Characterization of zero divisors in a polynomial ring

Let R be a commutative ring with 1 \neq 0. Let p(x) = \sum_{i=0}^n a_ix^i be an element of the polynomial ring R[x]. Prove that p(x) is a zero divisor in R[x] if and only if there is a nonzero b \in R such that bp(x) = 0.


If bp(x) = 0 for some nonzero b \in R, then letting b(x) = b, p(x) is a zero divisor.

Now suppose p(x) is a zero divisor; that is, for some q(x) = \sum_{i=0}^m b_ix^i, p(x)q(x) = 0. We may choose q(x) to have minimal degree among the nonzero polynomials with this property.

We will now show by induction that a_iq(x) = 0 for all 0 \leq i \leq n.

For the base case, note that p(x)q(x) = \sum_{k=0}^{n+m} \left(\sum_{i+j = k} a_ib_j\right) x^k = 0. The coefficient of x^{n+m} in this product is a_nb_m on one hand, and 0 on the other. Thus a_nb_m = 0. Now a_nq(x)p(x) = 0, and the coefficient of x^m in q is a_nb_m = 0. Thus the degree of a_nq(x) is strictly less than that of q(x); since q(x) has minimal degree among the nonzero polynomials which multiply p(x) to 0, in fact a_nq(x) = 0. More specifically, a_nb_i = 0 for all 0 \leq i \leq m.

For the inductive step, suppose that for some 0 \leq t < n, we have a_rq(x) = 0 for all t < r \leq n. Now p(x)q(x) = \sum_{k=0}^{n+m} \left( \sum_{i+j=k} a_ib_j\right) x^k = 0. On one hand, the coefficient of x^{m+t} is \sum_{i+j = m+t} a_ib_j, and on the other hand, it is 0. Thus \sum_{i+j=m+t} a_ib_j = 0. By the induction hypothesis, if i \geq t, then a_ib_j = 0. Thus all terms such that i \geq t are zero. If i < t, then we must have j > m, a contradiction. Thus we have a_tb_m = 0. As in the base case, a_tq(x)p(x) = 0 and a_tq(x) has degree strictly less than that of q(x), so that by minimality, a_tq(x) = 0.

By induction, a_iq(x) = 0 for all 0 \leq i \leq n. In particular, a_ib_m = 0. Thus b_mp(x) = 0.

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Comments

  • mam  On December 31, 2010 at 8:13 am

    i would like to sent note regarding polynomial rings to me!!!!!

    • nbloomf  On January 1, 2011 at 1:03 am

      I am confused.

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