Let $\mathbb F_2$ be the field of two elements. Consider the factor ring $$R=\mathbb F_2[x, y]/\langle x^2, y^2\rangle.$$ I want to find all ideals of $R$.
Note that $R=\{a_0+a_1x+a_2y+a_3xy\;:\;a_0, a_1, a_2, a_3\in\mathbb F_2\}$. Hence, the ring $R$ has order 16. So, there are ideals of orders 1, 2, 4, 8 and 16. By brute force method I find the following:
a) Ideals of order 1: There is only one $I_1=\{0\}$.
b) Ideals of order 2: There is only one $I_2=\{0, xy\}$.
c) Ideals of order 4:
$I_3=\langle x\rangle=\{0, x, xy, x+xy\}$;
$I_4=\langle y\rangle=\{0, y, xy, y+xy\}$;
$I_5=\langle x+y\rangle=\{0, x+y, xy, x+y+xy\}$
d) Ideals of order 8: There is only one (This is because the set of all zero divisors is an ideal) $I_6=\{0, x, y, x+y, x+xy, y+xy, xy, x+y+xy\}$
e) Ideals of order 16: $I_7=R$
My problem is in the ideals of order 4. There are only three ideals of order 4? how to prove that there are no others?
$R$ is generated as a ring by any $2$ non-zero elements of the additive subgroup, $A$ say, with elements $\{0, x, y, x + y\}$. If $I$ is an ideal of $R$ with $4$ elements the quotient ring $R/I$ must be isomorphic to the ring $S = \mathbb{F}_2[z]/\langle z^2 \rangle$. The image of $A$ in $S$ generates $S$ and does not contain $1$ (since the elements of $A$ are all zero-divisors in $R$). It follows that $I$ must meet $A$ in a subgroup of order $2$ and must be generated by that subgroup: there are just $3$ such subgroups and hence $R$ has at most $3$ ideals of order $4$.