Polynomials over $\mathbb{Z}/p\mathbb{Z}$

Proposition 1.1   Let $f\in(\mathbb{Z}/p\mathbb{Z})[x]$ be a nonzero polynomial over the ring $\mathbb{Z}/p\mathbb{Z}$. Then there are at most $\deg(f)$ elements $\alpha\in\mathbb{Z}/p\mathbb{Z}$ such that $f(\alpha)=0$.

Proof. We proceed by induction on $\deg(f)$. The cases $\deg(f)=0,1$ are clear. Write $f = a_n x^n + \cdots a_1 x + a_0$. If $f(\alpha)=0$ then

$$f(x) = f(x) - f(\alpha)$$

$$= a_n(x^n-\alpha^n) + \cdots a_1(x-\alpha) + a_0(1-1)$$

$$= (x-\alpha)(a_n(x^{n-1}+\cdots + \alpha^{n-1}) + \cdots a_1)$$

$$= (x-\alpha)g(x),$$

for some polynomial $g(x)\in(\mathbb{Z}/p\mathbb{Z})[x]$. Next suppose that $f(\beta)=0$ with $\beta\neq \alpha$. Then $(\beta-\alpha) g(\beta) = 0$, so, since $\beta-\alpha\neq 0$ (hence $\gcd(\beta-\alpha, p)=1$, we have $g(\beta)=0$. By our inductive hypothesis, $g$ has at most $n-1$ roots, so there are at most $n-1$ possibilities for $\beta$. It follows that $f$ has at most $n$ roots. $\qedsymbol$

theorem_type[example][theorem][][remark][][] Find the roots¹ of displaymath f = x^4 - x^3 + x^2 +x +1 &isin#in;(Z/3Z)[x]. Solution: tex2html_wrap_inline$f(0) = 1$, tex2html_wrap_inline$f(1)=0$, tex2html_wrap_inline$f(2)=0$, so the roots are tex2html_wrap_inline${1,2}$. Also, displaymath f = (x+1)(x+2)(x^2+2x+2). Incidentally, PARI can factor polynomials over finite fields very quickly: tex2html_preform ? factormod(x^4 - x^3 + x^2 +x +1,3) %1 = [Mod(1, 3)*x + Mod(1, 3) 1] [Mod(1, 3)*x + Mod(2, 3) 1] [Mod(1, 3)*x^2 + Mod(2, 3)*x + Mod(2, 3) 1]

Proposition 1.2   Let $p$ be a prime number and let $d$ be a divisor of $p-1$. Then $f(x) = x^d-1\in(\mathbb{Z}/p\mathbb{Z})[x]$ has exactly $d$ solutions.

Proof. Let $e$ be such that $de=p-1$. We have

$$x^{p-1} - 1 = (x^d)^e - 1$$

$$= (x^d - 1)((x^d)^{e-1} + (x^d)^{e-2} + \cdots + 1)$$

$$= (x^d - 1)g(x),$$

where $\deg(g(x)) = p-1-d$. Recall that Fermat's little theorem implies that $x^{p-1}-1$ has exactly $p-1$ roots in $\mathbb{Z}/p\mathbb{Z}$. By Proposition 1.1, $f(x)$ has at most $p-1-d$ roots and $x^d-1$ has at most $d$ roots, so $g(x)$ has exactly $p-1$ roots and $x^d-1$ has exactly $d$ roots, as claimed. $\qedsymbol$

WARNING: The analogue of this theorem is false for some $f\in(\mathbb{Z}/n\mathbb{Z})[x]$ with $n$ composite. For example, if $n=n_1\cdot n_2$ with $n_1, n_2\neq 1$, then $f=nx$ has at least two distinct zeros, namely 0 and $n_2\neq 0$.