In a commutative ring with 1, every proper ideal is contained in a maximal ideal. and we prove it u

Yahir Tucker 2022-06-26 Answered
In a commutative ring with 1, every proper ideal is contained in a maximal ideal.
and we prove it using Zorn's lemma, that is, I is an ideal, P = { I A A  is an ideal }, then by set inclusion, every totally ordered subset has a bound, then P has a maximal element M.
My question is why M must contain I?
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Answers (2)

iceniessyoy
Answered 2022-06-27 Author has 27 answers
Because M is a maximal element of P, it is in particular an element of P, or in symbols, M P.
By definition, P is the collection of ideals that contain I.
Therefore, M contains I.

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cazinskup3
Answered 2022-06-28 Author has 6 answers
Since M P hence I M

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