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Abstract algebra

  1. Jun 21, 2015 #1
    1. The problem statement, all variables and given/known data
    Ring homomorphism.png

    Hello guys
    So I have the following problem, given the mapping above I have to check weather it's ring homomorphism, and
    maybe monomorphism or epimorphism.

    3. The attempt at a solution

    So the mapping is obviously well defined, and I have proven it's homomorphism, and it's obviously not monomorphism because a polynomial P(x)= 5 - x2 is in the kernel so kernel is not trivial.
    I am not sure how to prove if the function is surjective or not, obviously if the codomain were integers for every integers C , I could just use constant function p(x) = C and function would be surjective.
     
  2. jcsd
  3. Jun 21, 2015 #2
    It is just an intuition, but I doubt there are polynomials of ##\mathbb{Z}[X]## such that ##p(\sqrt{5}) \in \mathbb{Q}-\mathbb{Z} ##.
     
  4. Jun 22, 2015 #3

    Zondrina

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    Showing the map is a homomorphism shouldn't be too difficult. Simply show the map satisfies the properties of a homomorphism.

    The rest of the question is asking if the map is also a bijection, or something more specific. Try applying the first isomorphism theorem if you know it, and use the fact that ##x^2 - 5## is the minimal polynomial.
     
  5. Jun 22, 2015 #4

    Dick

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    Isn't it pretty obvious that the range is contained in the set ##a+b \sqrt{5}## where ##a## and ##b## are integers? Why isn't that all of ##R##?
     
  6. Jun 23, 2015 #5
    I managed to show that function is not surjective with the hint that every element in the range is of the form ## a+b \sqrt{5} ## because for example ##\sqrt{2}## doesn't get hit by any element in domain. Is it also valid argument that function can't be surjective because ##\mathbb{R}## is uncountable whilst ##\mathbb{Z}[X]## is countable?
     
  7. Jun 23, 2015 #6

    Dick

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    Both of those arguments are good. The second makes it obvious if you know cardinality.
     
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