Prove that there exists a $p$-root for every member of $GF(q)$ and/or existence of power in $GF(q)$

Question

I've been baffled by this question: "Let $\mathrm{GF}(q)$, where $q$ is a prime power of $p$ ($q=p^h$), be a finite Galois field, prove that for each member of the field there exists a $p$ root of it."

I think I have a solution, but there's one thing I'm unsure about, if for every $b$ in $\mathrm{GF}(q) \rightarrow b^{(ap^-1)}$ is in $\mathrm{GF}(q)$ ($a$ is a random number) ?

Hint: $x^{q}=x$ in $\forall x\in\mathrm{GF}(q)$. – Thomas Andrews Aug 12 '17 at 00:02 — Thomas Andrews, Aug 12 '17 at 00:02

score 0 · Accepted Answer · edited Aug 12 '17 at 12:15

0

In the answer I'm going to use the following theorems:

every Galois-field contains a primitive element
for all $x$, $x^q = x$
every field is closed in relation to it's multiplication

Let $a$ be our primitive element, we define $b=a ^{p^h-1}$. Then $b^p = ( a^{p^h-1} ) ^p = a^q = a$.

Hence, since $a$ is our primitive element, for each $c$ in the field

$$c=a ^ l = (b^p) ^l = (b^l) ^ p$$

So every element has a $p$-root.

edited Aug 12 '17 at 12:15

José Carlos Santos

427,504

answered Aug 12 '17 at 11:54

Thomas

38

Welcome to MSE. Please use MathJax. – José Carlos Santos Aug 12 '17 at 12:03
Thanks for the edit – Thomas Aug 12 '17 at 12:16
Hi, click edit to see the latex José typed. Why do you need a primitive element ? It hides the map is $x \mapsto x^{p^{k-1}}$ (and not $x^{p^k-1}$). And that $x^{q-1}= 1$ for $x \in K^\times$ because it is a group with $q-1$ elements, so you don't need theorems. – reuns Aug 12 '17 at 12:52

Prove that there exists a $p$-root for every member of $GF(q)$ and/or existence of power in $GF(q)$

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