Minimal ideal in commutative finite rings Let $R$ be a commutative finite ring with identity, and let $I$ be a minimal ideal of $R$, that is, a non-zero ideal that there is no ideal strictly between $I$ and $0$. Now let $\{I_i\}_{i\in A}$ be a family of ideals of $R$ such that $I\subseteq \sum_{i\in A} I_i$. How can we show that there exists $j\in A$ such that $I\subseteq I_j$? If the statment is not true is there any condition under which it is true?
 A: This is relatively easy to prove if $R$ is an integral domain. Without loss of generality, we may suppose that $I_j \neq \{0\}$ for all $j\in A$.
Suppose by way of contradiction that $I\not\subseteq I_j$ for any $j\in A$, then by minimality of $I$ it holds that
$$
I\cap I_j = \{0\}, \qquad \forall j\in A
$$
Now, fix $j\in A$ and $r\in I_j\setminus\{0\}$ and note that $rI$ is an ideal contained both in $I$ and in $I_j$. To see that $rI\subseteq I_j$, note that
$$
rI = \{ri \mid i\in I\}= \bigcup_{i\in I} \{r i\} \subseteq \bigcup_{i\in I} I_j i \subseteq I_j
$$
But then $rI \subseteq I\cap I_j = \{0\}$. In particular, $ri = 0$ for all $i\in I$ which contradicts the assumption that $R$ is an integral domain.
Note that this proof actually shows that $I\subseteq I_j$ for all $j\in A$.
A: It is not possible to show in general: there are counter examples. For example, from this recent question, you can adapt the counterexample given to be $F[x,y]/(x,y)^2$ where $F$ is the field of two elements.
If this property held for a particular minimal right ideal $T$, then there could not be any other copies of $T$ in $R$, because you can construct a third copy of $T$ such that the original $T$ is contained in the sum of the second two, but is not equal to them.  So a necessary condition would be for the minimal ideal to be nonisomorphic with all other minimal ideals.
I think it may also be sufficient, but haven’t sat down to think it through on paper yet.
