# What does “under inclusion” mean in: $R$ is Noetherian ring $\iff$ Every nonempty set of ideals of $R$ contains a maximal element under inclusion.

$$R$$ 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.)

• See e.g. the following post : it is the maximal element in a chain of inclusions. – Mauro ALLEGRANZA Dec 21 '18 at 10:55
• See the Poset formed by the set of subsets of a given set (its power set) ordered by inclusion. – Mauro ALLEGRANZA Dec 21 '18 at 10:59
• 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. – Adam Higgins Dec 21 '18 at 10:59
• @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) – MUH Dec 21 '18 at 12:04

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.