Question

If an ideal is contained in union of two ideals then show that it is wholly contained in one of them

Answer 1

Suppose $I,J,K$ are ideals of a ring $R$ such that $K\subseteq I\cup J$. We want to prove that $K\subseteq I$ or $K\subseteq J$.

We will suppose the contrary : suppose that $K \nsubseteq I$ nor $K \nsubseteq J$, in this case we should have $x,y\in K$ with $x\in I\setminus J$ and $y\in J\setminus I$. By definition of the ideal, we have $x+y\in K \subseteq I\cup J$, so we should have $x+y \in I$ or $x+ y \in J$ (by definition of a union). Suppose it's in $I$ without loss of generality. We then should have $(x+y)+(-x)=y\in I$ by definition of an ideal (as $x+y\in I, (-x)\in I$). This contradicts the hypothesis $y\in J\setminus I$. By contradiction, we conclude that $K$ is contained in $I$ or in $J$.