# The Frobenius Automorphism

If a finite field $F$ has characteristic $p$, prove that every element $a\in F$ can be expressed as $a=b^{p}$ for some $b\in F$?

Hint: Frobenius Automorphism.

Isn't the Frob Automorphism about $a$? Why does it have to do with $b$?

Thanks!!!

• Your first paragraph does not make sense, really. In every field, independently of its characteristic, and for all integer $p$, it is the case that «$a=b^p$ for some $a\in F$ and $b\in F$». What you wrote is probably not what you wanted to ask. (I wonder what question people are answering...) – Mariano Suárez-Álvarez Mar 21 '14 at 5:03

If you say Frobenius "Automorphism" then you should be done...

If you just want to say Frobenius map then,

Hint :

• Prove that map is injective
• any injective map between two finite sets of same cardinality is ???
• Why am I done? Isn't the Frobenius Automorphism about mapping $a$ to $a^{p}$? What does it have to do with $b$? – user113255 Mar 21 '14 at 4:56
• what is an automorphism by the way? – user87543 Mar 21 '14 at 4:56
• I still don't understand... – user113255 Mar 21 '14 at 5:04
• It is a ring homomorphism mapping to itself and is bijective? – user113255 Mar 21 '14 at 5:07
• do you see ring homomorphism and field homomorphism to be same?? – user87543 Mar 21 '14 at 5:08

The Frobenius map is just a map: $$\phi: F \rightarrow F$$ where $\phi(a) = a^p$ and char($F$) $=$ $p$

So if you know what an automorphism is, it just follows.

• Negative! For saying automorphism, he should prove what he is asked to.... – user87543 Mar 21 '14 at 4:59
• @Mark I am actually supposed to prove that $a=b^{p}$ if char($F$)=$p$... Would you elaborate on why it just follows from the definition of an automorphism? Thanks! – user113255 Mar 21 '14 at 4:59
• An automorphism is an isomorphism onto itself, it's a bijective map. Of course, you're supposed to prove it. :P – Mark Mar 21 '14 at 5:04
• @Mark And what am I supposed to do after proving it is bijective? That still doesn't solve the $b^{p}$ issue... Isn't it supposed to be $a^{p}$? Thanks! – user113255 Mar 21 '14 at 5:10

There is no constructive reason; it's just the pigeonhole principle. The Frobenius map is injective since the polynomial $x^p-1=(x-1)^p$, so since it acts on a finite set, it must be surjective.

Note that there can be no constructive reason, since there exist non-perfect fields: those for which the Frobenius map is not surjective. An example is $\mathbb{F}_p(t)$, where by degree, $t$ is not a value of the Frobenius map.