Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Ring theory question

  1. May 4, 2012 #1
    Hello,

    I wanted to know something regarding the quotient ring Z[x]/pZ[x], where Z[x] is the set of all polynomials with integer coefficients and pZ[x], for a prime number p, is the set of all polynomials with integer coefficients divisible by p. I'm currently working through some notes on Applying Basic Abstract Algebra to Problems of Number Theory, and I don't understand the proof of Eisenstein's criterion. My confusion hinges upon this issue. The author of the notes says that if p(x) is a reducible polynomial with integer coefficients, then p(x)+Z[x]={q(x)+Z[x]}{r(x)+Z[x]}, where q(x) and r(x) are elements of Z[x] such that q(x)r(x)=p(x). Now, I know this seems really basic, but I have a question: Doesn't this hold for general integers rather than prime numbers? Must p be prime?

    Thanks,
    Mathguy
     
    Last edited: May 4, 2012
  2. jcsd
  3. May 4, 2012 #2


    Where in what you quoted is the word "prime" written? If [itex]\,\,f(x)\in\mathbb Z[x]\,\,[/itex] is reducible then [itex]\,\,\exists q(x)\,,\,r(x)\in\mathbb Z[x]\,\,[/itex] s.t.

    [itex]f(x)=q(x)r(x)[/itex] . Why you add to each term in this equation the [itex]\,\,+\mathbb Z[x]\,\,[/itex] is beyond my comprehension, but I can try to guess that

    what the book/you is/are really trying to convey is that from the last equality it follows at once that [tex]f(x)+p\mathbb Z[x]=\left(q(x)+p\mathbb Z[x]\right)\left(r(x)+p\mathbb Z[x]\right)[/tex] which is an equality in the quotient ring [itex]\,\,\mathbb Z[x]/p\mathbb Z[x]\cong \left(\mathbb Z/p\mathbb Z\right)[x][/itex] .


    DonAntonio
     
  4. May 4, 2012 #3
    Sorry, I meant to add pZ[x] to each term. Regarding the equality you stated, doesn't that hold more generally, or is it special to primes because pZ[x] is a prime ideal?

    Thanks,
    mathguy
     
  5. May 4, 2012 #4


    If you meant the equality [itex]\,\,f(x)=q(x)r(x)\,\,[/itex] then this is the meaning of "the element f(x) is reducible" in any ring , or when

    it is given that the element q(x) divides the element f(x)...and this much is true for any kind of elements in a ring, not only polynomials.

    DonAntonio
     
  6. May 4, 2012 #5
    Oh, sorry again, I'm being awfully slow today. I meant the equality f(x)+pZ[x]={q(x)+pZ[x]}{r(x)+pZ[x]}. Doesn't this hold more generally for integers rather prime numbers?
     
  7. May 4, 2012 #6


    Oh, this is only the definition of product in the quotient ring [itex]\,\,\mathbb{Z}[x]/p\mathbb{Z}[x]\,\,[/itex] ...

    I don't understand your question about integers and primes,

    DonAntonio
     
  8. May 4, 2012 #7
    Oh, well, I understand the argument used to prove Eisenstein's criterion now. My original question is now irrelevant.

    Thanks for the help, though.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Ring theory question
  1. Ring theory (Replies: 3)

  2. Ring theory (Replies: 3)

  3. Ring theory (Replies: 11)

  4. Ring Theory Question (Replies: 2)

  5. Ring Theory Question (Replies: 1)

Loading...