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Given a finite field F with c elements, how would one prove that there are infinitely many polynomials that represent the zero function over the field F? I know we have infinitely many polynomials over this field, isn't it straightforward that by choosing all the coefficients to be zero all those infinitely many polynomials are zero functions?

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    $\begingroup$ Do you understand the difference between polynomial and polynomial function? $\endgroup$
    – Git Gud
    Commented Mar 15, 2017 at 0:10
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    $\begingroup$ Those polynomials all coincide. I think you want examples like $x^p-x, x^{p^2}-x$ and so on. $\endgroup$
    – lulu
    Commented Mar 15, 2017 at 0:10

2 Answers 2

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There are finitely many functions over a finite field. There are infinitely many polynomials over a finite field. By the pigeonhole principle, it follows that there are infinitely many polynomials which represent the same function.

The difference of any two such polynomials will represent the zero function.

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Denote by $f_1, f_2, \dots, f_n$ the elements of the finite field $\mathbb{F}$. Then $$ f(x) = (x-f_1)(x-f_2)\cdots (x-f_n) \not = 0 \in \mathbb{F}[x], $$ even though the function $\mathbb{F} \ni a \mapsto f(a)$ is identically zero.

The powers $f(x)^n$, $n\in \mathbb{N}$, provide you with infinitely many different non-zero polynomials representing the zero function.

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    $\begingroup$ Nice answer! I suppose, if the finite field is $\mathbb F_{p^n}$, then your polynomial is $f(x) = x^{p^n} - x$. $\endgroup$
    – Kenny Wong
    Commented Mar 15, 2017 at 0:50

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