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I have seen this lemma elsewhere.

Suppose $A$ is a domain, and $f \in A[X]$. Prove that

$$a - b \mid f(a) - f(b)$$


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

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    $\begingroup$ What's $f$? The statement clearly does not hold for all possible functions $f$... $\endgroup$ – 5xum Mar 20 '15 at 11:03
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    $\begingroup$ What is $f$? What are $a$ and $b$? $\endgroup$ – Andrea Mori Mar 20 '15 at 11:04
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    $\begingroup$ Does $(5-1)|(5!-1!)$ ? Anyway, the property is true for any polynomial of integer coefficients. $\endgroup$ – Yves Daoust Mar 20 '15 at 11:06
  • $\begingroup$ does OP mean for which f is this true? $\endgroup$ – JMP Mar 20 '15 at 11:12
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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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This follows by specializing the Polynomial Factor Theorem (below), namely if $\,f\in A[x]\,$ then

$$\begin{align} x\!-\!b\ \ \ \mid\ &\ f(x)-f(b)\ \ \ {\rm in}\ \ A[x]\\[.2em] \Rightarrow\ \ g(x)\, (x\!-\!b) =&\ f(x)-f(b)\,\ \ {\rm for\ some}\ \ g\in A[x]\\[.2em] \Rightarrow\ \ g(a)\, (a\!-\!b)\, =&\ f(a)-f(b)\,\ \ {\rm by\ eval\ at}\ \ x=a\\[.2em] \Rightarrow\qquad\ \ \ \ \, a\!-\!b\ \ \ \,\mid\ &\, f(a)-f(b)\ \ \ {\rm in}\ \ A\ \end{align}\ \ \ $$

Polynomial Factor Theorem $\ $ If $A$ is a commutative ring, $\,b\in A\,$ and $\,f(x)\in A[x]\,$ then

$$\ x\!-\!b\,\mid\, f(x)-f(b)\ \ {\rm in}\ \ A[x]\quad $$

Proof $\ \ {\rm mod}\,\ x\!-\!b\!:\,\ \color{#c00}{x\equiv b}\,\Rightarrow\, f(\color{#c00}{x})\equiv f(\color{#c00}b)\ $ by the Polynomial Congruence Rule.

Remark $\ $ The linked proof of the congruence rule is given in $\,\Bbb Z\,$ but it is in fact valid in any commutative ring, since the proofs use only commutative ring axioms.

Alternatively, use the Polynomial Division Algorithm to write $\,f(x)-f(b) = (x\!-\!b)g(x) + r\,$ then conclude $\,r = 0\,$ by evaluation at $\, x = b.\,$

Alternatively, use linearity and the high-school formula for $\,(x^n-a^n)/(x-a)$

The universal Polynomial Factor Theorem $\ x\!-\!y\,\mid\, f(x)-f(y)\in \Bbb Z[x,y]\,$ is the special case $\, A=\Bbb Z[y]\,$ and $\,b = y.\,$ This allows us to deduce all "number" instances as specializations of "function" (polynomial) cases. It is a simple prototypical example of deducing number divisibility as a special case of polynomial divisibility. See here for more on universality of polynomial identities.

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I'll assume that with $f$ you mean a polynomial with integer coefficients.

By setting $k=a-b$, it is sufficient to prove that $f(b+k)-f(b)$ is divisible by $k$. Now, consider the polynomial in two variables $$ g(x,y)=f(x+y)-f(x) $$ Then $g(x,0)=0$, which means that $g(x,y)$ is divisible by $y$ in the polynomial ring $\mathbb{Z}[x,y]$; thus $g(x,y)=yh(x,y)$ for some $h\in\mathbb{Z}[x,y]$.

Therefore $$ f(b+k)-f(b)=g(b,k)=kh(b,k) $$ is divisible by $k$ as required.

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The universal property of the polynomial ring $A[X]$ asserts that for every $A$-algebra $C$ and $c \in C$, there exists a unique ring homomorphism $\text{ev}_c : A[X] \to C$ which sends $X$ to $c$.

If $f \in A[X]$ is a polynomial, we commonly denote $\text{ev}_c(f) = f(c)$.

Let $\pi : A \to A/(a-b)$ be the projection.

Take $B = A$ with $c = a$ and $B = A$ with $c = B$. This gives us two ring homomorphisms $\pi \circ \text{ev}_a, \pi \circ \text{ev}_b$, which both send $X$ to $\pi(a) = \pi(b)$. From the uniqueness in the universal property, it follows that $$\pi \circ \text{ev}_a = \text{ev}_{\pi(a)} = \pi \circ \text{ev}_b$$ That is, $\pi(f(a)) = \pi(f(b))$ for all $f \in A[X]$, so that $f(a) \equiv f(b) \pmod{a-b}$.

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