Posts Tagged ‘zero-divisor’

Problem. (McCoy) Let R be a commutative ring with identity and let f(x)= \sum_{i=0}^n a_ix^i \in R[x].

1) Prove that f(x) is a zero-divisor if and only if there exists some 0 \neq c \in R such that cf(x) = 0.

2) Prove that if R is reduced and f(x)g(x)=0 for some g(x)=\sum_{i=0}^m b_ix^i \in R[x], then a_ib_j=0 for all 0 \leq i \leq n and 0 \leq j \leq m.

Solution. 1) If there exists 0 \neq c \in R such that cf(x)=0, then clearly f(x) is a zero-divisor of R[x]. For the converse, let g(x)=\sum_{i=0}^m b_ix^i, \ b_m \neq 0, be a polynomial with minimum degree such that f(x)g(x)=0. I will show that m = 0. So, suppose to the contrary, that m \geq 1. If a_jg(x)=0, for all j, then a_jb_m=0, for all j, and so b_mf(x)=0 contradicting the minimality of m because \deg b_m = 0 < m. So we may assume that the set \{j: \ a_jg(x) \neq 0 \} is non-empty and so we can let

\ell=\max \{j : \ a_jg(x) \neq 0 \}.

Then

0=f(x)g(x)=(a_{\ell}x^{\ell} + \cdots + a_0)(b_mx^m + \cdots + b_0).

Thus

a_{\ell}b_m=0 and so a_{\ell}g(x)=a_{\ell}b_{m-1}x^{m-1} + \cdots + a_{\ell}b_0.

Hence \deg a_{\ell}g(x) < m=\deg g. But we have f(x)(a_{\ell}g(x))=a_{\ell}f(x)g(x)=0, which is impossible because g(x) was supposed to be a polynomial with minimum degree satisfying f(x)g(x)=0.

2) The proof of this part is by induction over i+j. It is obvious from f(x)g(x)=0 that a_0b_0=0. Now let 0 < \ell \leq m+n and suppose that a_rb_s=0 whenever 0 \leq r+s < \ell. We need to show that a_rb_s=0 whenever r+s=\ell. So suppose that r+s=\ell. The coefficient of x^{\ell} in f(x)g(x) is clearly

0=\sum_{i < r, \ i+j=\ell}a_ib_j+ a_rb_s + \sum_{i > r, \ i+j=\ell} a_ib_j,

which after multiplying both sides by a_rb_s gives us

\sum_{i < r, \ i+j=\ell} a_rb_sa_ib_j+ (a_rb_s)^2 + \sum_{i > r, \ i+j=\ell} a_rb_sa_ib_j=0.

Call this (1). Now in the first sum in (1), since i < r, we have i+s < r+s=\ell and hence by the induction hypothesis a_ib_s=0. Thus a_rb_sa_ib_j=0. So the first sum in (1) is 0. In the second sum in (1), since i > r and i+j=r+s=\ell, we have j < s. Therefore by the induction hypothesis a_rb_j=0 and hence a_rb_sa_ib_j=0. So the second sum in (1) is also 0. Thus (1) becomes (a_rb_s)^2=0 and so, since R is reduced, a_rb_s=0. \ \Box