Is the Polynomial $X^{p^n}-X$ the zero polynomial in characteristic $p$?

Question

Suppose that $p$ is a prime number and that $\mathbb{F}_{p^n}[X]$ is the ring of polynomials with one variable, with coefficients in the field $\mathbb{F}_{p^n}$ for some $n\in\mathbb{N}$. Then

$p(X):=X^{p^n}-X\in \mathbb{F}_{p^n}[X]$

is a polynomial, which appears at many places in number theory, for example in the definition of the Carlitz Exponential. (In books about function field arithmetics this polynomial is often called $[1]$, where in general $[k]:=X^{p^{n\cdot k}}-X$)

Now the question is: Since $p(X)=0$ for any $X\in\mathbb{F}_{p^n}$, isn't $f$ just the zero polynomial? The same should be true for all $[k]$, but as this is not mentioned in any book on function field arithmetic, I think I might overlook something.

Maybe someone has something enlightening to say here.

Answer 1

No it is not. The zero polynomial has degree $-\infty$ whereas this polynomial has degree $p^n$. Moreover, $x^{p^n}-x$ has a unique factorization as it is a polynomial over a field, whereas $0$ is infinitely divisible by anything. The key disconnect here is that you are only considering solutions in $\Bbb F_{p^n}$ and you are thinking that a polynomial is determined by its values on too small a set--remember that polynomials of degree $n$ are determined by their values on sets of size $n+1$, so you have a bunch of zeros, but you are missing a final point to uniquely identify this polynomial, and if you take any field extension of $\Bbb F_{p^n}$ you get a non-zero output for that input, which is different from the zero polynomial.

Answer 2

$X^{p^n}-X$ is clearly not the zero polynomial because it has nonzero coefficients.

However, $X^{p^n}-X$ does induce the zero function.

In fact, the polynomials that induce the zero function are exactly the multiples of $X^{p^n}-X$.

In general, if $K$ is a field, then there is a homomorphism of $K$-algebras $K[X] \to K^K$ that transforms polynomials into functions. This map is injective iff $K$ is infinite. When $K$ is finite with $q$ elements, the kernel is generated by $X^q-X$. This is a consequence of Lagrange's theorem in group theory applied to $K^\times$.

Answer 3

In abstract algebra, by (univariate) (abstract) polynomial (over a ring $R$) one means a formal linear combination $$p(X) := a_n X^n + a_{n - 1} X^{n - 1} + \cdots + a_1 X + a_0$$ of $1, X, \ldots, X^{n - 1}, X^n$, where $X$ is an indeterminate and $a_0, \ldots, a_n \in R$.

Any (univariate) polynomial over $R$ determines a polynomial function $p : R \to R$ defined by $$x \mapsto p(x) .$$ In the most familiar setting, where $R$ is $\Bbb Q$ or $\Bbb R$ or $\Bbb C$, this map from abstract polynomials to polynomial functions (in fact, it is an $R$-module homomorphism) is a bijection, but this is not true for general rings. In particular, for any finite ring $R$, which includes the case of finite fields $\Bbb F_{p^n}$, the (abstract polynomial) $$q(X) := \prod_{r \in R} (X - r)$$ induces the zero function, as for any $r \in R$, the product $q(r)$ contains the factor $r - r = 0$. At least for fields $\Bbb F$, the ideal generated by $q(X)$ is the kernel of the about $\Bbb F$-module (in fact, $\Bbb F$-algebra) homomorphism, so we can identify the ring of polynomial functions with $\Bbb F[X] / \langle q(X) \rangle$.