If an ideal is contained in union of two ideals then it is wholly contained in one of them.
Let $A,B,C$ are ideals of a ring $R$ such that $C$ is contained in $A\cup B$. Then required to prove that either $C\subset A$ or $C\subset B$.
I can't understand how to do it. If $C\subset A$, then nothing to prove so assume $C \not\subset A$ then there exists some $c \in C$ such that $c \notin A$.
Now we have to prove that all $x \in C$ are in $B$. How to proceed next?
Please someone give me hint.
Best Answer
Assume by way of contradiction, that $C$ is contained neither in $A$ nor in $B$. Then there are $c_{i} \in C$ for $i = 1, 2$ such that and $c_{1} \in A \setminus B$ and $c_{2} \in B \setminus A$.
Then $c = c_{1} + c_{2} \in C \subseteq A \cup B$.
If $c \in A$, then $c_{2} = c - c_{1} \in A$, a contradiction.
If $c \in B$, then $c_{1} = c - c_{2} \in B$, a contradiction.