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In positive charcteristic $p$, we know that for every field element $x\in\mathbb{F}_{p}$ we get $x^p = x$.

Then I think (and I might be wrong, but I don't see how) monomials of the form $t^{p^i}\in\mathbb{F}_p[t]$ for arbitrary $i\in \mathbb{N}$ are all the same, since the functions $p_0(t)=t$, $p_1(t)=t^p$, $p_2(t)=t^{p^2}$ and so on are all actually the same functions. Not? I mean certainly they are equal on all elements, that is

$$p_i(x)= p_j(x)$$ for all $x\in\mathbb{F}_p$ and $i,j\in\mathbb{N}_0$. But in textbooks on finite function field and such, they are treated as if they where different. But this seems to contradict the (poinwise) definition of a function in terms of evaluation on elements.

So would be great to get some clarification here.

Bobby
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1 Answers1

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There are indeed two different notions of a polynomial over a field $K$: the first is that of formal expressions of the form $\sum_{i=0}^k a_iX^i$ where $a_i \in K$, and the second is that of functions $f : K \to K$ where there exist $a_i \in K$ such that for all $x \in K$ we have $$ f(x) = \sum_{i = 0}^k a_ix^k. $$ If we write $K[X]$ for the formal expressions, and $\mathrm{Pol}(K)$ for the functions, then there is a mapping $K[X] \to \mathrm{Pol}(K)$ given by interpreting the formal expression as a function. In fact, it is a homomorphism whose kernel is precisely those polynomial expressions which evaluate as the zero function.

Note that if $K$ is infinite, then the mapping $K[X] \to \mathrm{Pol}(K)$ is a bijection, and we tend to identify the two sets without too much worry.

When $K$ is finite, identifying the sets makes little sense: the set $\mathrm{Pol}(K)$ is a finite set -- in fact the set of all functions $K \to K$ -- that does not catch the structure we find interesting about polynomials over $K$. For a concrete example, $\mathrm{Pol}(\mathbb F_2)$ only has four elements, and they can be written (as polynomial expressions) as $0, 1, X, X - 1$. There are no polynomials of degree at least $2$ to form any splitting fields!

Mees de Vries
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  • Nice! That clears everything exactly. Unfortunately this is rarely mentioned. Likely because of the identification in char zero. Good answer. – Bobby Apr 29 '19 at 13:45
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    @Bobby: Note that the relevant property is not "characteristic zero", but the field being infinite. All fields of characteristic zero are infinite, but some fields of positive characteristic are also infinite. For example, for any field $K$, the field of formal rational functions over $K$ is infinite, but has the same characteristic as $K$ itself. (It looks like perhaps you think the fields $\mathbb F_p$ are the only fields of nonzero characteristic?) – hmakholm left over Monica Apr 29 '19 at 16:04