A Confusion over Krull's Theorem

Question

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.

Answer 1

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 :(

Answer 2

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?