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I was given this proposition but I was never able to prove it. Does anyone know how to solve this?

Let $f$ be a polynomial in $\mathbb{Z}/p\mathbb{Z}[x]$, where $p$ is prime. Then $f$ has at most $\deg f$ roots.

Bill Dubuque
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Chris
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  • The coefficient ring is contained in a field, $\Bbb Q_p$, (since $p$ is prime, $\Bbb Q_p$ is a field) of characteristic $0$ when $p$ is prime, and all characteristic $0$ polynomials are separable. (I'm using the characterization of separable as irreducible factors have no multiple roots) – Adam Hughes Jul 16 '14 at 00:17
  • Alternatively, we know that there are roots out there "somewhere" (i.e. in an algebraic closure) so we can use the fact that polynomials over a field are a Euclidean domain, hence have unique factorization, and the irreducibles in an algebraically closed field are exactly the monomials. – Adam Hughes Jul 16 '14 at 00:29
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    @AdamHughes: $\mathbb{Z}_p$ is already a field when $p$ is a prime. – hardmath Jul 16 '14 at 01:56
  • What? Oh you mean $\Bbb Z/p\Bbb Z$, not $\Bbb Z_p$, the former is the field with $p$ elements, the latter is the $p$-adic numbers, though I know some use the latter notation for the former some times. – Adam Hughes Jul 16 '14 at 02:06

2 Answers2

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Hint $ $ If $\,f\ne 0\,$ has $\:\!\ge n =\deg f\,$ distinct roots $\,r_i$ then inductively applying the Factor Theorem (as below) writes $\,f = c(x\!-\!r_1)\cdots (x\!-\!r_n),\,$ so $\ r\ne r_i\,\Rightarrow\, f(r)= c(r\!-\!r_1)\cdots (r\!-\!r_n) \ne 0\,$ since every factor is $\ne 0,\,$ and $\,\Bbb Z_p\,$ is an integral $\rm\color{#c00}{domain}$. Thus $\,f\,$ has at most $\,n\,$ roots.

Alternatively $\,f(r_i)= 0 \,\Rightarrow\, x-r_i \mid f(x).\,$ Being divisible by the nonassociate $\rm\color{#0a0}{primes}$ $\,x-r_i\,$ it follows $\:\!f\,$ is divisible by their lcm = product, so there can be at most $\,\deg(f)\,$ factors ($\leftrightarrow$ roots). (recall $\,x-r\,$ is $\rm\color{#0a0}{prime}$ in $\,R[x]\!\iff\! R[x]/(x-r)\cong R\,$ is a $\rm\color{#c00}{domain}$).

Bifactor Theorem $\ $ Let $\rm\,a,b\in R,\,$ a commutative ring, and $\rm\:f\in R[x]\:$ a polynomial over $\,\rm R.\,$
If $\rm\ \color{#0a0}{a\!-\!b}\ $ is $\,\color{#c00}{\rm cancellable}\,$ in $\rm\,R\,$ (i.e. not a zero-divisor, i.e. regular) $ $ then

$$\rm f(a)\! =\! 0\! =\! f(b) \iff f = (x\!-\!a)(x\!-\!b)\ h, \ for\ some\ h\in R[x],$$

so by induction $\rm \,(x\!-\!a)^k\mid f,\ g\mid f\,\Rightarrow\, (x\!-\!a)^k g\mid f\ $ if $\rm\,\color{#0a0}{g(a)}\,$ is $\rm\color{#c00}{cancellable}$.

Proof $\,\ (\Leftarrow)\,$ clear. $\ (\Rightarrow)\ $ Applying Factor Theorem twice, while canceling $\rm\: \color{#0a0}{a\!-\!b},$

$$\begin{eqnarray}\rm\:f(b)= 0 &\ \Rightarrow\ &\rm f(x)\, =\, (x\!-\!b)\,g(x)\ \ for\ \ some\ \ g\in R[x]\\[.1em] \rm f(a) = (\color{#0a0}{a\!-\!b})\,g(a) = 0 &\color{#c00}\Rightarrow&\rm g(a)\, =\, 0\,\ \Rightarrow\ g(x) \,=\, (x\!-\!a)\,h(x)\ \ for\ \ some\ \ h\in R[x]\\[.1em] &\Rightarrow&\rm f(x)\, =\, (x\!-\!b)\,g(x) \,=\, (x\!-\!b)(x\!-\!a)\,h(x)\end{eqnarray}$$

The claimed induction: $ $ let $\,\rm P_{j}\,$ denote $\rm\,\color{#08f}{(x\!-\!a)^{j}\:\!g}\mid f.\,$ Base case: $\,\rm P_0\,$ is true by hypothesis. Inductive step: suppose $\rm P_n$ is true, and $\,\rm n <\color{darkorange} k.\,$ Then $\rm\,(x\!-\!a)^{n+1}\mid (x\!-\!a)^k\mid f,\,$ i.e.

$\rm (x\!-\!a)^{n+1}\!\mid \color{#08f}{(x\!-\!a)^ng}\,h\Rightarrow x\!-\!a\mid g\:\!h\Rightarrow \color{#0a0}{g(a)} h(a)\!=\!0\color{#c00}\Rightarrow h(a)\!=\!0\Rightarrow x\!-\!a\mid h$

so $\rm\,h = (x\!-\!a)\bar h\,$ so $\rm\,f = (x\!-\!a)^{n+1}g\bar h,\,$ i.e. $\,\rm P_n\Rightarrow P_{n+1},\,$ so $\rm \,P_n\,$ is true for $\,\rm n={\color{darkorange} k}$.

Remark $\ $ The theorem may fail when $\rm\ \color{#0a0}{a\!-\!b}\ $ is not $\rm\color{#c00}{cancelable}$ e.g.

$$\rm mod\ 8\!:\,\ f=x^2\!-1\,\Rightarrow\,f(3)\equiv 0\equiv f(1)\ \ but\ \ x^2\!-1\not\equiv (x\!-\!3)(x\!-\!1)\equiv x^2\!-4x+3\qquad$$

Bill Dubuque
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  • See here for the simpler case when the coefficient ring is an integral domain (e.g. a field), and further elaboration. – Bill Dubuque Jun 09 '21 at 01:14
  • i get all of it except for the deduction $(x - a)^{n + 1} \mid (x - a)^n g h \implies x - a \mid gh$. how do you cancel $(x - a)^n$ from both sides? edit i guess this follows from the fact that $(x - a)^n$ is monic? – BD107 Mar 05 '24 at 03:34
  • @BD107 It's standard divisibility cancellation (of a cancellable element $,a),,$ i.e. $,ab\mid ac \Rightarrow a\mid c,$ by cancelling $,a,$ from $,abd = ac.,$ Monic polynomials $,a,$ are cancellable since multiplying by them can't decrease the degree, so $,ad=0\Rightarrow d=0,,$ so $,ab=ac\Rightarrow a(b!-!c)=0\Rightarrow b!-!c=0\Rightarrow b=c.\ \ $ – Bill Dubuque Mar 05 '24 at 03:55
  • yup; got it; that's what i was expecting; thanks. – BD107 Mar 05 '24 at 04:08
  • @BD107 When $,g(a),$ is a unit this implies $,g(x),$ is coprime to $,x!-!a,$ by $,(x!-!a,g(x))=(x!-!a,g(a)) = (1),,$ thus $,g(x),$ is coprime to $,(x!-!a)^k,$ so by Euclid's Lemma $,(x!-!a)^k!\mid gh!=!f\Rightarrow (x!-!a)^k!\mid h\Rightarrow (x!-!a)^kg\mid f.,$ We can reduce the Theorem to this case by passing to the total fraction ring, which adjoins inverses for all cancellable elements. – Bill Dubuque Mar 05 '24 at 04:26
  • i.e. Euclid's Lemma $\Rightarrow$ LCM = product for coprimes $\ $ @BD107 $\ \ $ – Bill Dubuque Mar 05 '24 at 04:47
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A polynomial of degree $n$ with coefficients in any field can have at most $n$ roots, because $X-\alpha$ is a factor of $f$ whenever $\alpha$ is a root, and the polynomial ring is a UFD.

As a consequence, a polynomial with coefficients in any integral domain can have at most $n$ roots: just consider the roots in the fraction field. (In fact, this property characterizes integral domains.)

Bruno Joyal
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    This doesn't require using that the polynomial ring is a UFD, but only the much simpler fact that $,x - \alpha,$ is prime $(!\iff!$ the coefficient ring is a domain) - see my answer. – Bill Dubuque Jul 16 '14 at 00:55