若设不可约多项式p是f‘的p-1重因式,记f'=gp^(k-1),且(p,g)=1
若f'|f,则可设f=hf'=hgp^(k-1) => f'=(hg)'p^(k-1)+(k-1)hgp^(k-2)
=[(hg)'p+(k-1)hg]p^(k-2),∴p|(hg)'p+(k-1)hg
即p|hg => p|h,可记h=pq,则f=qgp^k,若p|q,则p^(k+1)|f
∴p^k|f',这与p为f'的k-1重因式矛盾,∴(p,q)=1,即(p,qg)=1
即p为f的k重因式
若设不可约多项式p是f‘的p-1重因式,记f'=gp^(k-1),且(p,g)=1
若f'|f,则可设f=hf'=hgp^(k-1) => f'=(hg)'p^(k-1)+(k-1)hgp^(k-2)
=[(hg)'p+(k-1)hg]p^(k-2),∴p|(hg)'p+(k-1)hg
即p|hg => p|h,可记h=pq,则f=qgp^k,若p|q,则p^(k+1)|f
∴p^k|f',这与p为f'的k-1重因式矛盾,∴(p,q)=1,即(p,qg)=1
即p为f的k重因式