Is Frobenius the only magical automorphism?

Solution 1:

Here is a stronger statement. Let's weaken the requirement to asking for natural endomorphisms of commutative $k$-algebras ($k$ an integral domain - we need this but not the assumption that $k$ is a field), by which I mean endomorphisms of the forgetful functor $k\text{-Alg} \to \text{CRing}$.

Theorem: If $k$ has characteristic $p$, then the only natural endomorphisms of commutative $k$-algebras are the powers of the Frobenius map. Otherwise, the only natural endomorphism is the identity.

Proof. Any such endomorphism is in particular an endomorphism of the forgetful functor $U : k\text{-Alg} \to \text{Set}$. This functor is representable by the free $k$-algebra $k[x]$, and hence its endomorphisms can be canonically identified with $\text{Hom}_{k-\text{Alg}}(k[x], k[x]) \cong k[x]$ by the Yoneda lemma. (This is the most important step in the proof!) Thus any candidate natural endomorphism necessarily has the form

$$R \ni r \mapsto \sum f_n r^n \in R$$

for some polynomial $f(x) = \sum f_n x^n \in k[x]$. (This already rules out a large class of possible examples, including but not limited to complex conjugation.) Now, this endomorphism must preserve addition and multiplication. It preserves multiplication iff it does so in the universal example, namely we must have

$$k[x, y] \ni xy \mapsto \sum f_n (xy)^n = (\sum f_n x^n)(\sum f_n y^n) \in k[x, y].$$

Using the fact that $k$ is an integral domain and comparing coefficients, we conclude that this is possible iff $f(x)$ has at most one nonzero coefficient, which is an idempotent and hence must be $0$ or $1$. In the first case $f$ doesn't preserve the multiplicative unit. In the second case we must have $f(x) = x^n$ for some $n$. Again, our endomorphism preserves addition iff it does so in the universal example, namely we must have

$$k[x, y] \ni x + y \mapsto (x + y)^n = x^n + y^n \in k[x, y]$$

which is true iff $n = 1$ or ${n \choose m} = 0$ in $k$ for all $1 \le m \le n-1$. In the second case, in particular $n = 0$ in $k$, so $k$ must have characteristic $p | n$. Now, by Kummer's theorem, $p | {n \choose m}$ iff the process of adding $m$ and $n - m$ in base $p$ involves at least one carry, but if $n$ is not a power of $p$ we can always find $m$ for which this is not the case by writing $n$ as a sum of powers of $p$ using its base-$p$ expansion and distributing some of this sum to $m$ and the rest to $n - m$. Hence $n$ must be a power of $p$ as desired. $\Box$

See also this MO question involving a very similar argument and this blog post for some discussion of natural transformations between forgetful functors in general.

One can rephrase this argument as being about endomorphisms of $\text{Spec } k[x]$, regarded as a commutative ring object in affine schemes over $k$, but I don't think this adds much to the argument.