# Talk:Ring homomorphism

WikiProject Mathematics (Rated C-class, High-importance)
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Field: Algebra

## f(1)=1

Correct me if I'm wrong, but the condition f(1)=1 on the ring homomorphism is wrong/misleading. A ring homomorphism is a group homomorphism between the additive subgroups of R and S s.t. it preserves multiplication; hence, f(0_R)=0_S, but f(1_R) does not necessarily map to 1_S. I'm not sure what definition of a ring homomorphism is being used here, but it's much stronger than the usual definition I've seen. 131.215.172.175 09:33, 10 January 2007 (UTC)

The rings in this article are assumed to have unity. It is very natural I would think that any ring homomorphism maps 1 to 1. Oleg Alexandrov (talk) 16:43, 10 January 2007 (UTC)
A ring homomorphism f: R -> S need not map 1_R to 1_S. Take R to be the integers, S to be 2 x 2 matricies with real entries, and the map f: x --> [[ x,0 ][ 0,0 ]] as an example. Mascdman 05:55, 12 January 2007 (UTC)
Yeah, I am aware of that. You could even simpler take f(r)=0 for all r. But I doubt such homomomrphisms are that important. In your example, for instance, you could restrict f to make it surjective, then its image would be a ring with unity and f would be a homomorphism mapping 1 to 1. Oleg Alexandrov (talk) 13:51, 12 January 2007 (UTC)
Why are all rings assumed to have unity? This doesn't sound standard to me. One would like to have 0 as a ring, and talk about the zero map, for example. Also, one would like to have the linear map $\times n \colon \mathbb{Z} \to \mathbb{Z}$ to be a ring map, for all integers n, not just for n= 1! Actually, I just looked up the page on subrings, and I noted that $n \mathbb{Z}$ is declared there not to be a subring of the integers!! This is getting weirder and weirder. In particular, this says ideals need not be subrings. Isn't this silly? Turgidson 17:21, 12 January 2007 (UTC)
See Wikipedia:WikiProject Mathematics/Conventions. Also, to me personally, it sounds very standard that rings must have unity. I never heard ideals be called "subrings". Oleg Alexandrov (talk) 17:23, 12 January 2007 (UTC)
OK, thanks for pointing out that convention. Sorry, I'm relatively new to Wikipedia, so I don't know in detail all the various mathematical conventions adopted at Wikipedia. Just to make sure: are these conventions set in stone, for all eternity, or are they still open to debate? I maintain that this particular convention is not standard, at least in the mathematical world. In fact, I'm teaching right now a course on Rings and Fields, and the (absolutely canonical) textbook takes what I believe to be the standard convention, namely, that rings need not have 1, and even if they do, ring homomorphisms need not respect 1. Now, I could give many more references to that effect (in case my argument above was not convincing enough), but just to give an existence proof that this convention I am talking about may well be adopted, even on wikipedia-related sites, see [1] and [2]. Turgidson 17:45, 12 January 2007 (UTC)
I don't know if the conventions are set in stone. The forum for discussing things is Wikipedia talk:WikiProject Mathematics. You could give it a try. Oleg Alexandrov (talk) 04:12, 13 January 2007 (UTC)
The condition f(1) = 1 is not natural, because it would exclude the trivial homomorphism f(x) = 0 that exists between any two rings. Also, many other interesting homomorphisms will be excluded, for example one that takes any 2x2 matrix, and turns it into a 3x3 matrix by filling with zeroes. Albmont 11:43, 5 April 2007 (UTC)
The matrix ring example is not very good. Just map 2x2 matrices into 3x3 matrices by putting a 1 in the lower right corner, and then it's a unital ring homomorphism. This is the natural embedding anyway, IMO, since it's the natural embedding into the first factor of the direct sum of 2x2 and 1x1 matrix rings. I also don't know why it's good to have a trivial homomorphism between two arbitrary rings. With f(1) = 1, you still always have a unique homomorphism from any ring to the 0 ring (where 1=0). Do you have an example where more is needed? 173.250.203.231 (talk) 00:33, 16 November 2013 (UTC)
If you just put a 1 in the lower right, then it won't respect addition, so it's not a ring homomorphism. Anyway, I agree with you that the zero map should not be considered a ring homomorphism (except for the zero ring). For instance, there is a theorem that the inverse image of a prime ideal under a ring homomorphism is a prime ideal; this fails if you allow the zero map from Z to Z, for instance. Ebony Jackson (talk) 01:08, 16 November 2013 (UTC)
Turgidson and Mascdman have a point there re rings w/o a unit and non-unital homomorphisms. article oughtta be modified. Mct mht 08:15, 28 March 2007 (UTC)
My introductory abstract algebra textbook, Fraleigh's A First Course in Abstract Algebra, makes a distinction between "unital" and "non-unital" homomorphisms. By Fraleigh's definition, a "ring homomorphism" does not normally require f(1)=1. But if f(1)=1, then he calls it a "unital ring homomorphism". My lecturer doesn't care for the distinction and only talks about unital homomorphisms, but as long as we're being flexible about unity I think Fraleigh's approach on this is reasonable. It avoids artificially excluding interesting homomorphisms, while still letting us talk specifically about the f(1)=1 case when/if we want to. Crispy 11:26, 22 October 2007 (UTC)

## rofl

Very funny, User:66.92.14.198.

• The purpose of ring homomorphisms is to flip out and ring elements [3]

[4] --Zarel (talk) 03:28, 12 February 2010 (UTC)

## Further on f(1) = 1

There are evidently two important definitions of "ring homomorphism" in use, both of them notable. Both of these must be captured by this article. Currently, only one of these is given any airtime in this article, against WP policy. The structure that is preserved by these two notable definitions is respectively:

1. (R,+RR,0R) → (S,+SS,0S)
2. (R,+RR,0R,1R) → (S,+SS,0S,1S)

Homomorphisms are named by the structure they preserve. The first is not required to preserve the identity element, but the second requires that f(1R) = 1S. Unfortunately, the term ring is ambiguous, as some define it as unital, and others not. Therefore, we could name these categories of homomorphisms respectively either

• rng (or pseudoring) homomorphisms and ring homomorphisms, or
• ring homomorphisms and unital ring homomorphisms.

There is even a false statement with the current definition: "For every ring R with unity, there is a unique ring homomorphism ZR. This says that the ring of integers is an initial object in the category of rings." This is only true for case 2, not presently addressed at all by the article. The statement "The image of f, Im(f), is a subring of S" is also incompatible with the definition of subring in that article. In line with WP convention, I propose to follow the first of the two bullets above, and to redo the article to describe both these types of homomorphism, possibly splitting it into two articles. Note that this is in line with the definition of rng homomorphism and ring homomorphism under Pseudo-ring#Formal definition. See also Pseudo-ring#Unital homomorphism. Chime in now or... — Quondum 21:14, 30 September 2013 (UTC)

I agree completely. Not only is this in keeping with the modern mathematical literature, but it is also in line with the current article on rings, and with the current Wikipedia mathematical convention that rings are unital, as mentioned in an earlier comment. Your suggestion to define rng homomorphisms and ring homomorphisms separately is excellent. I think that they should both fit in this article. In fact, this would fix several inconsistencies currently in the article in addition to the false statement you mentioned. Another one is the statement that if R and S are commutative and S has no zero divisors, then the kernel is a prime ideal. Since no one objected to Quondum, I will implement a fix. Ebony Jackson (talk) 05:14, 12 November 2013 (UTC)
Thank you for doing that. I've made a tweak: pseudo-ring hommormisms on unital rings are a useful class of homomorphism. —Quondum 05:01, 13 November 2013 (UTC)

## Ring isomorphism

Currently, ring isomorphism is defined to mean "bijective ring homomorphism". It would be better to use the definition at the ring page, namely that a ring isomorphism is a ring homomorphism having an inverse that is also a ring homomorphism. Of course the definitions are equivalent, but the latter is in line with the uniform definition that one has in any category. The statement that a bijective ring homomorphism is a ring isomorphism should be a theorem. Ebony Jackson (talk) 16:21, 16 November 2013 (UTC)

Some care may be needed here. See the general definition in the the lead of Isomorphism, which defines an isomorphism as a "bijective homomorphism". Definitions in category theory evidently differ in respects from definitions in more "concrete" branches of mathematics such as ring theory, and should not automatically be regarded as "the" definitions. It may be appropriate to change the definition in Ring (mathematics) to be a "bijective ring homomorphism". —Quondum 17:09, 16 November 2013 (UTC)
The lede of isomorphism is simply wrong. The correct definition of an isomorphism is a morphism which admits a two-sided inverse. That this is the correct definition was only recognized with the development of category theory, but nevertheless it is correct: There are situations where a bijective morphism should not be considered an isomorphism (e.g., continuous maps of topological spaces), but it is always true that a morphism with a two-sided inverse should be considered an isomorphism. Ozob (talk) 00:31, 17 November 2013 (UTC)
I'll have to take your word for it. I see a hint of an explanation at Homeomorphism#Notes, though as a rookie I'm tempted to suggest where I see the discrepancy lying with the definition of a homomorphisms on topological spaces, particularly with what it means to be structure-preserving, rather than with the definition of an isomorphism as a bijective structure-preserving mapping. That said, I have to acknowledge that my topology is nonexistent. Several articles will probably have to be updated to bring this into line with what you are saying here, but Isomorphism would be a good start. —Quondum 02:26, 17 November 2013 (UTC)
I agree with Ozob. Best would be to give the category-theoretic definition of isomorphism, and then to say that for many specific kinds of objects arising in abstract algebra (groups, rings, etc.) it is a theorem that bijective homomorphisms are the same as isomorphisms. I also agree with Quondum that Isomorphism would be a good place to start fixing things. Ebony Jackson (talk) 04:10, 17 November 2013 (UTC)

## Integral domain

I note that this edit weakened the statement from applying to noncommutative rings (a ring with no [nonzero] zero divisors) to commutative rings (integral domains). Is this intentional? —Quondum 14:58, 17 November 2013 (UTC)

No, the rings are assumed to be commutative to begin with. -- Taku (talk) 16:00, 17 November 2013 (UTC)