Can every set (of sets) be ordered by inclusion?

Question

Related question in question

The statement is (one of) Cohen's theorems:

Let $R$ be commutative and have $1_R$ and all its prime ideals finitely generated, then $R$ is Noetherian

The proof starts with the set

$$\Theta = \{I : I \; \text{not finitely generated}\}$$

Q1: There is a preceding lemma that say the set is non-empty, I am not sure why, but maybe the answer in the next question will answer this. i don't have this book

Q2: It says this can be ordered by inclusion. How? If I have two arbitrary $I_1,I_2$, the only possible element they might have in common is additive identity $0_R$. How can we ask if $I_1 \subset I_2$ or $I_2 \subset I_1$? The application of Zorn's Lemma is used a lot, but I don't understand how we know can apply the ordering.

It is used in the proof that all commutative ring with $1_R$ as at least 1 maximal ideal. Doesn't this mean all commutative ring satisfies the maximal condition and therefore it is Noeteria

Note $I_js$ are not prime by assumption. Or is this ordering independent of this assumption? Note I suspect the answer to (1) are any non-maximal elements.

Answer 1

Q1: Proving that statement by contradiction is common. Thus I guess one assumes that $R$ is not noetherian, such that the set $\Theta$ is non-empty (the statement that you are refering to is probably that a ring $R$ is noetherian iff all ideals $I \subset R$ are finitely generated), which one needs for Zorn's lemma.

Q2: They did not say that all ideals have to be comparable. An order is not the same as a total order. A partial order on a set $X$ is a relation $R \subset X \times X$, such that certain three axioms are fullfilled (reflexivity, antisymmetry and transitivity). As it simply is a subset of $X \times X$, not all pairs of elements need to be comparable.

Yes, every non-zero ring has at least one maximal ideal which follows from Zorn's lemma. That does not mean that the ring is noetherian though:

For example $(x_1,x_2,\dots) \subset k[x_1,x_2,\dots]$ is maximal, but not finitely generated, such that the ring $k[x_1,x_2,\dots]$ is not noetherian.

Note that rings can have many maximal ideals (for example $p\mathbb{Z} \subset \mathbb{Z}$ for every prime $p$). Here the problem you have is once again that you think about total orders.

Answer 2

Per the comments, the underlying issue seems to be:

Why can partial orders have incomparable elements?

(This is really the whole point of partial orders, as opposed to total or linear orders.)

The key point here is that in a definition, anything not forbidden is permitted - even if it seems counterintuitive. Let's take the definition of partial order, as presented in wikipedia: a partial order (on a given set) is a relation $R$ which is reflexive, antisymmetric (in the sense that [$aRb$ and $bRa$] iff $a=b$), and transitive.

Now let's check that on any family of sets $\mathcal{S}$, the relation "$\subseteq$" is a partial ordering.

• Reflexivity: Any set is a subset of itself, so $\subseteq$ is reflexive.

• Transitivity: If $A\subseteq B$ and $B\subseteq C$, then $A\subseteq C$.

• Antisymmetry: In the left-to-right direction, if $A\subseteq B$ and $B\subseteq A$ then $A$ and $B$ have the same elements - that is, $A=B$. Conversely, the right-to-left direction follows from reflexivity.

But of course in general $\subseteq$ is not a total ordering.