Ring Homomorphisms on Integral Domains THIS PROBLEM WAS TAKEN FROM A CURRENTLY UNPUBLISHED MANUSCRIPT OF JOSEPH SILVERMAN FROM BROWN UNIVERSITY
Characterize all integral domains $R$ for which the map $f: R \rightarrow R$ given by $f(a) = a^{pq}$ for distinct primes $p,q$ is a ring homomorphism.
If $pq=6$, then there's only one such integral domain ($\mathbb{F}_2$, and for $pq = 15$, the integral domains are $\mathbb{F}_2, \mathbb{F}_3$), but beyond that I can't make much meaningful progress. Could we maybe separate into cases based on the characteristic of $R$? This seems related to the Frobenius endomorphism, but that deals with a map to a single prime power. Perhaps if we could somehow prove a congruence with a power of a prime so as to permit iteration of the mapping?
 A: More generally: Let $f:x \mapsto x^n$ with $n \ge 2$ be a ring homomorphism on an integral domain $R$. Then either

*

*$n$ is a power of the (prime) $\mathrm{char}(R)$, and no further restrictions on $R$, or

*$R =\mathbb F_{\ell^r}$ and $n \equiv \ell^\nu$ (mod  $\ell^r-1$) for some $\nu \in \{0, ..., r-1\}$.

Conversely, it is easily checked that in all these cases, $f$ is a ring homomorphism on $R$.

Note that both cases can occur for a given $n$, e.g. $x \mapsto x^5$ is a homomorphism both on any integral domain containing $\mathbb F_5$, and also on $\mathbb F_{2^2}$ (where it is equal to the Frobenius $x \mapsto x^2$).
Note that $R=\mathbb F_2$ works for all $n$ (obviously). A little less obviously, $R= \mathbb F_3$ works for all odd $n$; $\mathbb F_5$ works for all $n \equiv 1$ (mod $4$); $\mathbb F_7$ works for all $n \equiv 1$ (mod $6$) etc.; then, $R= \mathbb F_{2^2}$ works for all $n$ not divisible by $3$; $R= \mathbb F_{3^2}$ works for all $n \equiv 1,3$ (mod $8$); etc.
Note further that since you have $n=pq$ for distinct primes $p,q$, you are automatically in the second case, and you only have to check those prime powers $\ell^r$ which are $\le n$. In particular, for $n=6$ we must have $R=\mathbb F_2$; for $n=10$ and $n=14$, we have $R=\mathbb F_2$ or $\mathbb F_{2^2}$; for $n=15$, we have $R=\mathbb F_2$ or $R=\mathbb F_3$; for $n=21$, we have $R=\mathbb F_2$ or $R=\mathbb F_3$ or $R=\mathbb F_5$ or $R=\mathbb F_{11}$; for $n=22$, we have $R=\mathbb F_2$ or $R=\mathbb F_{2^2}$ or $R=\mathbb F_{2^3}$; for $n=26$, we have $R=\mathbb F_2$ or $R=\mathbb F_{2^2}$; etc.

Proof: Let $f:x \mapsto x^n$ with $n \ge 2$ be a ring homomorphism on an integral domain $R$. Then $\mathrm{char}(R) = \ell$ for some prime $\ell$, because otherwise $\mathbb Z \subseteq R$, and $f$ is obviously not a homomorphism on $\mathbb Z$. So $\mathbb F_\ell \subseteq R$.
Since $(x+1)^n=x^n+1$ it follows that all $x \in R$ are roots of the polynomial $$(X+1)^n-X^n-1 = \sum_{k=1}^{n-1} \binom{n}{k}X^k\in \mathbb F_\ell[X]$$
If $n$ is a power of $\ell$, this polynomial is identically $0$, and we cannot conclude anything more (and indeed should not: any $R$ which contains $\mathbb F_\ell$ does the job.)
If $n$ is not a power of $\ell$, then for $k_0:=$ the highest power of $\ell$ which divides $n$, we have $1\le k_0 \le n-1$ and $\binom{n}{k_0} \not \equiv 0 \;\mathrm{mod}\; \ell$, hence the above polynomial is non-trivial, meaning that all elements of $R$ must lie in a finite extension of $\mathbb F_\ell$, say $\mathbb F_{\ell^r}$.
Now it's well-known that $\mathbb F_{\ell^r}^*$ is cyclic of order $\ell^{r}-1$, meaning that for all $x \in \mathbb F_{\ell^r}$, we have $x^n = x^s$ where $s$ is the residue class of $n$ mod $\ell^r-1$, but different $s$ give different maps. But finally it's well-known that all automorphisms of $\mathbb F_{\ell^r}$ can be written in the form $x \mapsto x^{\ell^\nu}$ where $0 \le \nu \le r-1$.
