10
$\begingroup$

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.

$\endgroup$
9
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ haha oops. Thanks, I fixed it. $\endgroup$ – Nate Apr 17 '17 at 17:59
  • $\begingroup$ I'd write $ (a-b)^p=a^p-b^p=0$ or $0= a^p-b^p = (a-b)^p$ instead. $\endgroup$ – lhf Apr 17 '17 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.