This is what I want to prove:
Every family $\mathcal{C}$ of ideals in $R$ has a maximal element in $\mathcal{C}$ implies that every ideal in $R$ is finitely generated.
I found this answer here Every subset of ideals of $R$ has maximal element $\implies$ Every ideal of $R$ is finitely generated, if $R$ is Noetherian.
Proof. We take an ideal $I$ of $R$ and we define the set
$$F:=\{J: J\text{ finitely generated ideal of } R \text{ and } J\subseteq I\}.$$
We can see that $F\neq \emptyset$ (because $\langle 0_R \rangle \in F$). So, from hypothesis, there exists a maximal element $M\in F$.
We want to show that $I=M$.
Obviously, $M\subseteq I$, because $M\in F$.
We suppose now that $M\subsetneq I$. Then, there exists an element $a\in I\backslash M\iff a\in I$, with $a\notin M$.
We consider the ideal
$$J:=M+\langle a\rangle.$$
Then, if $M=\langle m_1,…,m_n \rangle$, we have $J=M+\langle a\rangle=\langle m_1,…, m_n,a \rangle \implies J\in F$
But I am not fully understanding the idea of the solution. specifically,
1- Why we took $J\subseteq I$ in the set we defined? i.e. why we need all the f.g. ideals to be inside our arbitrary ideal $I$? Are we going to use Zorn's Lemma in our proof?
2- why we need to show that $M= I$?
3- why we are considering the case $M \subsetneq I$ in the proof?
4- Why $M+\langle a\rangle=\langle m_1,…, m_n,a \rangle$?
could anyone explain this to me please?
Zorn's Lemma which states that:
A partially ordered set in which every chain (i.e.totally ordered subset) is upper bounded has a maximal element.
Best Answer
You are not using Zorn's Lemma. In fact, no choice is necessary for this proof, since we already start by assuming that a maximal element exists, and choice is necessary exactly when trying to prove the existence of maximal elements.
The idea behind the proof is simple, but it might be clearer if we prove the following theorem of Tarski:
Proof. One direction is trivial, if $A$ is finite. In the other direction, let $J$ be the set of finite subsets of $A$, and let $M$ be a maximal element in $J$. If $M\neq A$, then there is some $a\in A\setminus M$, and then $M\cup\{a\}$ is a finite subset of $A$ which is a proper superset of $M$. But $M$ was maximal, so no such $a$ can exist, so $A=M$. $\square$
The idea is to look at the set of all finitely generated subobjects, and are that if the maximal, which exists by assumption, is not equal to the whole object, then we derive a contradiction.