(\Bbb{Z}/4\Bbb{Z})[x]/(x^2-1)$. The number of all elements of $J$ is just 16, so my text says it is easy to find all ideal by hand.
I can understand $J$ has just 16 elements, but I cannot go further, I would be appreciated if you show me how to classify all ideal of $J$, Thank you!
On
Here is another idea: Note that $J$ is isomorphic to
$$ R = \left\{ \begin{pmatrix} a & b \\ b & a \end{pmatrix} : a, b \in \mathbb{Z}/4\mathbb{Z} \right\} $$
via the correspondence $a+bx \leftrightarrow \begin{pmatrix} a & b \\ b & a \end{pmatrix}$. From this, we observe:
As an element of $R$ via the above correspondence, the determinant of $a+bx$ is $a^2-b^2$. So, $a+bx$ is a unit of $J$ if and only if $a$ and $b$ have different parities, i.e., $a+b$ is odd.
The remaining 8 non-invertible elements may be partitioned into 5 different classes according to whether they only differ by multiplication by units: \begin{gather*} \{0\}, \quad A := \{ 2, 2x \}, \quad B := \{1+x, 3+3x\}, \\ C := \{1+3x, 3+x\}, \quad D := \{2+2x\}. \end{gather*}
Now using the fact that
$$ A+B = B+C = C+A, \qquad AB = BC = CA = D, $$
we have total 7 different ideals:
Notation Denote $z=ax+b\in J$ for $a,b\in\{0,1,2,3\}$, $J^+$ as the additional group on $J$, $[z_1,\dots,z_n]$ and $\langle z_1,\dots,z_n\rangle$ as the subgroup of $J^+$ and the ideal of $J$ generated by $z_1,\dots,z_n\in J$ respectively.
First comes the conclusion, the $7$ ideals: $$\{0\},I_0=\langle 2x+2\rangle=[2x+2],I_1=\langle 2\rangle=[2],I_2=\langle x+1\rangle=[x+1],I_3=\langle x+3\rangle=[x+3],I_4=I_1\bigcup I_2\bigcup I_3,J$$are all ideals in $J$ that we are looking for.
To find it, we can start with ideals like $\langle z\rangle,z\in J$. We know $$\langle z\rangle=\{jz\mid j=ax+b\in J\}=[z,xz]$$and $x^2=1$ in $J$, so, for example, $\langle x+2\rangle=J$. This is because $(x+2)$ and $x(x+2)$ is linear indepedent entries of order $4$ in $J^+$ and then the group generated by them are a subgroup with $16$ entries in $J^+$, which is undoutedly $J^+$ or $J$ itself. In other words, $\langle x+2\rangle=[x+2,2x+1]=J$. $\langle z\rangle=J$ holds for $z=1,3,x,x+2,2x+1,2x+3,3x,3x+2$ according to the same reason above as well.
To use the same methods, it's not difficult to find the ideals generated by the others. Here is another example:$\langle x+3\rangle=[x+3,3x+1]=[x+3,3(x+3)]=[x+3]=\{0,x+3,2x+2,3x+1\}$.
Up to now, we've found, in fact, all the ideals in $J$ but $I_4$ mentioned above, which is generated by more than one element.
Since $\langle z_1,z_2\rangle$ is trivial if $\langle z_1\rangle=J$ or $\langle z_2\rangle=J$, we assume $z_1,z_2\in I_1\bigcup I_2\bigcup I_3$($I_0$ is included in any of them). Noticed that $(2x)+(x+1)+(x+3)=0$, so any two of these three generate $I_4$, and further any two of $I_i-I_0,i=1,2,3$ generate $I_4$(why?). Since $\forall z\in J,2x+2\in\langle z\rangle$, the case for $\langle z_1,z_2\rangle$ is ended and the proof is ended, too.
What's more, if there is any other new ideal, one must satisfies $I_4<I'=\langle z_1,\dots,z_n\rangle(n\ge 2)<J^+$ and then $8=\mid I_4\mid<\mid I'\mid<\mid J^+\mid =16$, which is impossible for a subgroup of $J^+$.