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Let I be a proper ideal of a commutative ring R with 1. Prove that I is a maximal 3. (10 ideal of R if and only if for every

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Answer #1

I is a proper ideal of R and let ag I

Then I(a) is also an ideal of R such that I\subsetneq I+\left \langle a \right \rangle\subseteq R which means we must have I(a) R because I is maximal

So if I is a proper maximal ideal, then for every ag I we must have I(a) R

On the other hand if for proper ideal I, and for every ag I we have I(a) R , we will show that I must be maximal

Suppose it is not maximal. Then there exists some ideal J such that うrう1

Let EJ- (such an element must exist as I\subsetneq J)

Then I(r) must equal R

But we also have ГСЛ (т)С so that R=I+\left \langle x \right \rangle\subseteq J

And so we must have J=R which contradicts our assumption that うrう1

Thus, our assumption that I is not maximal is incorrect

So that I must be maximal

Therefore, for a proper ideal, I is maximal if and only if for every ag I we have I(a) R

\blacksquare

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