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Why do morphisms in category of rings respect identity

  1. Jun 10, 2012 #1
    Hi, I'm looking for intuition and/or logic as to why we would want or need morphisms according to axioms in category theory, to imply that in the category of rings, they must preserve the identity (unless codomain is "0").

    Thank you very much.
  2. jcsd
  3. Jun 11, 2012 #2
    Okay, I think I thought it was more complicated then it was. I'm sure I was interested for some specific reason, but I guess morphism in concrete categories are just structure preserving functions (I do not have an in depth source to check yet). So the identity part is as Artin says, there aren't necessarily a lot of inverses, so we need to explicitly mention that 1 maps to 1. f(x)=f(x)f(1)→f(x)(1-f(x))=0. For instance we are not in an integral domain so we should require that f(1)=1. Okay, now that I write that, I don't get it. Are all morphisms by definition, in categroy of rings with identity, necessarily taking identity to identity, or is this a stronger restriction that Artin has requested, as opposed to definition of morphism.
  4. Jun 11, 2012 #3
    The way I understand it, if the category itself demands that all its members have multiplicative identity, then the morphisms will have to respect that. Otherwise, the image of your morphism may not be in the category any longer. If you are in the category of rings without 1 (Pseudo-rings, Rngs), then you aren't restricted this way.

    I am still working on categories myself, so others may be able to give you a deeper explanation.
  5. Jun 11, 2012 #4
    I'm still not 100% certain what your question is, so first I'll give my interpretation of the question.

    I believe you are asking why the axioms of general ring homomorphism (f(x)f(y)=f(xy) and f(x)+f(y)=f(x+y)) imply the additional axiom of ring homomorphism for unital rings (f(1)=1), in the case that both rings have a unit.

    If this is your question, then the simple answer is that they don't. A general ring homomorphism R→S actually does not need to be a unital ring homomorphism just because R and S are unital. For instance, the zero function between any two rings is a (pseudo-)ring homomorphism, but it is never a unital ring homomorphism unless the codomain is the zero ring (since we would need 0=1).

    Put in other words, the category of rings is a subcategory of the category of pseudo-rings. However, it is not a full subcategory.
  6. Jun 11, 2012 #5
    Another interpretation of your question is "Why do we include this additional axiom f(1)=1 explicitly?"

    If that's what you meant, then the answer is essentially the same. It's because it's not implied by the other ones.

    When we're talking about a concrete category of algebraic objects, a full subcategory basically corresponds to the case that no new axioms of homomorphism are necessary. Since the "extra structure" (the unit) is not automatically preserved, we need two things:
    1. On the algebra side, we add an extra rule to our homomorphism.
    2. On the categorical side, we cut out the morphisms that didn't satisfy that rule, making our subcategory no longer full.
  7. Jun 11, 2012 #6
    Thank you!!
  8. Jun 12, 2012 #7
    Yes, very nicely said.
  9. Jun 12, 2012 #8
    To be sure, I was thanking you equally Sankaku!!
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