If $a^n = a$ for all $a \in K$, what is $\operatorname{char} K$?

Question

Question: If $a^n = a$ for all elements $a$ in a field $K$ (where $n \in \Bbb N, n\ge 2$), what is $\operatorname{char} K$?

The above problem with $n=4$ is an exercise in Juliusz Brzeziński's Galois Theory through Exercises. The solution for all even $n$ is similar and quite straightforward. Suppose $n \in \Bbb N, n\ge 2$ is even and the hypothesis holds. Put $a = -1$ to get $(-1)^n = -1$, i.e., $1 = -1$. This gives $2 = 0$ and it follows that $\operatorname{char} K = 2$. I'd like to explore the possibilities in case $n$ is odd. We can rewrite the condition $a^n = a$ as $$_{\forall a\in K} \ \, a(a^{\frac{n-1}{2}} - 1)(a^{\frac{n-1}{2}} + 1) = 0$$ What can we say about $\operatorname{char} K$ in this case?

Answer 1

I answered the more general question Ring Homomorphisms on Integral Domains a while ago. That one asks what we can say about the characteristic of an integral domain $R$ if, for some natural $n$, we know the map

$$x \mapsto x^n$$

to be a ring homomorphism (i.e., additive).

Here we have the special case that ($K$ is a field, and) the map is the identity. Specializing the more general answer there to this, we see that we must have $K \simeq \mathbb F_{\ell^r}$ for some prime $\ell$ and natural $r$. Further, the first case there specializes to $n = \ell^r$ itself; the second case there specializes to $n \equiv 1 \mod \ell^r-1$ (as opposed to the more general $n \equiv \ell^\nu$ there). Now we see the specialized first case has become included in the second, and the answer here is (consistent with Brian Moehring's comment under the other answer):

$$a^n = a \text{ for all } a\in K \Leftrightarrow K \simeq \mathbb F_{\ell^r} \text{ and } (\ell^r-1) \vert (n-1)$$

To effectively find the possibilities for $K$:

• List the numbers $1+d$ for all divisors $d$ of $n-1$.
• Discard the items in the list that are not prime powers.
• For the remaining ones, if $\ell^r$ is the highest power of the prime $\ell$ that is on your list, then $K= \mathbb F_{\ell^r}$ is the biggest field of characteristic $\ell$ that will do (i.e. all its subfields are solutions).

(Since e.g. $d=1$ divides every $n-1$, this explains why $K=\mathbb F_2$ is an option for all $n$.)

Of course, if you just want to list possibilities for $\mathrm{char}(K)$ itself, since always $(\ell-1) \vert (\ell^r-1)$, you can just scan your list for primes.

This confirms your result that if $n$ is even (so $n-1$ is odd so all $1+d$ are even again), the characteristic must be $2$. It also easily gives e.g.

$$n=9 \implies K \in \{\mathbb F_2, \mathbb F_3, \mathbb F_5, \mathbb F_{3^2} \} \text{ (up to isomorphism) }$$

or

$$n=15 \implies K \in \{\mathbb F_2, \mathbb F_3, \mathbb F_{2^3} \} \text{ (up to isomorphism) }$$

or

$$n=31 \implies K \in \{\mathbb F_2, \mathbb F_3, \mathbb F_{2^2}, \mathbb F_7, \mathbb F_{11}, \mathbb F_{2^4}, \mathbb F_{31} \} \text{ (up to isomorphism) }$$

etc.

I'll leave it as an exercise to conversely find the $n$ which work for a given finite field $K$.

Answer 2

If $K=\Bbb F_2$ then this equation is true for all $n\in\Bbb N$. Therefore you cannot make deductions without further hypotheses e.g. $n$ is the least $n>1$ for which this equation always holds.

If $K=\Bbb F_3$ then the equation is true for $n$ a power of three. Thus the characteristic need not always equal two.

Answer 3

Here are the main points:

• $K$ is finite because it is the set of roots of $x^n-x$ and a polynomial of degree $n$ has at most $n$ roots in a field.

• $a^n=a$ for all $a \in K$ implies $a^{n-1}=1$ for all $a \in K^*$ and so $n-1$ is an exponent for the multiplicative group $K^\times$.

• $K^\times$ is cyclic because the multiplicative group of a finite field is cyclic (as mentioned in a comment by @mr_e_man).

• Therefore, $n-1$ is a multiple of $|K^\times|=|K|-1$. In other words, $n-1$ is a multiple of $p^k-1$ for some prime $p$ and $k \in \mathbb N$.

• Finally, $\operatorname{char}(K)$ is a prime $p$ such that $p^k-1$ divides $n-1$. There may be several possiblibities, depending on $n$.