Showing an element in a Finite Field can be written as a power.

Question

I had a question that I'm stuck with:

Show that every element in $GF(p^n)$ can be written in the form of $a^p$ for some unique $a\in GF(p^n)$.

So this field is the splitting field for the polynomial $f(a) = a^{p^n} - a$ which is what I can understand, but I really don't have any idea how to progress further with the problem.

Answer 1

This is the Frobenius automorphism $\phi:K \rightarrow K$, where $K$ is some finite field extension of $GF(p)$.

Hint: Show this map is injective by showing that $ker(\phi)={{0}}$. This will instantly imply that $\phi$ is an isomorphism, and hence that every element of $K (=GF(p^n))$ can be written as $a^p$ for some $a \in K$.

Answer 2

$\frac{a^p}{b^p}=1 \iff \big(\frac{a}{b}\big)^p = 1.$ But the multiplication is a cyclic group of order $p^n-1,$ and that showed $\frac{a}{b}$ has order a factor of $p$. $p \vert p^n$ implies $\gcd(p^n-1,p)=1,$ and the order of $\frac{a}{b}$ must be a factor of both numbers and thus the gcd, so $\frac{a}{b}$ must be $1$. Therefore each element can be written in at most one way as a $p^{\text{th}}$ power.

But then the order of the image of $a \mapsto a^p$ is the same as the order of the domain, $|\{a^p | a \in \mathbb{F}_{p^n}\}| = |\{a | a \in \mathbb{F}_{p^n}\}|$. Therefore this is a surjection, so every element can be written in this way.

Answer 3

Is "Better late than never" still true? Anyway, here is my $US \$0.02$ worth, from Robert Kenneth "Can't Stop Typing" Lewis:

I think the key is to realize that the map $\phi:GF(p^n) \to GF(p^n)$, $\phi(a) = a^p$ is actually an automorphism of $GF(p^n)$; it is easy to see that $\phi(xy) = \phi(x)\phi(y)$ since $\phi(xy) = (xy)^p = x^py^p = \phi(x)\phi(y)$; to see that $\phi(x + y) = \phi(x) + \phi(y)$, note that the binomial theorem holds in $GF(p^n)$ as it does in any commutative ring with unit; examining the coefficients of $(x + y)^p = x^p + px^{p - 1} y + \ldots + y^p$ one sees that all except those of $x^p$ and $y^p$ have $p$ as a factor; thus they reduce to zero in characteristic $p$, yielding $\phi(x + y) = x^p + y^p = \phi(x) + \phi(y)$. Thus $\phi$ is a ring endomorphism of $GF(p^n)$; its kernel is an ideal in $GF(p^n)$. We also have $\phi(1) = 1^p = 1$, so $1 \notin \ker \phi$. But the only ideals of any field are $\{0\}$ and the field itself; thus $\ker \phi = \{0\}$ and $\phi$ must be injective. Since $GF(p^n)$ is finite, $\phi$ must also be surjective; thus for any $z \in GF(p^n)$ there is a $y \in GF(p^n)$ with $y^p = z$; $y$ is unique since $\phi$ is injective. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!