How is association between polynomial expression and function unique modulo a prime when its exponents are less than $p-1$?

Question

My course slides say that the association between a polynomial expression $a_d x^d+a_{d-1} x^{d-1}+\cdots+a_1 x+a_0$ and a function isn't unique modulo a prime, because due to Fermat's little thrm, exponents $p-1$ or greater can be reduced. It goes on to say that this association is unique modulo a prime when all of the exponents are less than $p-1$. However, how is this the case when the coefficients of $x$ may be greater than $p$ and can also be reduced? I am not understanding this statement. Am I misinterpreting the sense in which a unique association is meant?

Answer 1

Since you are not familiar with rings we can instead work with polynomial congruences. Define $\,\Bbb Z[x]\,$ to be the set of all polynomials with integer coefficients with the usual sum and product operations. If $\,f,g\in\Bbb Z[x]\,$ then we define $\,f\equiv g\pmod{\!p}\iff p\mid f-g\,$ in $\Bbb Z[x]\iff$ all coef's of $\,f-g\,$ are divisible by $\,p\iff$ corresponding coef's of $\,f,g\,$ are congruent $\!\bmod p.\,$

Then $\,x^p\not\equiv x\,$ since $\,x^p-x\,$ has lead coef $\,1\,$ not divisible by $\,p.\,$ But by little Fermat $\,x^p\,$ and $\,x\,$ are equal functions $\!\bmod p,\,$ i.e. their values are congruent for $\,x\equiv 0,1,2,\ldots,p\!-\!1.\,$ But if $\,f\not\equiv g\,$ both have degree $< p\,$ then so too does their difference $h\not\equiv 0\,$ so it is not the zero function since it has at most $\,\deg h < p\,$ roots $\!\bmod p\,$ (as follows by iterating the Factor Theorem as here).