## Zero-divisors in polynomial rings

Posted: November 1, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , , ,

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, $a_rb_s=0. \ \Box$