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Recently, while solving a problem in Ring Theory, I came accross a theorem called, "Krull's Theorem". I searched the internet and nearly every source had this written:

"$(\text{Krull's Theorem}:)\space$Let $R$ be a ring, and let $I$ be a ideal of $R$ that is distinct from $R.$ Then there exists a maximal ideal $M$ of $R$ containing $I$"

Now, I recently got to know that nowadays, a ring is defined to be commutative having an identity, $1\neq 0.$

The problem is, I mostly study Ring Theory from the book, "Topics in Algebra" by IN Herstein and there a ring may not necessarily be commutative and even might not have an identity. So, my general notion about rings is that it may not be commutative and may not have an identity unless otherwise specified.

Because of all these reasons, I am unable to figure out or, am rather confused whether the Zorn's Theorem which I quoted above implicitly assumes that $R$ is a commutative ring with identity?

I know that $R$ can be guranteed to have a maximal ideal if it has a unit element. So, I can understand that in the quoted text of the theorem, $R$ is implicitly assumed to be commutative and thereby, I modify the phrasing of the theorem as follows:

"$(\text{Krull's Theorem}:)\space$Let $R$ be a commutative ring, and let $I$ be a ideal of $R$ that is distinct from $R.$ Then there exists a maximal ideal $M$ of $R$ containing $I$"

But the confusion still remains whether or not "$R$ has a unit element" is implicitly assumed.

Any clarification regarding this issue will be greatly appreciated.

Thomas Finley
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  • You possibly strayed into the domain of Commutative Algebra, which just consider commutative rings with identity. – rushusuixing Dec 19 '23 at 06:10
  • @ΑΘΩ No, "ring" without any context is ambiguous. It has various denotations - including those mentioned above. Try applying your argument with "associative" replacing "commutative". – Bill Dubuque Dec 19 '23 at 21:45
  • @BillDubuque Agreed as to the lack of a consensus regarding what a "ring" should refer to in terms of unit elements (accordingly, I deleted my previous comments). As to your second remark, commutativity or lack thereof for a given species of associative algebraic structures is -- if you permit the loose metaphor -- as the difference between birds that can become airborne and those which can't, nevertheless they are still birds...whereas the difference between associative and non-associative structures is tremendous and of unparalleled scale, I would say... – ΑΘΩ Dec 20 '23 at 11:17

2 Answers2

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First of all, it is crucial for the ring to have unit element. The proof supports this idea, because, while using Zorn’s Lemma, you need the unit element to prove that a certain ideal is not the entire ring. In fact, there are examples of rings without unit elements and without maximal ideals: counter example for "every ideal is contained in a maximal ideal" in non-unital case?

And talking about the non-conmutative case, the result is “true”, but being careful with the notion of ideal. In non-conmutative rings you don’t have just ideals but left, right and two-sided ideals, and the result is true in each case: Why doesn't every ring have a maximal ideal?

user26857
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Earnur
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  • So, Krull's theorem precisely says : " $(\text{Krull's Theorem}:)\space$Let $R$ be a commutative ring with identity $1\neq 0,$ and let $I$ be a ideal of $R$ that is distinct from $R.$ Then there exists a maximal ideal $M$ of $R$ containing $I$", correct? – Thomas Finley Dec 19 '23 at 06:26
  • @ThomasFinley Not entirely correct. In general form, Krull's theorem can be taken to say: Let $A$ be an arbitrary non-zero ring (commutativity not required). Then the set of proper* (left/right/bilateral) ideals of $A$ is inductive with respect to the order given by inclusion*. As in general any element of an inductively ordered set sits below a certain maximal element, the version which you present follows as an immediate consequence. – ΑΘΩ Dec 19 '23 at 06:49
  • @ΑΘΩ $(\text{Krull's Theorem}:)\space$Let $R$ be a ring containing a unit element, and let $I$ be a ideal of $R$ that is distinct from $R.$ Then there exists a maximal ideal $M$ of $R$ containing $I$----- Is this correct? I think it's not. Commutativity is required to gurantee existence of a maximal ideal. Like I said, I don't define rings to have a identity necessarily. – Thomas Finley Dec 19 '23 at 07:06
  • @ThomasFinley You are inclined to think wrongly about a fundamental matter: commutativity has nothing to do with the existence of maximal ideals. What is essential in order for these maximal ideals to exist is the assumption of finite generation: explicitly, under the assumption that your Thomas Finley ring (a priori non-unital as you wish to define them) is actually unital (hence a ring in the traditional sense), it follows that the subjacent set is a finitely generated left/right/bilateral ideal. Krull's theorem -- with its specific focus on maximal ideals -- (to be continued) – ΑΘΩ Dec 19 '23 at 09:02
  • @ThomasFinley (cont.) -- actually occurs as the very particular application of an extremely general result concerning the existence of maximal substructures in finitely generated non-trivial closure systems. Another direct application of the forementioned general result is Nakayama's lemma. The notion of closure system is pervasive throughout the vast majority of branches of mathematics and in particular the left/right/bilateral ideals of a given ring form typical examples of closure systems. – ΑΘΩ Dec 19 '23 at 09:13
  • @ΑΘΩ I am sorry, looks like I wrote something wrong. I rechecked and yes you are correct:" Given a ring R there is no guarantee that it has any maximal ideals! If the ring has a unit element this can be proved, assuming a basic axiom of mathematics, the so-called axiom of choice." – Thomas Finley Dec 19 '23 at 09:14
  • @ΑΘΩ Can I say that Zorn's lemma precisely says: " If R is a ring (not necessarily commutative) with identity then it has a maximal ideal" in the context of ring theory? – Thomas Finley Dec 19 '23 at 09:15
  • Let us continue this discussion in chat. – ΑΘΩ Dec 19 '23 at 09:16
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I think that a unit (different from 0) is necessary if we want to apply Zorn's lemma for the existence of the maximal ideal. In fact if $S$ is the set of (proper) ideals of $R$ which contain $I$, let us consider an ascending chain of elements of $S$. If we follow the idea of the "commutative proof" we can look at the union of these elements of $S$ to get a majorant and then we use Zorn's lemma. In this case the union is of course an ideal, but is it proper? I.e. is it in $S$? Or equivalently is it different from $R$? If $R$ has a unit it is so.

P.S. I do not know if there are other ways to prove it in your hypothesis and I don't even know any counterexample :(

BernyPiffaro
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