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) \equiv 0 \pmod{a-b} basically.


Let, a - b = c


f(a) - f(b)/(a-b) = f'(\xi) for Some \xi \in (a, b).


But I dont see it showing divisibility


Answer



Let A be a domain with field of fractions K and let f(X)=a_0+a_1X+a_2X^2+\cdots+a_nX^n be a polynomial in A[X]. Then for all a\neq b\in A, \begin{eqnarray} f(b)-f(a)&=&(a_0+a_1b+\cdots+a_nb^n)-(a_0+a_1a+\cdots+a_na^n)\\ &=& (b-a)\left(a_1+a_2\frac{b^2-a^2}{b-a}+\cdots+a_n\frac{b^n-a^n}{b-a}\right) \end{eqnarray} is certainly valid in K. But now note that both factors actually belong to A since b-a divides b^k-a^k for all k\geq0.


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