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