# Is F a field automorphism?

Let $\mathbb{F}_q$ be a finite field of order $q=p^n$ for some prime $p$ and let $f:\mathbb{F}_q\to \mathbb{F}_q$ be a bijection. If

1. $f(0)=0$ (additive identity),
2. $f(1)=1$ (multiplicative identity),
3. $f(a)f(b)=f(ab)$ for all $a,b\in \mathbb{F}_q$,

then is $f$ a field automorphism ($i.e.$ does it follow that $f(a+b)=f(a)+f(b)$)?

A source with a proof or counterexample would be very helpful.

Updated Question:

1. Is every such bijection above that is not an automorphism of the form $f(x)=x^k$ for some k?
2. Is there another property that $f$ could satisfy to force it to be an automorphism?
• Hint: Consider a map that raises everything to a fixed power. – Tobias Kildetoft May 19 '16 at 16:37
• Of course, silly question...my apologies. – Mark Greer May 19 '16 at 16:40
• To your updated question: All maps satisfying the conditions are of that form, whether or not they respect addition, even if they are not assumed bijective. This is because they are endomorphisms of the multiplicative group of the field, which is cyclic. – Tobias Kildetoft May 19 '16 at 17:11
• Excellent, thank you! – Mark Greer May 19 '16 at 17:13

The map $a\mapsto a^{-1}$ (for $a\ne0$) and $0\mapsto0$ satisfies the requirement, but is not generally a field automorphism: $$a^{-1}+b^{-1}=\frac{a+b}{ab}$$ so the equality $$(a+b)^{-1}=\frac{a+b}{ab}$$ should hold as soon as $a+b\ne0$; this means $$(a+b)^2=ab$$ that is $$a^2+ab+b^2=0$$ in particular $3=0$.

So if $p\ne2$ and $p\ne3$ we're done. Try your hand for the other cases.

• I've updated my question now, since (as you pointed out), there are plenty of such bijections! – Mark Greer May 19 '16 at 16:52

There are $n$ field automorphisms of $\mathbb{F}_q$, since it is a Galois extension of degree $n$ of $\mathbb{F}_p$.

On the other hand, your $f$ corresponds to any group automorphism of $\mathbb{F}_q^* \simeq \mathbb{Z}/(q-1)\mathbb{Z}$ (extended by $0\mapsto 0$). Now there are $\phi(q-1)$ such automorphisms (with $\phi$ the Euler function).

So your question amounts to whether $\phi(p^n-1)>n$. I'm not sure if it always holds, but at least it does for big enough $p$ and $n$.

• That's a nice way to think of the problem. Thanks! – Mark Greer May 19 '16 at 16:53