Cohomology theory with only one Adams operation?

Question

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.

Answer 1

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.