# if an ideal is contained in union of two ideals then it is wholly contained in one of them.

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.

• This is not true for the union of three ideals. Dec 11, 2017 at 2:47
• @Improve: Can you please provide an example? Dec 7, 2020 at 8:14
• @Saikat Sure. Let $R = \mathbb{F}_2 \times \mathbb{F}_2$. Then $R = \mathbb{F}_2 \cdot (0,1) \cup \mathbb{F}_2\cdot (1,1) \cup \mathbb{F}_2 \cdot (1,0)$ Mar 29, 2021 at 13:11

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.