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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?

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    $\begingroup$ A related question: math.stackexchange.com/questions/16168/…. The survey article on commutative rngs by D.D. Anderson may also be of interest: springerlink.com/content/p684h666156n0151 $\endgroup$
    – Jonas Meyer
    Commented Jun 30, 2011 at 2:23
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    $\begingroup$ How does one translate the pun into other languages? $\endgroup$
    – André Nicolas
    Commented Jun 30, 2011 at 2:49
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    $\begingroup$ 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. $\endgroup$
    – Jyrki Lahtonen
    Commented Jun 30, 2011 at 6:01
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    $\begingroup$ @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". $\endgroup$
    – Pete L. Clark
    Commented Jun 30, 2011 at 9:31
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    $\begingroup$ 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. $\endgroup$
    – KCd
    Commented Jun 30, 2011 at 10:20

3 Answers 3

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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.)

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    $\begingroup$ Herstein also defines rings that way, and I'm pretty sure Herstein predates Hungerford. $\endgroup$
    – Zev Chonoles
    Commented Jun 30, 2011 at 2:44
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    $\begingroup$ 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. $\endgroup$
    – Qiaochu Yuan
    Commented Jun 30, 2011 at 2:52
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    $\begingroup$ @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. $\endgroup$
    – Brian M. Scott
    Commented Jun 30, 2011 at 2:57
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    $\begingroup$ "I cannot help this bad notation. I did not create this part of the Universe." -- Hendrik Lenstra. $\endgroup$
    – Arturo Magidin
    Commented Jun 30, 2011 at 3:16
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    $\begingroup$ @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.) $\endgroup$
    – Qiaochu Yuan
    Commented Jun 30, 2011 at 4:10
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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.

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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.)

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    $\begingroup$ 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 =) $\endgroup$
    – Adrián Barquero
    Commented Jun 30, 2011 at 3:15
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    $\begingroup$ @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.) $\endgroup$
    – Pete L. Clark
    Commented Jun 30, 2011 at 10:14
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    $\begingroup$ @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). $\endgroup$
    – Akhil Mathew
    Commented Jun 30, 2011 at 18:09
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    $\begingroup$ 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... $\endgroup$
    – KCd
    Commented Jun 30, 2011 at 18:31
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    $\begingroup$ 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. $\endgroup$
    – Pete L. Clark
    Commented Jun 30, 2011 at 23:35

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