Let $X$ be a smooth projective variety over $\mathbf{F}\_p$, call $\overline{X}$ the base change to $\overline{\mathbf{F}}\_p$, and denote by $F$ the base change to $\overline{X}$ of the absolute Frobenius of $X$ over $\mathbf{F}\_p$.


Call $A$ the Chow ring of cycles up to homological equivalence (defined using, say, $\ell$-adic cohomology, i.e. the image of the cycle map $Z^\*(\overline{X}\times \overline{X})\otimes\_{\mathbf{Z}}\mathbf{Q}\to H^{2\*}(\overline{X}\times \overline{X},\mathbf{Q}\_{\ell})(\*)$).


Call $R$ the $\mathbf{Q}$-subalgebra of $A$ generated by the classes of the graphs of the endomorphisms $\{F^n, n\ge 1\}$ of $\overline{X}$.



> 
> Is anything at all known about $R$?
> 
> 
> 


Examples of questions I'd be interested in are


* is $R$ known to not be $A$?
* is $R$ a domain, or even a field?
* is $R$ normal?
* does $R$ contain the Lefschetz class?
* is $R$ expected to contain the inverse of the Lefschetz class, perhaps assuming some standard conjectures.



Subrings of Chow rings
Plenty!


$R$ is generated, as a ring, by $F$. So its structure as a ring is going to be $\mathbb Q(\alpha)/f(\alpha)$, where $f$ is the minimal polynomial of $F$. Because you are using homological equivalence, $f$ is just the least common multiple of the minimal polynomials of the action of $F$ on the various cohomology groups of $X$.


By Deligne's RH, these polynomials are coprime, so it is in fact the product of the minimal polynomials on the different cohomology groups.


Thus $R$ is never a domain unless $X$ is a point.


Since the graphs of powers of $F$ have dimension $n$, $R$ is not equal to $A$ unless every homologically nontrivial cycle on $X \times X$ occurs in degree $n$, i.e. unless $X$ is a point.


Since normal rings are by definition domain, it's not normal.


Because the Lefschetz class lives in dimension $n-1$ and its inverse lies in dimension $n+1$, neither one is contained in this ring since neither has dimension $n$.


