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When working with polynomials in mod $p$ (restricting the coefficients and $x$-values to numbers in $\{0,\ldots, m-1\}$ where $p$ is prime), I am getting that the number of polynomials of at most degree $d$ where $d>p$ is greater than $p^{d+1}$, and am wondering what I am doing wrong.

I am considering polynomials as objects rather than as functions here. First of all, I know that the number should be $p^{d+1}$ as there are $d+1$ coefficients with $p$ possibilities each. However, when I count by value representation of a polynomial (specifying its value at $d+1$ points to uniquely determine a polynomial of at most $d$), I am seeing that the number should be greater. I see that each polynomial can be evaluated at $d+1$ points to uniquely determine the function corresponding to this polynomial object- when considering the number of value representations given fixed $x$ (such as $x=0,1,\ldots d$) there are $p^{d+1}$ possibilities for the value representation. However, each of these value representations corresponds to a unique polynomial function (as a function can't have 2 different values at the same input), and we know there is a polynomial of at most degree $d$ corresponding hereto because of Lagrange Interpolation- so there are at least $p^{d+1}$ polynomial functions in this field. However, each polynomial function corresponds to at least one polynomial object, and there exists polynomial functions which correspond to different polynomial objects, as by Fermat $x^p \equiv x \pmod p$, which we comprehend by having $d>p$. Therefore there are more polynomial objects of at least $d$ than polynomial functions in this field, meaning more than $p^{d+1}$ of them. What have I done wrong?

Princess Mia
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  • $m$ needs to be prime or none of what you've written is true. With that assumption, $x^d \equiv x^0 \bmod m$ only holds when $x \neq 0$. When $x = 0$ they differ. So they actually are different and $m^m$ is the correct count, of polynomial functions (again, when $m$ is prime). One has to distinguish between polynomials and polynomial functions here, also. – Qiaochu Yuan Sep 09 '23 at 06:08
  • @QiaochuYuan did you mean to say $m^{d+1}$ rather than $m^m$? Also, combinatorically how do we count coefficients when sometimes $x^d$ and $x^0$ are the same, but sometimes they are not? Thanks – Princess Mia Sep 09 '23 at 06:18
  • Well, $d = m-1$ so those are the same thing, I don't know why you're using two different variables here. Formally you can show that the polynomial functions $1, x, x^2, \dots x^{m-1}$ are linearly independent $\bmod m$; this amounts to showing that if a polynomial function of degree $\le m-1$ vanishes then its coefficients are identically zero. – Qiaochu Yuan Sep 09 '23 at 06:37
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    It is essential to understand the difference between formal polynomials and polynomial functions, e.g. see here. Many of the stumbling blocks in your recent questions are because you are trying to run before you have learned to walk, i.e. you have not yet studied ring theory but the questions are essentially ring theoretic. These matters will be much clearer once you learn rudimentary ring theory. – Bill Dubuque Sep 09 '23 at 06:44

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For Lagrange interpolation, you need to specify the values of the polynomial at $d+1$ points for a polynomial of degree $d$. But if $d > p$ (or in fact, even if $d \ge p$), then even if you specify at all the elements in the field of integers mod $p$ , then you have only specified $p$ elements which is not enough. So Lagrange interpolation doesn't apply.

Ted
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