Does $X^p = X$ over a field of characteristic $p$? Since $\mathbb F_p*$ is of order $p-1$, for all $a\in F_p$ we have $a^{p-1} = 1$ and so $a^p = a$. I feel like this should imply that $X^p = X$ but everything I see differentiates the two. I think it's because the multiplicative subgroup of the polynomial ring over $\mathbb F$ doesn't have order $p-1$ but I'm not sure.
If $X^p\neq X$, then are the equal as functions $\mathbb F \to \mathbb F$? Thank you!
 A: If $x$ is a variable, then  by definition it is transcendental over a field $\mathbb{F}$. So  when you talk about polynomials in $\mathbb{F}_p[x]$, even though $x^p$ and $x$ are equal as functions from  $\mathbb{F}_p$ to $\mathbb{F}_p$, they are different elements of $\mathbb{F}_p[x]$. There are valid reasons for  this. In particular there are some larger fields $\mathbb{K} \supseteq \mathbb{F}_p$ such that $x^p$ and $x$ are not equal as polynomials over those fields. Can you create an example of such a field  $\mathbb{K}$?
A: I think you have mostly answered your own questions, but let me clarify the two things.
Given any (commutative) ring $R$, we may form the ring of polynomials $R[X]$, which is, by definition, nothing but finite length sequences $(a_0, \dots, a_d)$ with addition and multiplication defined in the usual way. We usually write it as $a_0 + a_1 X + \dots a_d X^d$ because this notation is more convenient and more intuitive, but you really should think about them just as sequences of elements in $R$.
The other thing is all maps from $R$ to $R$.
If $f=a_0 + \dots + a_d X^d$ is any polynomial in $R[X]$, then $f$ induces to a map $\overline f: R\rightarrow R$ by sending every element $r\in R$ to $a_0 + a_1 r + \dots + a_d r^d$.
Therefore the answer to your question is: $X^p$ and $X$ are different polynomials in $\Bbb F_p[X]$, but they induce the same map from $\Bbb F_p$ to $\Bbb F_p$.
A: Like any polynomial of degree $p$, $X^p - X$ can't have more than $p$ roots in any field.  So in any
extension of $\mathbb F_p$, all the elements that are not in $\mathbb F_p$ itself do not have $x^p = x$.
A: The indeterminates in $R[X]$ for any suitable $R$ are transcendental over $R$. This means that $a_nX^n + a_{n-1}X^{n-1} + \cdots + a_1X+a_0=0$ implies $a_n = a_{n-1} = \cdots = a_0 =0$. Applying this to $X^p-X$, we have $a_p=1, a_1=-1$, so $X^p-X\neq0$.
A: In abstract algebra, a polynomial is not thought of as a function but as a string of symbols. In characteristic zero, this is a distinction without a difference, since two distinct polynomials give different functions for any field (this is a neat exercise).
But in characteristic $p$, this matters for precisely the reason you stated: $X^p$ and $X$ are the same function on the field $\mathbb{F}_p$, but are different polynomials.
