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The statement is:

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=\{A\mid A\text{ is a proper ideal s.t. } I\subset A\} $, 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 $I$?

Nothing special
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annimal
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    Actually, you must define P by subset of set of $proper$ ideals, not just ideals. Your current definition always gives the whole ring as the maximal element. In the proof, the fact that the upper bound does not contain 1 plays a crucial role. – Scream Dec 25 '14 at 23:29
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    Where is commutativity of the ring being used? – R_D Dec 04 '17 at 06:30

3 Answers3

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Having read this (which is now technically correct), it makes sense to fill in some missing details, which could help other readers.

To be completely correct, one needs:

  1. Assume $I \neq R$
  2. Define $P=\{I \subset A : A \text{ is a proper ideal in } R\}$

Assumption 1 is necessary because the result is not true if $I=R$. A maximal ideal is by definition a proper subset of the ring.

For $P$ in assumption 2 above to exist, $P$ must contain at least one element. This follows #1 because $I$ is contained $P$.

To prove that $P$ contains a maximal ideal (which would clearly contain $I$) requires invoking the Zorn's lemma. Since set inclusion is a partial ordering on any collection of sets, we use the set inclusion predicate as the partial ordering on $P$.

To apply Zorn's lemma we take an arbitrary chain $C$ in $P$ and show it has an upper bound in $P$. Define $U_C = \cup C$ (which exists by the axiom of union of set theory). You can show that $U_C$ is an ideal and that it contains $I$. This takes a little bit of work, but it is routine. To show that $U_C$ is in $P$, one notices that $1 \neq U_C$ because none of the elements in $C$ contain $1$ (else they would not be proper ideals in $R$ and would not belong to $P$). Finally, it obvious that $U_C$ is an upper bound for the chain $C$.

Since we have shown that every chain $C$ in $P$ has an upper bound in $P$, Zorn's lemma states that $P$ has a maximal element. The maximal element in in $P$. Hence, by the definition of $P$, it contains $I$.

One more note, the ring $R$ must have unity. The unity requirement is essential to show that $U_C$ is an ideal not equal to $R$. Without assuming $1 \in R$, one CANNOT show that $U_C$ is a proper ideal in $P$.

Pablo A Perez-Fernandez
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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 $I$.

Kez
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Since $M\in P$ hence $I\subset M$

usermath
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