充分性显然下证必要性.设f(x)是R[x]的零因子.若f(x)∈R则结论显然.故下设f(x)的次数≥1且 f(x)=a 0 +a 1 x+…+a m-1 x m-1 +a m x m . 由于f(x)是R[x]的零因子故存在R上多项式 g(x)=b 0 +b 1 x+…+b n-1 x n-1 +b n x n ≠0使g(x)f(x)=0其中0≠b n ∈R. 显然只需证明必有次数小于n的多项式与f(x)之积仍为0即可. 由于g(x)f(x)=0故由上知: a m b n =0 (1)但是a m g(x)f(x)=0而 a m g(x)=a m (b 0 +b 1 x+…+b n-1 x n-1 +b n x n ) =a m (b 0 +b 1 x+…+b n-1 x n-1 )因此若a m g(x)≠0则其次数小于n便已得证.若a m g(x)=0则由上知: a m b 0 =a m b 1 =…=a m b n-1 =0从而有 g(x)f(x)=(b 0 +b 1 x+…+b n x n )? (a 0 +a 1 x+…+a m-1 x m-1 )=0由此又得 a m-1 b n =0 (2)于是有 a m-1 g(x)f(x)=a m-1 (b 0 +b 1 x+…+b n-1 x n-1 )f(x)=0. 若a m-1 g(x)≠0则其次数小于n也已得证.若a m-1 g(x)=0则有 g(x)f(x)=(b 0 +b 1 x+…+b n x n )? (a 0 +a 1 x+…+a m-2 x m-2 )=0由此又得 a m-2 b n =0. (3) 如此继续下去或者有次数小于n的多项式与f(x)之积为0或者有 g(x)f(x)=g(x)a 0 =0即有a m g(x)=a m g(x)=…=a 1 g(x)=a 0 g(x)=0.于是由(1).(2)(3)…有 a m b n =a m-1 b n =…=a 0 b n =0.从而b n f(x)=0其中0≠b n ∈R. 充分性显然,下证必要性.设f(x)是R[x]的零因子.若f(x)∈R,则结论显然.故下设f(x)的次数≥1,且f(x)=a0+a1x+…+am-1xm-1+amxm.由于f(x)是R[x]的零因子,故存在R上多项式g(x)=b0+b1x+…+bn-1xn-1+bnxn≠0,使g(x)f(x)=0,其中0≠bn∈R.显然只需证明必有次数小于n的多项式与f(x)之积仍为0即可.由于g(x)f(x)=0,故由上知:ambn=0,(1)但是amg(x)f(x)=0,而amg(x)=am(b0+b1x+…+bn-1xn-1+bnxn)=am(b0+b1x+…+bn-1xn-1),因此,若amg(x)≠0,则其次数小于n,便已得证.若amg(x)=0,则由上知:amb0=amb1=…=ambn-1=0,从而有g(x)f(x)=(b0+b1x+…+bnxn)?(a0+a1x+…+am-1xm-1)=0,由此又得am-1bn=0,(2)于是有am-1g(x)f(x)=am-1(b0+b1x+…+bn-1xn-1)f(x)=0.若am-1g(x)≠0,则其次数小于n,也已得证.若am-1g(x)=0,则有g(x)f(x)=(b0+b1x+…+bnxn)?(a0+a1x+…+am-2xm-2)=0,由此又得am-2bn=0.(3)如此继续下去,或者有次数小于n的多项式与f(x)之积为0,或者有g(x)f(x)=g(x)a0=0,即有amg(x)=am,g(x)=…=a1g(x)=a0g(x)=0.于是由(1).(2),(3),…,有ambn=am-1bn=…=a0bn=0.从而bnf(x)=0,其中0≠bn∈R.