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I am using a text right now for abstract algebra ("A Concrete Introduction to Abstract Algebra" by Lindsay Childs) that seems to use a non-standard defn of ring homomorphism. I want to see if others agree and if the difference is a significant one or not.

Let (R,+,$*$) and (R',+',$*'$) be rings and f:R->R'. The standard defn for f to be a ring homomorphism (given in Birkhoff and McLane, Dummit and Foote, and Herstein) is that f preserves addition and mutliplication in the two rings. Childs, however also requires that f satisfy a third condition: if R and R' have multiplicative identity elts 1 and 1', resp, then f(1)=1'.

To me this seems like a non-trivial distinction b/c some mappings that would be considered as homomorphisms under the standard defn would not be a homomorphism under the definition given by Childs. Intuitively, to me this seems non-trivial b/c there are some slippery things that happens when f doesn't map 1 to 1' (i.e. f(1') is the unit in the subring f(R), but not necessarily a unit in R', not to mention the fact that the map of the multiplicative inverse of a unit is the multiplicative inverse of the map of that element, again in f(R), etc)

Not having much experience in abstract algebra, this naturally leads me to wonder if this is a difference which isn't really important and I should just ignore, so I was wondering what others (much more experienced and intelligent than myself) thought.

Thank-you, Matt

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    $\begingroup$ It seems that Childs defines homomorphism of unital rings rather than just rings. So now you have additional structure to preserve. You should check in the introduction where he defines rings whether or not he requires them to have an identity element. $\endgroup$ – Asaf Karagila♦ Oct 13 '13 at 19:12
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If you’re talking about his book A Concrete Introduction to Higher Algebra, the definition in the third edition, as displayed at Google Books, is the definition of a homomorphism of unital rings, i.e., of rings with identity. He doesn’t say that a homomorphism $h:R\to S$ must satisfy $h(1_R)=1_S$ if the rings have multiplicative identities: he defines the notion of homomorphism only for rings that do have multiplicative identities. I looked back to see how he defines ring, and it appears that he requires a ring to be unital.

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  • $\begingroup$ I'm still confused, b/c if you consider the example given by Dietrich, this is a case where both the domain and codomain of f have a unit element (both R and R' are unital rings, i.e. rings with units). However, f maps the unit element in the integers to an element which is not the unit in the codomain. My understanding is that Childs would say this is not a ring homomorphism but the other sources would say it is. Is this correct, or is there something I am still missing? $\endgroup$ – Matt Brenneman Oct 13 '13 at 19:37
  • $\begingroup$ @Matt: Your understanding is correct. He doesn’t say so explicitly, at least in the part of the book that I can see at Google Books, but Childs is apparently working in the category of unital rings, in which the multiplicative identity is part of the structure that must be preserved. $\endgroup$ – Brian M. Scott Oct 13 '13 at 19:42
  • $\begingroup$ Thanks for the clarification on the definition. I think I can see his motivation in doing this b/c later on he wants to talk specifically about homomorphisms from the set of integers to any non-trivial ring, and he needs this condition. It is just that when you start out, you want to have a clear understanding of the main concepts and sometimes I find in abstract algebra there are small things that are important and not so obvious. $\endgroup$ – Matt Brenneman Oct 13 '13 at 21:00
  • $\begingroup$ @Matt: You’re welcome. I have to say that if I were writing such a book, I’d include some discussion of the issue, or at least mention that others use a slightly different definition. $\endgroup$ – Brian M. Scott Oct 13 '13 at 21:03

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