Let $E$ be a multiplicative cohomology theory. Fix a prime p. Call a ring map $\psi^{p}:E\rightarrow E$ an *Adams operation* if it lifts the Frobenius map $E/p\rightarrow E/p$.


It is of course well-known that $K$-theory has Adams operations. (If it weren't for $K$-theory, these operations would have a different name.) In fact, it has an Adams operation for every prime $p$.


My question is: are there examples of multiplicative cohomology theories which have an Adams operation $\psi^{p}$ for only one prime $p$? 


By "only one prime" I don't mean that I require that operations at other primes *necessarily don't* exist, just that they *don't necessarily* exist. In other words, their (non)existence is much less obvious/clear/explicit than that of $\psi^{p}$.


Motivation: This would turn $E^{\*}(X)$ into a $\delta$-ring, which is something some people like to study.



Cohomology theory with only one Adams operation?
Adams operations exist in quite wide generality. For any even periodic ring spectra $E$ and $F$, we have associated formal groups $G\_E$ and $G\_F$ over base schemes $S\_E$ and $S\_F$. There is a moduli scheme $\text{Hom}(G\_E,G\_F)$ parametrising pairs $(f,\widetilde{f})$ consisting of a map $f\colon S\_E\to S\_F$ and a homomorphism $\widetilde{f}\:G\_E\to f^\*G\_F$. This contains an open subscheme $\text{Iso}(G\_E,G\_F)$ consisting of pairs where $\widetilde{f}$ is an isomorphism. There are natural comparison maps $\text{spec}(\text{Ind}(E\_0\Omega^\infty F))\to\text{Hom}(G\_E,G\_F)$ and $\text{spec}(E\_0F)\to\text{Iso}(G\_E,G\_F)$, both of which are isomorphisms when $E$ and $F$ are Landweber exact. By considering the case $(f,\widetilde{f})=(\text{id},k.\text{id})$ we see that $\psi^k$ exists as a ring automorphism of $E$ when $k$ is invertible in $\pi\_0(E)$, and as a ring endomorphisms of $\Omega^\infty E$ when $k$ is not invertible. In other words, in the first case $\psi^k$ is an additive and multiplicative stable operation, and in the second case it is an additive and multiplicative unstable operation. This remains true in many cases when $E$ is not Landweber exact, by various less systematic arguments. It will always be easier to produce $\psi^k$ in cases where $k$ is invertible.


