Results 1 to 5 of 5

Math Help - Polynomial rings - zero divisors

  1. #1
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    558
    Thanks
    2

    Polynomial rings - zero divisors

    Let R = Z/4Z

    Find all the zero divisors of R[x] ...

    ... and then, further, find all elements of R[x] that are neither units nor zero divisors.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,310
    Thanks
    687

    Re: Polynomial rings - zero divisors

    hint: what does the homomorphism used in another post suggest?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    558
    Thanks
    2

    Re: Polynomial rings - zero divisors

    I presume you mean use the following ring homomorphism:

     \phi:  (Z/4Z) [x] ---> (Z/2Z)[x] [then Z/2Z is a field and the units of Z/2Z[x] are simply the units of Z/2Z ... but zero divisors???)

    Then .... ??/

    If a(x) is a zero divisor in (Z/4Z)[x] then a(x)b(x) = 0 for some b(x) with a(x) \ne 0 and b(x) \ne 0

    Then  \phi (a(x)b(x)) = \phi (a(x)) \phi (b(x)) = \phi (0) = 0 with a(x) \ne 0 and b(x) \ne 0 ...


    But ... where to now??

    What can one do with a(x) \ne 0 and b(x) \ne 0 ... how to progress ... that is what does this imply about  \phi (a(x)) and  \phi (b(x))


    Maybe we could proceed by arguing that  \phi (a(x)) \phi (b(x)) = 0 implies that

     ([a_n] x^n + [a_{n-1} x^{n-1} + ... + a_1 x + a_0) ([b_m] x^m+ [b_{m-1} x^{m-1} + ... + b_1 x + b_0)  = 0

    But this would mean that

    [a_n] [b_m] = 0 [mod 2] which means  [a_n] = 0 or  [b_m] = 0 or both

    This seems to imply that Z/2Z[x] has no zero divisors .... but where to from here ... need some further clarification and help.

    Peter
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,310
    Thanks
    687

    Re: Polynomial rings - zero divisors

    of course it has zero divisors, for example:

    every polynomial of the form 2(f(x)) is a zero-divisor.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    558
    Thanks
    2

    Re: Polynomial rings - zero divisors

    Are you referring to Z/4Z which certainly has zero divisors of that form?

    I was referring to Z/2Z[x] which (I think) has no zero divisors [2Z is a prime ideal so Z/2Z has no zero divisors and thus Z/2Z[x] has no zero divisors.

    Is that correct?

    I would like your guidance on the use of the homomorphism  \phi Z/4Z[x] ---> Z/2Z[x]

    It seems that the kernel of the transformation is those polynomials of the form 2(f(x)) which are the zero divisors of Z/4Z. Can we use this somehow?

    Peter
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Rings and zero divisors
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 28th 2011, 08:58 AM
  2. Replies: 1
    Last Post: October 7th 2010, 07:38 PM
  3. Rings with unity and zero-divisors
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 8th 2009, 07:45 AM
  4. Rings, and zero divisors
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 2nd 2009, 10:27 PM
  5. Zero Divisors in Polynomial Ring
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 17th 2007, 12:36 PM

Search Tags


/mathhelpforum @mathhelpforum