 # 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, , 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 $I$?
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it iceniessyoy
Because $M$ is a maximal element of $P$, it is in particular an element of $P$, or in symbols, $M\in P$.
By definition, $P$ is the collection of ideals that contain $I$.
Therefore, $M$ contains $I$.

We have step-by-step solutions for your answer! cazinskup3
Since $M\in P$ hence $I\subset M$

We have step-by-step solutions for your answer!