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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.

user26857
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Lemon
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    "Ordered by inclusion" is often used as shorthand for "partially ordered by inclusion" - which any family of sets trivially is. – Noah Schweber Sep 16 '19 at 22:30
  • @NoahSchweber "which any family of sets trivially is" this is what I don't get. If my family only has two sets which have no nothing to do with each other, how can we put an ordering?" – Lemon Sep 16 '19 at 22:32
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    A partial ordering allows elements to be unrelated to each other, there's no problem. (Re-read the definition of partial ordering.) – Noah Schweber Sep 16 '19 at 22:33
  • @NoahSchweber am I reading this incorrectly? The diagram still agrees with me, ${x}$ is not connected to say ${y,z}$. Where i the definition say $a,b$ don't have to related to write $a \leq b$ (for example)? – Lemon Sep 16 '19 at 22:40
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    I don't understand what you're asking at this point. Do you understand that in a partial order, we can have incomparable elements? – Noah Schweber Sep 16 '19 at 22:41
  • @NoahSchweber "partial order, we can have incomparable elements" No I do not know. I don't see where in the definition says that. – Lemon Sep 16 '19 at 22:43
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    What part of the definition do you think prevents it (that is: demands that for any $x,y$, either $x\trianglelefteq y$ or $y\trianglelefteq x$)? If the definition doesn't prevent it, it allows it. – Noah Schweber Sep 16 '19 at 22:44
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    @Hawk It is rather that the definition does not force all elements to be comparable and not that it explicitly states that they do not have to be. – Con Sep 16 '19 at 22:44
  • I quote "Then ≤ is a partial order if it is reflexive, antisymmetric, and transitive, i.e., for all $a, b$ and $c$ in $P$, we have that:"

    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 22:46
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    @Hawk Yes exactly, for all elements in the relation $P$ and not for all elements of the given set. – Con Sep 16 '19 at 22:47
  • The preceding sentence "Orders are special binary relations. Suppose that $P$ is a set and that ≤ is a relation on $P$", $P$ is the set. Thus in this case $P$ is the set of sets and $a,b,c$ are the sets in question. – Lemon Sep 16 '19 at 22:49
  • A partial order on a set $X$ is a subset $R \subset X \times X$, such that ... As it is a subset, not all pairs of elements need to be comparable. Check that the wikipedia entry mentions $P$ but then continues "if ...". At that point they are restricting to the relation again. – Con Sep 16 '19 at 22:57
  • @ThorWittich so for a partial order set there may exist incomparable elements (much like my example), and if such elements exist, they form their own chain for the given $\leq$? – Lemon Sep 16 '19 at 23:04
  • @Hawk I am not sure whether you understand it properly due to the word "chain". But yes, incomparable elements will not be "directly connected" but can still be "connected" to other elements and thus form their own "connections". If you once again look at the image on the wikipedia page you can see that. I would not call that chain though as chains have a clear definition. – Con Sep 16 '19 at 23:08
  • So in this case I mean a chain of inclusions. For example let $\Sigma = $ideals $\neq (1)$ in $R$ as in my example.

    Let'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 23:14
  • If you chose the families to be comparable, then yes. They do not necessarily form chains. For Zorn's lemma you explicitly pick a chain and then you can do what you call the standard proof. Yes. You show that every chain has an upper bound. Well, as Zorn's lemma yields a maximal element for the entire set and not only for the chains, I would say no. It is rather once again that maximal elements just are not unique such that it would be too strong to demand that. Take maximal ideals as an example. The existence is also proved with Zorn's lemma, but most rings have more than one maximal ideal. – Con Sep 17 '19 at 06:32
  • " Well, as Zorn's lemma yields a maximal element for the entire set and not only for the chains, I would say no. "

    But the set of sets is $\sum$ of all proper ideals.

    – Lemon Sep 17 '19 at 23:53

2 Answers2

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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.

Con
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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.

Noah Schweber
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  • So according to wikipedia, total ordering requires connexity which is forcing all elements in the set (of sets) to be related (either $a\leq b$ or $b\leq a$). So for incomparable elements, are there multiple chains for the given order? – Lemon Sep 16 '19 at 22:57
  • I don't think you're interpreting that correctly, but you may mean something else by "chain" than I would. Here's a partial order on three elements, $A, B, C$. I'll denote the order by $\le$: (1) $A \le B$, (2) $A le C$ (3) $A \le A, B \le B, C \le C$. THe graph representing this partial order looks like a letter "V" with item $A$ at the bottom, and items $B$ and $C$ at the top. This graph has only a single connected component. – John Hughes Sep 16 '19 at 23:05
  • @Hawk Yes. In John Hughes' example, there are five chains: ${A}$, ${B}$, ${C}$, ${A,B}$, and ${A,C}$. (The first three are silly in some sense, but they're perfectly good chains.) – Noah Schweber Sep 16 '19 at 23:13
  • It's also a good idea to look at a slightly larger example: the partial order of subsets of ${1,2,3}$. This partial order has $2^3=8$ elements. The sub-family ${\emptyset,{1},{1,2},{1,2,3}}$ is a chain in this partial order, and indeed a maximal chain (nothing can be added to it while retaining chain-ness); there are five more maximal chains, gotten by permuting $1,2$, and $3$ in the above example. There are then a number of smaller chains, e.g. ${\emptyset,{1,2}}$. (Meanwhile, something like ${{1,2},{1,2,3},{2,3}}$ is not a chain.) – Noah Schweber Sep 16 '19 at 23:17
  • But each of these chains have a maximal element of their own. That is basically the point of the Zorn Lemma right? Like in the proof of rings have at least 1 maximal ideal. Because you can't say we have the maximal element when there might be incomparable elements. – Lemon Sep 18 '19 at 14:50
  • @Hawk So? What does that have to do with the original question? In a partially ordered set, elements don't need to be comparable. (And this can in particular lead to having multiple distinct maximal elements.) – Noah Schweber Sep 18 '19 at 15:07