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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.

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

1 Answer 1

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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.

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    $\begingroup$ One thing that has puzzled me for a while: can this fact be proved constructively, i.e. not by contradiction? $\endgroup$
    – Zhen Lin
    Sep 19, 2013 at 11:29

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