Tuesday, October 2, 2018

calculus - Show that abmidf(a)f(b)


I have seen this lemma elsewhere.


Suppose A is a domain, and fA[X]. Prove that



abf(a)f(b)



I need to prove this.


f(a)f(b)0(modab)

basically.


Let, ab=c


f(a)f(b)/(ab)=f(ξ)

for Some ξ(a,b).


But I dont see it showing divisibility


Answer



Let A be a domain with field of fractions K and let f(X)=a0+a1X+a2X2++anXn

be a polynomial in A[X]. Then for all abA, f(b)f(a)=(a0+a1b++anbn)(a0+a1a++anan)=(ba)(a1+a2b2a2ba++anbnanba)
is certainly valid in K. But now note that both factors actually belong to A since ba divides bkak for all k0.


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