# Show that $a - b \mid f(a) - f(b)$

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

• What's $f$? The statement clearly does not hold for all possible functions $f$... – 5xum Mar 20 '15 at 11:03
• What is $f$? What are $a$ and $b$? – Andrea Mori Mar 20 '15 at 11:04
• Does $(5-1)|(5!-1!)$ ? Anyway, the property is true for any polynomial of integer coefficients. – Yves Daoust Mar 20 '15 at 11:06
• does OP mean for which f is this true? – JMP Mar 20 '15 at 11:12

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 $k\geq0$.

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.

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.

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 $$f(a) \equiv f(b) \pmod{a-b}$$.