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Homomorphisms between two isomorphic rings ?

  1. Apr 6, 2013 #1
    1. The problem statement, all variables and given/known data
    True or False?
    Let R and S be two isomorphic commutative rings (S=/={0}). Then any ring homomorphism from R to S is an isomorphism.

    2. Relevant equations

    R being a commutative ring means it's an abelian group under addition, and has the following additional properties:

    i) a*(b+c)=a*b+a*c
    ii) ab=ba
    iii) a*(b*c)=(a*b)*c
    iv) there exists an element eR s.t. a*eR=a for all a in R.



    A "ring homomorphism" from R to S is a function f from R to S such that
    i) f(a)*f(b)=f(a*b)
    ii) f(a+b)=f(a)+f(b)
    iii) f(eS)=eR

    3. The attempt at a solution
    BAck of the book says false

    I thought to make f(a)=0S for all a in S which would have worked as a counterexample but but it implies f(eR)=0S which by property (iii) of ring homomorphisms implies eR=0S which means a=a*eS=a*0S=0 so a=0 for all a in S but that means S={0} which is a contradiction.


    Thanks for reading
     
  2. jcsd
  3. Apr 6, 2013 #2

    jbunniii

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    Consider for example a polynomial ring, such as ##\mathbb{R}[x]##. Clearly it is isomorphic to itself. But there are many homomorphisms from ##\mathbb{R}[x] \to \mathbb{R}[x]## which are not isomorphisms. Can you find one?
     
  4. Apr 6, 2013 #3

    A homomorphism could me mapping every polynomial to its constant term (term without an x).
    Say ##P_{1}(x)=c_{n}x^n+...+c_{1}x+c_{0}## and
    ##P_{2}(x)=k_{m}x^n+...+k_{1}x+k_{0}##

    So ##f(P_{1}(x)P_{2}(x))=c_{0}k_{0}=f(P_{1}(x)f(P_{2}(x))##
    Also, ##f(P_{1}(x)+P_{2}(x))=c_{0}+k_{0}=f(P_{1}(x)+f(P_{2}(x))##
    And lastly, ##f(1)=1##.

    So indeed this is a ring homomorphism, but obviously it is not surjective or injective.


    Is this right?
     
  5. Apr 7, 2013 #4

    jbunniii

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    Yes, that's the example I had in mind.
     
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