Definition of Ring Vs Rng

Question

When I took abstract algebra I learned that a ring was a set that is an abelian group under addition and monoid under multiplication (along with the distributive property).

In preperation to tutor someone in algebra I've noticed that some books present a ring as what I know as a "rng" or an abelian group under addition and a semi-group under multiplication.

Is there any reason to prefer one as the definition for a ring vs the other?

EDIT And a very related question, is there any math authority or consensus that has dictated/specified that it is more correct to use the ring/ring with unity or the rng/ring definition?

A related question: http://math.stackexchange.com/questions/16168/applications-of-rings-without-identity. The survey article on commutative rngs by D.D. Anderson may also be of interest: http://www.springerlink.com/content/p684h666156n0151/ — Jonas Meyer, Jun 30 '11 at 02:23
Check out this note by Keith Conrad: http://www.math.uconn.edu/~kconrad/blurbs/ringtheory/ringdefs.pdf — Corey, Jun 30 '11 at 03:11
As evidence that there is no authority, note that one of the classic texts in (noncommutative) Ring Theory, Lam's "First Course...", explicitly states that in the book "ring" will mean "ring with $1$" and homomorphisms will be required to take $1$ to $1$. If Lam feels the need to clarify it by stating it explicitly, I don't think anyone can assume much either way... — Arturo Magidin, Jun 30 '11 at 03:11
@Corey: I'm not sure I agree with the strong language in that note; and I know for sure the many ring theorists I know that work extensively with radical theories (theories involving radicals of rings) do not agree with it. — Arturo Magidin, Jun 30 '11 at 03:13
@Corey For an opinion contrary to Conrad's see the quote here. — Bill Dubuque, Jun 30 '11 at 03:23
@user6312: It's usually really hard, if not impossible. I remember when I was a kid reading a lot of "Translator note" footnotes in books (in Spanish) where it said "juego de palabras en el original" (pun in the original), because there just wasn't any way of doing it. It made Asimov's story "The Shah Guido G." a complete waste of space... — Arturo Magidin, Jun 30 '11 at 04:29
Please do correct me if I am wrong but I believe Herstein does not include the assumption that a ring has a unity in his textbook "Noncommutative Rings". — Amitesh Datta, Jun 30 '11 at 05:24
May be it is my representation background (and the fact that I was raised by Jacobson's Basic Algebra), but I have felt that the remark 'all the rings in this book are assumed to have a unit element' at the beginning of all the books on my shelf was just to accomodate some old-timers, as this view was not always universally accepted :-). We always have the identity mapping, and the constant function 1. True, the latter does not often have a compact support, but the analysts can include the exception in their books. Well, it is pointless to argue matters of faith. — Jyrki Lahtonen, Jun 30 '11 at 06:01
@Arturo: Agree with 'The Shah Guido G." (even though I shudder at the thought of a ring without '1'). Asimov has several similar short stories (Death of a Foy,...) that were a waste of my time for the reason that the word play reference was only accessible to educated native speakers of the English language. No sane person would ever attempt to translate such a piece. — Jyrki Lahtonen, Jun 30 '11 at 06:33
@Amitesh: Herstein's book was written over 40 years ago. So it's not an example of contemporary use of terminology. I have, for instance, never met a live, working algebraist who uses "ring" to mean "not necessarily unital ring". — Pete L. Clark, Jun 30 '11 at 09:31
Didn't really expect to start such debate. I haven't studied enough algebra to see the benefit of using one definition over the other. I didn't realize it was such a contentious issue. — crasic, Jun 30 '11 at 09:59
The note of mine that Corey cited was written because Dummit and Foote's algebra book (a text used in my department) does not insist rings contain 1, and my experience is that this causes unnecessary confusion when it's time to learn commutative algebra. Someone who thinks of 2Z inside Z as a subring [sic] instead of as an ideal has to unlearn things later. As for the quote Bill linked to, its ending says "Thus, in many, maybe most, branches of ring theory the requirement of the existence of a unity element is not sensible, and therefore unacceptable." This is wrong for commutative algebra! — KCd, Jun 30 '11 at 10:15
Along the same lines, consider tensor products of modules over a ring. I think it is better to first study them when the ring is commutative, where there are already a lot of basic examples, techniques, and results to absorb. Once a student is comfortable with that, the transition to tensor products over a noncomm. ring takes less time than the other way around (noncomm. case first). The noncomm. case is important (e.g., representation theory of groups uses tensor products over the group ring), but the comm. case doesn't lack for lots of examples on its own first. — KCd, Jun 30 '11 at 10:20
I'm not suggesting that one should study abelian groups exclusively before learning about nonabelian groups. But I think comm. rings and their modules have enough meat at a basic level that the intuitions learned there can be adapted when you later need to learn about noncomm. rings. You shouldn't study not-necessarily-comm. rings first, and likewise you shouldn't study not-necessarily-unital rings first. That's all I wish to say. — KCd, Jun 30 '11 at 10:25
I have to say that I have encountered both definitions of a Ring, and provided it is clear which applies I don't ever seem to have encountered a problem. My basic course had Rings with 1, and an Ideal was not a subring, but a submodule of R (considered as a module over itself). In representation theory decomposing 1 into central idempotents was also seen from a 'Module' perspective rather than a 'Ring' one. — Mark Bennet, Jun 30 '11 at 10:52
Dear Pete, the point I had in mind (although I did not explicitly say so) is that Herstein's book actually uses the lack of this assumption productively in his book. I must admit that I have not read Herstein's book beyond chapter 1 and therefore cannot comment further. (I read Lam instead.) However, in chapter 1, at least, there are genuine results regarding rngs. (Please do correct me if I am wrong as I have last looked at Herstein's book two years ago.) — Amitesh Datta, Jun 30 '11 at 12:26
Dear KCd, I disagree to some extent. For example, Martin Isaacs treats noncommutative rings before he treats commutative rings in his textbook "Algebra: A Graduate Course". Also, some basic results in general (not necessarily) commutative rings such as the Jacobson density theorem are really at about the same conceptual level as results such as primary decomposition in commutative algebra. But I am likely biased since I began learning about noncommutative rings at least a year before I studied any results that "belong to commutative algebra". — Amitesh Datta, Jun 30 '11 at 12:32
@Amitesh: I'm not saying that rngs never occur in nature or are somehow unworthy of study. In fact I just wrote up notes on a theorem of Wedderburn concerning nilpotent subrng of a finite dimensional $k$-algebra: note there are no nonzero nilpotent rings! It's just that when you're studying something systematically, you need to decide what class of objects you're interested in, for instance (but only this) you want to look at the category of rings. If you don't include an identity as part of your definition, then even when you study rings which happen to have an identity... — Pete L. Clark, Jun 30 '11 at 19:32
...you will be looking at not-necessarily-unital homomorphisms of rings. If you happen to be studying unital rings more than 90% of the time, then this will be rather inconvenient, not in some deep mathematical way but simply because so many individual results will require you to assert that your rings are unital and your homomorpihsms are unit-preserving.... — Pete L. Clark, Jun 30 '11 at 19:34
...As an analogy, why do we require rings to be associative? Some older texts do not (and in fact the recent text that Bill Dubuque quotes does not, which IMO is not to its credit), and everyone agrees that the study of "nonassociative rings" is extremely important: e.g. Lie algebras. But it's a matter of terminology: because non-associative things behave so much differently, it's inefficient to call them rings, so we call them algebras instead. (And yes, it depends on context: in many of the circles I run in, when someone says "for any ring $R$" they probably mean "commutative ring"...) — Pete L. Clark, Jun 30 '11 at 19:38
Dear Pete, I absolutely agree, needless to say. Once you told me about the result in your notes, I just remembered a few points: @crasic There seems to be three main instances of rngs: (1) One wishes to prove that certain rngs are actualy rings under certain assumptions; this is done in Herstein. (2) It is convenient to phrase problems about rings as problems about rngs in some instances; for example, I believe the validity of the following assertion is open: if $R$ is a nil rng, then $R[x]$ does not have a (proper) maximal ideal. (3) Rngs have an interesting theory. — Amitesh Datta, Jul 01 '11 at 01:20
@PeteL.Clark I listen to a research seminar whose participants are mostly ring theorists. Prof. Jan Krempa is one and he never forgets to mention that he is speaking about associative rings when he is. I was also recently told in another seminar in algebra that in radical theory rings are usually not required to be unital in order to be able to call the radicals (and ideals in general) subrings. (Arturo said that earlier too.) — , Mar 24 '12 at 12:34

score 12 · Accepted Answer · answered Jun 30 '11 at 02:40

12

There are no math authorities. There are just conventions, and as far as I can tell the convention that "ring" means "ring without identity" can only be traced back to people who learned algebra using Hungerford.

The main reason to prefer "ring" to mean "ring with identity" is that I am pretty sure it is the statistically dominant convention, although I don't have the statistics to actually back that up. (Unless this is not what you mean by "reason," in which case I'll guess another possible meaning: for most applications, your rings will have identities.)

answered Jun 30 '11 at 02:40

Qiaochu Yuan

419,620

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Herstein also defines rings that way, and I'm pretty sure Herstein predates Hungerford. – Zev Chonoles Jun 30 '11 at 02:44
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I might've been thinking of Herstein. The point is it seems like whichever convention you learned first has a way of sticking to you. Myself, I learned algebra from PROMYS, then from Artin, and rings had identities both times. – Qiaochu Yuan Jun 30 '11 at 02:52
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@Qiaochu Yuan: I've never seen a definition of ring that requires the existence of a multiplicative identity. Checking my shelves, I see that neither Birkhoff & MacLane nor Herstein has such a requirement, and both books were around and influential well before Tom Hungerford's. – Brian M. Scott Jun 30 '11 at 02:57
Noether didn't require an identity either, see my answer here (under 2. modern usage) for the definition she used (at least in one seminal paper). And by the way, speaking of authorities: I'd elect this self-proclaimed candidate for Emperor of Notation. – t.b. Jun 30 '11 at 03:01
Well, I do require rings to have identities. – Mariano Suárez-Álvarez Jun 30 '11 at 03:04
@Mariano: I just added something to my previous comment :) – t.b. Jun 30 '11 at 03:06
:) ${}{}{}{}{}$ – Mariano Suárez-Álvarez Jun 30 '11 at 03:09
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"I cannot help this bad notation. I did not create this part of the Universe." -- Hendrik Lenstra. – Arturo Magidin Jun 30 '11 at 03:16
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@Brian: like I said, I learned algebra from Artin's Algebra, which does require rings to have multiplicative identities. I don't really know what algebra books the kids are reading these days, but I don't think it's Birkhoff and MacLane. (Also, confession: the main reason I included that comment was so people would respond with counterexamples.) – Qiaochu Yuan Jun 30 '11 at 04:10
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@Qiaochu: We used Papantonopoulou the last 2-3 times, I think. I'm sure that you're right about B&MacL: it already felt old-fashioned to me back in 1964 or 1965 (when the price stamped in the hardcover was all of $7.95!). – Brian M. Scott Jun 30 '11 at 05:42

score 6 · Answer 2 · answered Jun 30 '11 at 06:47

Wikipedia had a large discussion of this from 2003 to 2008 including an analysis of publications, and comments about both Bourbaki and Cambridge University changing to require a 1.

There does not seem to be a consensus, but there does seem to be a trend towards more modern and more advanced texts being more likely to require a 1.

score -2 · Answer 3 · answered Jun 30 '11 at 03:01

-2

The standard definition of a ring $R$ is that $R$ is an abelian group under addition and a semi-group under multiplication. The existence of multiplicative identity is generally not required. If rings are defined in the "monoid under multiplication" way we can not consider the set $\{0\}$ as the smallest possible ring. (I see Artin defines rings in the "monoid under multiplication" way where Dummit & Foote, Herstein do not. So there is inconsistency in the standard definition.)

answered Jun 30 '11 at 03:01

Sayantan

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Then if there's inconsistency in the definitions, how come you refer to the one without multiplicative identity as "the standard" one? Seems somehow contradictory to me =) – Adrián Barquero Jun 30 '11 at 03:15
Simply because it is more general. – Sayantan Jun 30 '11 at 05:46
I don't understand why you claim that ${0}$ is not the smallest possible ring. It is, under either convention. – Pete L. Clark Jun 30 '11 at 09:33
@Pete I've also encountered some textbooks that state $1\neq0$ as a ring axiom specifically to exclude the trivial ring. Why they do that is beyond me. – crasic Jun 30 '11 at 09:56
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@crasic: IMO that is much worse than any of the conventions being discussed here. If you do that then how do you know whether $R/I$ is a ring or not? (To anyone who answers: "we won't allow $I = R$ as an ideal of $R$", then I say that they must not have worked with ideals very much, because in general it can be very hard to tell whether the ideal one has written down is proper or not. So you have a bunch of elements of a ring, and you don't know whether they generate an ideal?? Not good.) – Pete L. Clark Jun 30 '11 at 10:14
@crasic: actually I would be interested to know examples of such books. I would be interested to know whether anyone who follows such a dubious practice could actually write a good algebra text! – Pete L. Clark Jun 30 '11 at 10:16
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@crasic: Incidentally, one reason to include the zero ring is to have a terminal object in the category of rings (so an initial object in the category of schemes). – Akhil Mathew Jun 30 '11 at 18:09
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Note, however, that in a field one always assumes $1 \not= 0$, so the one-element ring {0} is not a field. Which is not to say people aren't interested in the idea of a field with one element... – KCd Jun 30 '11 at 18:31
@Pete: W.E. Deskins, Abstract Algebra, and Elbert A. Walker, Introduction to Abstract Algebra, both require $1\ne0$ in their definition of a ring with unity. (Neither requires a ring to have a $1$.) – Brian M. Scott Jun 30 '11 at 21:52
@Brian: Thanks, that's interesting. I am not familiar with either book, but here's some preliminary detective work: Deskins' book was published in 1966 and identically reprinted by Dover in 1995: there are no MathScinet citations. Walker's book does not appear in MathSciNet at all, but I found a free copy online. I hope it is fair to say that these are not the leading brands... – Pete L. Clark Jun 30 '11 at 23:30
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Okay I looked in Walker's book on p. 111 where this occurs. For whatever it's worth, I do not approve of the way he does things: I'll just leave it at that. – Pete L. Clark Jun 30 '11 at 23:35
@Pete the Deskins book (well the dover reprint, I obsessively buy dover reprints for dirt cheap at used book stores) is the one I remember seeing, I was trying to track it down again but @Brian got it before me. – crasic Jul 01 '11 at 19:44
There is no reason not to admit the trivial ring as a ring. It is, after all, nothing but the ring of functions on the empty set. (Exercise: is the trivial ring an integral domain?) – Qiaochu Yuan Jul 01 '11 at 19:44

Definition of Ring Vs Rng

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