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Show whether two quotient rings are isomorphic

  1. Apr 25, 2017 #1
    1. The problem statement, all variables and given/known data
    Sorry for the multiple postings. I actually solved the other problems, so I have this last one:

    Are ##\mathbb{Z}_3[x] / \langle x^2 + 2x + 1 \rangle ## and ##\mathbb{Z}_3[x] / \langle x^2 + x + 2 \rangle## isomorphic

    2. Relevant equations


    3. The attempt at a solution
    It seems like they wouldn't be, because one ideal is maximal while the other isn't, but I am not sure if this is correct.
     
  2. jcsd
  3. Apr 25, 2017 #2

    fresh_42

    Staff: Mentor

    Which is maximal? Have you factored the polynomials?
     
  4. Apr 25, 2017 #3
    ##x^2 + 2x + 1 = (x+1)^2## but ##x^2 + x + 2## can't be factored because it has no roots in the field, hence it is maximal
     
  5. Apr 25, 2017 #4

    fresh_42

    Staff: Mentor

    Yes, you're right. I thought it was ##x^2+x-2##. So one quotient ring is a field and the other one not even an integral domain.
     
  6. Apr 25, 2017 #5
    Also, one more thing. How does one list out the elements of quotient rings such as these? Also, is there a general way to count the number of elements in ##
    \mathbb{Z}_n[x] / \langle p(x) \rangle##, where ##p## is an mth degree polynomial?
     
  7. Apr 25, 2017 #6

    fresh_42

    Staff: Mentor

    Yes, the equivalence classes of ##{R}[x]/\langle p(x) \rangle## are all of the form ##[r(x)]_{p(x)} = r(x) + \{s(x)\cdot p(x)\,\vert \,s(x) \in R[x]\}## which means, one can push all monomials of degree ##m := \operatorname{deg}p## and higher into the ideal. However, it gets a little bit more complicated, if the highest coefficient isn't a unit (and I would have to do the math first). Let's therefore assume the highest coefficient is ##1## and ##p(x) = x^m +\ldots + p_1x+p_0##. Then ##[x^m]_{p(x)}= -[p_{m-1}x^{m-1}]_{p(x)} - \ldots - [p_{1}x]_{p(x)} - [p_{0}]_{p(x)}## which means all elements ##x^n## with ##n \geq m## can be reduced to lower powers. Now multiply the possible coefficients with the degree and you have the number of elements. Or better: ## \{[1]_{p(x)}, \, \ldots \,, [x^{m-1}]_{p(x)} \}## forms a basis of the ##R-## module (or vector space in case ##R## is a field). Of course if ##R## isn't finite, you will still get a finite dimension, but infinitely many elements.
     
  8. Apr 25, 2017 #7
    So in this case, for both of the quotient rings, would we have ##3^2 = 9## elements?
     
  9. Apr 25, 2017 #8

    fresh_42

    Staff: Mentor

    I get six.
     
  10. Apr 25, 2017 #9
  11. Apr 25, 2017 #10

    fresh_42

    Staff: Mentor

    Because (for short without the annoying ##[.]_{p(x)}##) we have a basis ##\{1,x,x^2,x^3,\ldots , x^{m-1}\}##. These are all monomials of lower degree than ##m##. All higher can be reduced with subtractions by ##p(x)## to a lower one. And ##p(x)=0## in this quotient ring, so the subtractions do no harm to the ring element. This means ##m## basis vectors (or elements) and each can carry ##n## elements.

    And I have to correct myself. Your ##9## was correct, I mistakenly added the ##n##'s instead of multiplying them. I should get some sleep ...
     
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