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Math Help - A commutative Ring R has no zero divisors

  1. #1
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    A commutative Ring R has no zero divisors

    Show that if R has no zero divisors, then R[x] has no zero divisors.

    Since R has no zero divisors, then \forall a,b\in R such that a,b\neq 0, ab\neq 0.

    Is this next part correct (not sure what to do next)?

    R[ab]=R[a]R[b]\neq 0
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    MHF Contributor Siron's Avatar
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    Re: A commutative Ring R has no zero divisors

    Is it not easier to say, if a,b are no zero divisors, then \forall a,b \in R: a.b=0 such a=0 or b=0.
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    Re: A commutative Ring R has no zero divisors

    Quote Originally Posted by Siron View Post
    Is it not easier to say, if a,b are no zero divisors, then \forall a,b \in R: a.b=0 such a=0 or b=0.
    My question isn't how to define a zero divisor. Regardless of how you or I want to define it, how can I use that fact to do the proof?
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    MHF Contributor Siron's Avatar
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    Re: A commutative Ring R has no zero divisors

    Yes indeed, I agree with your proof. Because if R[X] is a polynomial ring, his coefficients are \in R.
    Last edited by Siron; July 24th 2011 at 04:23 AM.
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    Re: A commutative Ring R has no zero divisors

    Quote Originally Posted by dwsmith View Post
    Show that if R has no zero divisors, then R[x] has no zero divisors.

    Since R has no zero divisors, then \forall a,b\in R such that a,b\neq 0, ab\neq 0.

    Is this next part correct (not sure what to do next)?

    R[ab]=R[a]R[b]\neq 0
    This doesn't even make sense. R[x] is the ring of polynomials with variable x, with coeffcients in x. I have no idea what you mean by "R[a]" where a is a member of R. The best I can think of is that it is just a complicated way of saying "R"!
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    MHF Contributor Siron's Avatar
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    Re: A commutative Ring R has no zero divisors

    @HallsofIvy:
    You're absolutely right, but is it not enough to say that the coefficients of the polynomial ring R[X] are \in R?
    And I think dwsmith means with:
    R[a]=\sum_{i=0}^{n} a_iX^{i}
    R[b]=\sum_{i=0}^{n} b_iX^{i}

    That's how I interpreted it.
    Last edited by Siron; July 24th 2011 at 01:28 PM.
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    Re: A commutative Ring R has no zero divisors

    Quote Originally Posted by dwsmith View Post
    Show that if R has no zero divisors, then R[x] has no zero divisors.

    Since R has no zero divisors, then \forall a,b\in R such that a,b\neq 0, ab\neq 0.

    Is this next part correct (not sure what to do next)?

    R[ab]=R[a]R[b]\neq 0

    The first thing that comes to my mind is to prove the contrapositive statement. That is I'm going to prove that if R[X] has any zero divisors the so does R. Here goes:

    Let P,Q \in R[X] such that P,Q \neq 0 and PQ = 0 (the zero polynomial). We will write

    P = \sum_{k=0}^{n} a_k x^k and Q= \sum_{i=0}^{m} b_i x^i..

    Now since P and Q are nonzero then there exist some coefficients a_{k_1} and b_{i_1} such that a_{k_1},b_{i_1} \neq 0 (these coefficients are in R). Now we multiply P and Q to get

    PQ = \sum_{k=0}^{n} \sum_{i=0}^{m} a_k b_i x^{k+i}=0.

    But we know that the only way for PQ to be the zero polynomial is if all its coefficients are 0, and since all the coefficients are zero then a_{k_1}b_{i_1} = 0, but a_{k_1},b_{i_1} \neq 0 so a_{k_1} and b_{i_1} must be zero divisors. qed
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    MHF Contributor Drexel28's Avatar
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    Re: A commutative Ring R has no zero divisors

    Quote Originally Posted by dwsmith View Post
    Show that if R has no zero divisors, then R[x] has no zero divisors.

    Since R has no zero divisors, then \forall a,b\in R such that a,b\neq 0, ab\neq 0.

    Is this next part correct (not sure what to do next)?

    R[ab]=R[a]R[b]\neq 0
    The simple way to do this is to note that if R is an integral domain then \deg(pq)=\deg(p)+\deg(q). So, break the problem in two cases, either p,q\in R or \max\{\deg(p),\deg(q)\}\geqslant 1 from where it follows that \deg(pq)\geqslant 1 and so pq\ne 0.


    Perhaps simpler to say, is just look at the leading coefficient of pq and you tell me why the conclusion follows.
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    Re: A commutative Ring R has no zero divisors

    A little simpler than obd2 but it's the same idea :

    P,Q \in \mathcal R [X]\backslash\{0\} such that PQ=0

    Let k_P = \max \{k\in\mathbb N / X^k | P\} and k_Q = \max \{k\in\mathbb N / X^k | Q\} (those exist because of the non nullity of P and Q). Let now R = \frac P {X^{k_P}}, S = \frac Q {X^{k_Q}}.

    One have 0 = PQ = X^{k_P}X^{k_Q}RS.
    So r_0s_0 is the coefficient of X^{k_P+k_Q}. And so it is null : r_0s_0 = 0. So is r_0 = 0 or s_0 = 0.

    If r_0 = 0, it's a contradiction with the definition of k_P. If s_0 = 0, it's a contradiction with the definition of k_Q.

    QED.

    P.S. : I'm new on this forum, and I love the \LaTeX integration. It's really more efficient and beautiful than in other forums.
    Last edited by pece; August 1st 2011 at 07:07 AM. Reason: [TEX]\LaTeX[/TEX] fix
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