Which fields satisfy the Freshman's Dream?

It is well-known that the Freshman's Dream, $(a+b)^p = a^p + b^p$, holds in fields of characteristic $p$. For $p = 2$, in fact those are the only fields; for, $(a+b)^2 = a^2 + b^2 \Rightarrow 2ab = 0$ for each $a, b$, so in specific the field has characteristic $2$. But $(a+b)^3 = a^3 + b^3$ is satisfied by every element of $\mathbb{F}_2$, as well as in fields of characteristic $3$. In fact this is the only field of characteristic $\ne 3$ in which this identity holds; the identity is equivalent to $3ab(a+b) = 0$, which in characteristic $\ne 3$ means $ab(a+b) = 0$ or, letting $b = 1$, $a = 0,-1$.

In which fields does the identity $(a+b)^p = a^p + b^p$ hold for every $a, b$ (with $p$ fixed)?

Solution 1:

Claim: Let $n \geq 2$ be a natural number and $F$ be a field. Then $F \to F, a \mapsto a^n$ is additive iff $F$ has positive characteristic $p$ and

• either $F$ is finite, say $|F|=p^e$, and $p^k \equiv n \bmod p^e-1$ for some $k$,
• or $F$ is infinite and $n$ is a power of $p$.

Proof. First we note that if $a \mapsto a^n$ is additive, it is actually a field homomorphism (since it is multiplicative anyway). It follows that if $F=\mathbb{F}_p$ for a prime $p$ the map is additive iff it is the identity iff $a^{n-1} = 1$ for all $a \in \mathbb{F}_p^*$ iff $p-1\mid n-1$, using that $\mathbb{F}_p^*$ is cyclic.

If more generally $F=\mathbb{F}_{p^e}$ is a finite field, the map is additive iff it equals one of the elements of $\mathrm{Gal}(\mathbb{F}_{p^e}/ \mathbb{F}_p)$. This group is cyclic, generated by the Frobenius. Hence, the condition becomes $\forall a \in F : a^n = a^{p^k}$ for some $k$. We may assume $a \in F^*$ here. Using that $F^*$ is cyclic, this reduces to $p^k \equiv n \bmod p^e -1$.

Now let $F$ be an infinite field, and consider the equation $(a+b)^n=a^n + b^n$ in $F$. It is trivial for $b=0$. If $b \neq 0$, we may as well assume $b=1$ by dividing $b^n$ on both sides. Thus, we reduce to $(a+1)^n = a^n+1$, which becomes $P|_F=0$ for $P(x)=\sum_{k=1}^{n-1} \binom{n}{k} x^k$. Since $F$ is infinite, this implies $P=0$, i.e. $\binom{n}{k}=0$ for all $0 < k < n$. In particular, $n=\binom{n}{1}=0$ in $F$, and we see that $F$ has positive characteristic $p$. But then these equations already hold in $\mathbb{F}_p$, and we get $\binom{n}{k} \equiv 0 \bmod p$ for $0<k<n$. Now it is well-known that $n$ is a power of $p$: Write $n = p^v m$ with $p \nmid m$. Since $a \mapsto a^{p^v}$ is injective and additive, it follows that $(a+b)^m=a^m+b^m$. By the same reasoning as above, it follows $\binom{m}{k} \equiv 0 \bmod p$ for all $0<k<m$. But since $\binom{m}{1}=m \not\equiv 0 \bmod p$, it follows that $m=1$. $\square$

The condition is in both cases that $n \bmod \exp(F^*)$ is a power of $p$. Here, we define the exponent $\exp(G)$ of a group to be the nonnegative generator of the ideal $\{z \in \mathbb{Z} : \forall g \in G : g^z=1\}$. If $F$ is a finite field, $\exp(F^*)=|F|-1$, and if $F$ is an infinite field, then $\exp(F^*)=0$.