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Is it true that when considering a degree $d$ polynomial $p(x)$ in a composite modulus $q$, it has more than $d$ roots (i.e., more than $d$ solutions to $p(x) \equiv 0 \pmod q$), and that when we consider $p(x)$ in a prime modulus $p$, it still satisfies the property of polynomials that it will have at most $d$ roots? If so, why? If not, is it only the case that with a composite modulus, we will sometimes but not always get more than $d$ roots?

I am asking because I have seen that $x^3 \equiv 0 \pmod 8$ has more than $3$ roots, but whenever I work over a prime it seems to always have at most $d$ roots, which has made me wonder what the relationship is between the modulus and how many roots there are, and most importantly why this relationship is the case.

Princess Mia
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  • As proved in the dupes, it is true in domains by iterating the factor theorem, and there are simple counterexamples in nondomains, e.g. if $,ab=0, a,b\neq 0,$ then $,ax,$ has at least $2$ roots $,b,0;,$ and $,(x-a)(x-b),$ has at least $,4,$ roots $,a,b,0,a+b, $ if $,a\neq b.\ \ $ – Bill Dubuque Sep 07 '23 at 23:47
  • @BillDubuque I understand how the iterating factor theorem shows that $p(x)$ has at most $d$ roots if we aren't doing modular arithmetic and just talking about polynomials in general. However, I am not seeing why this line of reasoning still holds when working modulo a prime, and why this line of reasoning fails when working modulo a composite. Why is it the case that $\ r\ne r_i,\Rightarrow, f(r)= c(r!-!r_1)\cdots (r!-!r_n) \ne 0,$ when we are working modulo a prime, but not necessarily when working modulo a composite? (I see why it works in general). thanks – Princess Mia Sep 07 '23 at 23:58
  • The proof by iterating the Factor Theorem needs all root differences to be cancellable, which may fail when there are zero-divisors - see the proof of the Bifactor Theorem in the first dupe. Note that generally for composite moduli we can count the number of roots using CRT (which combines multiplicatively - clarifying how the count can exceed the degree when the modulus has more than one prime factor, e.g. see here and here). $\ \ $ – Bill Dubuque Sep 08 '23 at 00:05
  • Note that it's not true that there must be more roots than the degree for composite moduli, e.g. $x(x-1)$ has only roots $,x=0,1,$ for prime power moduli $p^n$ cf. here. – Bill Dubuque Sep 08 '23 at 00:19
  • To gain insight I recommend that you trace the proof of the BiFactor Theorem for the roots $,x=1,3,$ of $,x^2-1,$ in $\Bbb Z_8$ to see how it breaks down (by not being able to cancel an even mod $8).\ \ $ – Bill Dubuque Sep 08 '23 at 00:36

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