$R$ is Noetherian ring $\iff$ Every nonempty set of ideals of $R$ contains a maximal element under inclusion.
What does the phrase "maximal element $\textbf{under inclusion}$" mean? I am having a hard time understanding the context of "under inclusion".
For e.g. assume that we are given any increasing chain of ideals $I_1\subset I_2\subset \cdots$. Since, every set (say $\mathcal{S}$) of ideals of $R$ contains a maximal element under inclusion, meaning that if $\mathcal{S}=\{I_1\}$, then there exist maximal ideal $M$ which satisfy $I_1 \hookrightarrow M$. Similarly, if $\mathcal{S}=\{I_1,I_2\}$, then there exist maximal ideal $M$ which satisfy $I_1 \hookrightarrow M$ and $I_2 \hookrightarrow M$. From here, it is easy to prove that $R$ is Noetherian.
Conversely, Assume that $R$ is Noetherian ring. Let $\mathcal{S}$ be any non-empty set of ideals of $R$ with no
maximal element. Since, $\mathcal{S} \neq \emptyset$, let $I_1$ be in $\mathcal{S}$.
Now, I have seen in some of the references that "Because $I_1$ is not maximal, we can choose $I_2$ in $\mathcal{S}$ with $I_1 \subset I_2$ and $I_1 \neq I_2$."
Doubt: Why does there exist such $I_2$ in the above statement. (Or you can give some other explanation. I guess this is where "under inclusion" comes to picture.)
Best Answer
The phrase "maximal element" has no meaning unless some partial order is indicated.
That's what "under inclusion" means: the partial order is the set containment relation $\subseteq$.
Because that is what "not maximal" means. If no such $I_2$ existed, $I_1$ would be maximal. I would encourage you to look at the definition of "maximal" and contemplate its negation, and you will understand.