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$R$ 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.)

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    $\begingroup$ See e.g. the following post : it is the maximal element in a chain of inclusions. $\endgroup$
    – Mauro ALLEGRANZA
    Commented Dec 21, 2018 at 10:55
  • $\begingroup$ See the Poset formed by the set of subsets of a given set (its power set) ordered by inclusion. $\endgroup$
    – Mauro ALLEGRANZA
    Commented Dec 21, 2018 at 10:59
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    $\begingroup$ If there did not exist such an $I_2$, then $I_1$ would be maximal in $\mathcal{S}$ with respect to subset inclusion, thus contradicting the supposistion. $\endgroup$
    – Adam Higgins
    Commented Dec 21, 2018 at 10:59
  • $\begingroup$ @MauroALLEGRANZA Can we say, equivalently, that If $R$ is Noetherian ring and if the set $\mathcal{S}$ contains ideals $I_1,I_2,I_3,\cdots$ under inclusion (that is to say that $I_1 \subset I_2 \subset I_3\subset \cdots$) then we have to prove that the maximal element exist for this inclusion? (I guess the point is that maximal element makes sense under some "orderedness" which "inclusion" provides here) $\endgroup$
    – MUH
    Commented Dec 21, 2018 at 12:04

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What does the phrase "maximal element under inclusion" mean?

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 $\subseteq$.

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.

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.

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