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Every maximal ideal is prime in a commutative ring with identity.
There were several posts on the site about analogues of the claim for rngs (rings with or without identity):

A maximal ideal is always a prime ideal?

and

Examples of a commutative ring without an identity in which a maximal ideal is not a prime ideal

It looks like they do not provide the correct extension of the claim onto rngs.

We call an ideal maximal if it is a maximal proper ideal in the poset of ideals.
This notion assumes that the only "bigger" ideal for a maximal ideal is the principal ideal of units in a ring with identity.

It looks like the correct extension of the notion of a maximal ideal onto rngs is not a maximal proper ideal, but a maximal non-unit ideal (a maximal ideal in the poset of ideals that are not generated by units).

For example, the ideal $2 \mathbb Z$ is a maximal non-unit ideal in the ring with identity $\mathbb Z$, and it is prime;
the ideal $2 \mathbb Z$ is a maximal non-unit ideal in the ring without identity $2 \mathbb Z$, and it is prime.

In this case the claim for maximal ideals in rngs should be formulated in the following way:
every maximal non-unit ideal is prime in a commutative rng.

Is this correct?
Is there any use of the term "maximal non-unit ideal"?

Alex C
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  • Rings which don't have identity don't have units, so defining maximality of ideals in rngs in terms of units is nonsensical. Those posts are completely correct as far as standard literature goes. – rschwieb Mar 13 '20 at 13:29
  • @rschwieb I am using term "rng" for rings with or without identity. Is there an appropriate term for this type of objects? – Alex C Mar 13 '20 at 13:36
  • The thing is that one may argue the definition of maximality must be strengthened to discuss them in rings without identity. For example, Jacobson's texts use maximal modular right ideals as the type of maximal right ideal that one should discuss, to eliminate bad cases like the ones given in the posts you linked. – rschwieb Mar 13 '20 at 13:38
  • No, I would say it is fair to call "ring not necessarily with identity" a rng. What I'm saying is that if you define maximal ideals in terms of "not generated by units" it will have no meaning for the ones which do not have identity (but it will amount to the same thing for rings with identity, yes) so nothing is gained. – rschwieb Mar 13 '20 at 13:39

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It looks like they do not provide the correct extension of the claim onto rngs.

It all looks standard to me.

It looks like the correct extension of the notion of a maximal ideal onto rngs is not a maximal proper ideal, but a maximal non-unit ideal (a maximal ideal in the poset of ideals that is not generated by units). [...] In this case the claim for maximal ideals in rngs should be formulated in the following way: every maximal non-unit ideal is prime in a commutative rng.

If a ring does not have an identity, then it does not have units either (the definition of a unit requires the existence of an identity.) So, the proposed "better" definition for maximal ideals in rings does not have any meaning in a ring without identity.

One can argue, however, that the definition of a maximal ideal (for rings with identity) should be elaborated on to make it work in rings without identity.

One way to do this, as Jacobson did, is to additionally require the ideal to be modular. To accurately state it, he called a right ideal $T$ of $R$ modular if there exists an element $e\in R$ such that $ex=x$ for all $x\in T$. Put another way, there is an element that acts like a left identity on $T$. Notice how when a ring has identity, $e=1$ works for all right ideals maximal in the poset of proper right ideals, so they are all modular. This is a "good" extension of the "absolute" definition of maximal right ideals.

He used these ideals to characterize the Jacobson radical of rings without identity as the intersection of maximal modular right ideals (and not the "absolutely" maximal right ideals.)

In the most common example given in the posts you linked, the rng in question is $R=2\mathbb Z/4\mathbb Z$. Now, the zero ideal is certainly a maximal proper ideal in the ring, but it fails to be modular, as you can see. For this reason, $J(R)=R$, and not the zero ideal.

rschwieb
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  • Cannot we use the "maximal non-unit ideal" in multiple claims common for rings with or without identity, e.g. "a generator of a maximal non-unit principal ideal is irreducible in a commutative rng"? – Alex C Mar 13 '20 at 13:52
  • @AlexC I can't make any sense of that comment. Can you please refine it to be clearer? – rschwieb Mar 13 '20 at 14:09
  • I am trying to find common terminology for all the rings regardless of existence of identity. And it looks like replacing the word "proper" onto "non-unit" in the definition of a maximal ideal makes most of the claims that are true for rings with identity also true regardless of the identity. Does it make sense? – Alex C Mar 13 '20 at 14:18
  • @AlexC Yes, thank you. The thing I don't understand at this instant is the distinction you are making between "proper" and "non-unit". As far as I know, "proper ideal=non-unit ideal". Can you explain how you believe they are different? – rschwieb Mar 13 '20 at 14:27
  • "proper ideal=non-unit ideal" only for rings with identity, isn't it? A ring without identity is a maximal non-unit ideal of itself, but not a maximal proper ideal. By "maximal" I mean the standard maximal element of a poset in this case. – Alex C Mar 13 '20 at 14:37
  • @AlexC The only definition of "unit ideal" i know says "the subset $R$ of $R$ is itself an ideal and can be called the unit ideal of $R$." What is your definition of a "unit ideal" of a ring without identity, if not that? – rschwieb Mar 13 '20 at 14:39
  • I would define a "unit ideal" as a principal ideal of a unit for all rings regardless of identity. A "non-unit ideal" is an ideal that is not a "unit ideal". – Alex C Mar 13 '20 at 14:53
  • @AlexC the set of “unit ideals” of a ring without unity is empty, in that case. That does not seem like a good outcome. – rschwieb Mar 13 '20 at 16:02
  • In case we define a ring as a commutative additive group + semigroup under multiplication + distributivity, such structure would combine both rings and rngs. I thought the term "non-unit ideal" could help to make common statements for the structure. Separating rings with units and rings without units seems does not follow the classification principle when we take a class and narrow it down with some constraints, or combine several classes into a single one. Why not separating rings with idempotents and rings without idempotents, for example? – Alex C Mar 13 '20 at 16:29
  • @AlexC I don't understand your last comment. Can you give an example of a rng where your definition works better than the old one? (Maximal among proper ideals.) An example that shows what you believe the distinction is. – rschwieb Mar 13 '20 at 16:49
  • Every maximal non-unit ideal is prime in any commutative ring/rng. A maximal proper ideal may not be prime in a commutative rng. Every generator of a maximal non-unit principal ideal is irreducible in a commutative ring/rng. A generator of a maximal proper principal ideal may not be irreducible in a commutative rng. We can use the example with $2 \mathbb Z$ from my question. – Alex C Mar 13 '20 at 17:14
  • Every maximal non-unit ideal is prime in any commutative ring/rng. This isn't obvious at all. Let $R=2\mathbb Z/4\mathbb Z$. If you think $R$ is "maximal nonunital", then you're saying $R$ is a prime ideal, which breaks the definition of prime, and fails to be proper (which seems like a bigger flaw to me.). If you think that the zero ideal of $R$ is maximal nonunital, then it fails to be prime because $2^2=0$ but $2\neq 0$. So it doesn't appear to work like that at all for the example at hand. – rschwieb Mar 13 '20 at 18:35
  • We can use the example with 2ℤ from my question. But that is an example of a maximal ideal in a ring with identity. I thought we agreed that we are trying to prove that this definition has importance for rings without identity. For rings with identity, I think what you're saying is equivalent but not obviously useful. – rschwieb Mar 13 '20 at 18:54
  • You are correct. I missed the part that a prime ideal must not be the whole ring. Then my statement should be: if $a \cdot b$ belongs to a maximal non-unit ideal $M$, then $a$ is in $M$ or $b$ is in $M$. Is this right? But then again, the requirement for a prime ideal to not be the whole ring looks like we just don't want to include units into the picture. – Alex C Mar 13 '20 at 19:29
  • If we replace "proper" with "non-unit", then everything remains correct: a prime ideal in an rng may be the whole ring. – Alex C Mar 13 '20 at 19:37
  • @AlexC Only do one thing at a time. Right now I'm asking you to define "maximal" clearly and given an example for your definition. Now you are throwing "redefine prime" in also, which confuses the issue. I am still interested in an example of a rng 1) without identity; 2) with an ideal $I$ which is maximal with respect to your definition; 3) such that $I$ is not maximal with respect to the "maximal proper" definition; and 4) which generalizes maximal ideals in commutative rings with identity. – rschwieb Mar 13 '20 at 19:41
  • $2 \mathbb Z$ is an rng without identity 2) $2 \mathbb Z$ is a maximal non-unit ideal in $2 \mathbb Z$ 3) $2 \mathbb Z$ is not a maximal proper ideal in $2 \mathbb Z$.
    • – Alex C Mar 13 '20 at 19:44
  • @AlexC I don't think any definition of maximal that permits the entire ring to be a maximal ideal is acceptable. It also leads to the further complication of the definition of prime being further disturbed for rings without identity. As I mentioned earlier, a successful re-definition of maximal will add requirements to the existing one for rings with identity, in such a way that rings with identity already have the property you add, and eliminate "bad" cases for rings without identity. What you are suggesting is not a good generalization. – rschwieb Mar 13 '20 at 19:45
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    Yes. And I am trying to go through the complications. At this point I think you gave me a clear and detail description of the problem. Thank you. I really appreciate your help. – Alex C Mar 13 '20 at 19:53
  • Thanks for being patient and clarifying. – rschwieb Mar 13 '20 at 19:53