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.
It is when in the symmetry part that they write $a \leq b$, this is already assuming these elements are related and it make sense to write $\leq$
– Lemon Sep 16 '19 at 22:46Let's say I have two (sets) of ideals that I cannot compare $A_i$ and $B_i$ (so can't write $A_i \subset B_i$, then it is still comparable in their own family $A_1 \subset A_2 \dots$ and $B_1 \subset B_2 \dots$. They form their own two chains.
By the standard proof each chain has the maximal element $\cup_i A_i$ and $\cup_i B_i$. Is that why in the proof it says Zorn's lemma say there is A maximal element instead of THE maximal element?
– Lemon Sep 16 '19 at 23:14But the set of sets is $\sum$ of all proper ideals.
– Lemon Sep 17 '19 at 23:53