$p^{th}$ roots of a field with characteristic $p$

Question

This is problem 10.9 from the book "Error-Correcting Codes and Finite Fields by Oliver Pretzel".

The Question:

Show that in a field of characteristic $p$, any element $\alpha$ has at most one $p$-th root $\beta$ (i.e., an element $\beta\in F$ with $\beta^p = \alpha$). Show further that if $F$ is finite, then every element has exactly one $p$-th root

This my attempt at the second part of the question. From Fermat's little theorem $\beta^{{p^n}-1} = 1$, where $p^n$ is the size of the field. Now multiplying both sides by $\beta$ we get $\beta^{p^n} = \beta$. If there is $p$ elements then $n=1$ and we can see this is true for any non-zero element. For the general case, take the $p$-th root of both sides $\beta^{p^{n-1}} = \beta^{1/p}$ and we know from the multiplicative properties of a field if $\beta$ is a non zero element of the field then any multiple will be.

The first part of the question I'm not sure where to begin. Any help would be appreciated.

Answer 1

We are looking for a root of $x^p-\alpha$; the formal derivative of this polynomial is zero, which means that $x^p-\alpha$ has repeated roots.

Indeed, if $K$ is an extension of $F$ where the polynomial has a root $\beta$, we have $$ (x-\beta)^p=x^p-\beta^p=x^p-\alpha $$ which shows the root is unique.

For a finite field $F$, the map $$ \alpha\mapsto\alpha^p $$ is a field homomorphism, so it is injective. Finiteness yields surjectivity.

Answer 2

If $a^p = b^p$ then $a^p-b^p = (a-b)^p=0$, and since you are in a field this implies $a=b$. This shows that for a field of characteristic $p$ the map $a \to a^p$ is always injective, and an injective map from a finite set to itself is automatically bijective.