# Is Frobenius the only magical automorphism?

The Frobenius automorphism is special because the $p$-power map makes sense in any characteristic $p$ ring, which allows us to canonically extend the Galois-theoretic Frobenius to any such ring. I don't know of any other field automorphism having this property, and I wonder if there are any other examples.

In order to formalize my question, let $K$ be a field and $\sigma$ a nontrivial automorphism of $K$. Let $\mathcal C$ be the category of commutative, unital $K$-algebras. Given $R \in \mathcal C$, we can twist $R$ by $\sigma$ by pre-composing the structure map $K \to R$ with $\sigma$. The map $R\mapsto R_\sigma$ determines a functor $T_\sigma$ which is an automorphism of $\mathcal C$.

Let us say that $\sigma$ is magical if there exists a natural transformation $$M: \mathbf{1}_{\mathcal C} \to T_\sigma.$$

The data of $M$ is equivalent to the data of a $K$-algebra homomorphism $R \to R_\sigma$ for every $K$-algebra $R$, in a manner compatible with the morphisms in $\mathcal C$. By definition of the $K$-algebra structure on $R_\sigma$, this amounts to giving a $\sigma$-linear morphism $R \to R$, or, what is the same thing, an extension of $\sigma$ to $R$.

The following proposition illustrates that magical automorphisms are probably quite rare.

Proposition 1: Complex conjugation $\sigma \in \text{Gal}(\mathbf C/\mathbf R)$ is not magical.

I don't know if an easier proof is possible, but this one is fun:

Proof: I prove something stronger, namely that there exists a $\mathbf C$-algebra $R$ with no morphism $R\to R_\sigma$. Indeed, there exists an elliptic curve $E/\mathbf C$ such that $E$ is not isogenous to $E^\sigma$; any elliptic curve whose period lattice is not homothetic to a sublattice of its complex conjugate does the trick. Taking for $R$ the function field of $E$, we get a $\mathbf C$-algebra $R$ with no $\mathbf C$-algebra maps $R \to R_\sigma$.

Remark: In this case, giving an extension of $\sigma$ to $R$ could be viewed as putting an almost real structure on $R$. Indeed, any $\mathbf{C}$-algebra of the form $S \otimes_\mathbf{R} \mathbf{C}$, $S$ an $\mathbf{R}$-algebra, carries a natural extension of complex conjugation.

Proposition 2: If $K$ is a perfect field of characteristic $p$ and $\sigma$ is the $p$-power automorphism of $K$, then $\sigma^n$ is magical for every $n>0$.

Proof: For any $K$-algebra $R$, the $p^n$-th power endomorphism $R \to R$ is an extension of $\sigma^n$ to $R$, and it is obviously natural in $R$.

Questions: Does there exist an example of a magical automorphism which is not Frobenius? Are negative powers of Frobenius magical? Does there exist a magical automorphism in characteristic $0$? If $K/F$ is $\overline{\mathbf F_p}/\mathbf F_p$, and $\sigma \in G = \widehat{\mathbf Z}$ is not a positive integer power of Frobenius, is $\sigma$ magical? (My guess is that it is not.)

Remark: It seems that the fact that Frobenius is magical plays a basic role in mod $p$ and $p$-adic geometry, by giving rise to the notion of Frobenius morphism.

• In the last proof $R\to R$ should be $R\to R_{\sigma^n}$, right? – Grigory M May 14 '14 at 8:41
• @GrigoryM Yes if you want it to be $K$-linear; I have it written as $\sigma^n$-linear. – Bruno Joyal May 14 '14 at 8:58

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.

• Thanks Qiaochu! This answers the question perfectly. On the same topic, do you have a favorite way of thinking of Frobenius geometrically? – Bruno Joyal May 15 '14 at 5:54
• @Bruno: The geometric meaning of Frobenius is totally mysterious to me! – Qiaochu Yuan May 15 '14 at 6:08
• To me as well! And it bugs me a little that something could be so important yet so mysterious. – Bruno Joyal May 15 '14 at 22:05
• Sorry for the edits; I thought I had omitted the case of the zero endomorphism but I just realized that that doesn't preserve the multiplicative unit. That also rules out the idempotents in the non-integral domain case. – Qiaochu Yuan May 18 '14 at 4:59
• The geometric meaning of automorphisms of the ground field is rather obscure. For example, IIRC, automorphisms of number fields can change even the topology of the associated analytic spaces of affine varieties defined over those fields. – Mariano Suárez-Álvarez May 18 '14 at 5:02