Let $l$ be a prime number and $q$ be a power of a prime number different from $l$. Recall that a Weil number (relative to $q$) is an element $x$ of $\overline{\mathbb{Q}\_l}$ that is an algebraic integer and such that for any embedding $\sigma:\overline{\mathbb{Q}}\_l\to\mathbb{C}$ one has $|\sigma(x)|=q^{w/2}$ for some integer $w$ called the weight of $x$. These numbers were introduced by Deligne in his work on the Weil conjecture. If $X$ is a smooth and proper scheme over a finite field $k$ with $q$ elements, then the endomorphism of $H^i(X\times\_{k}\overline{k},\mathbb{Z}\_l)/torsion$ induced by the Frobenius has eigenvalues that are Weil numbers of weight $i$. For instance, the powers of $q$ are Weil numbers (with $q^n$ of weight $2n$). Weil numbers form a monoid under multiplication and the weight function is a morphism of monoids to $(\mathbb{Z},+)$.


I am interested in what happens to Weil numbers when one passes to the residue field of $\overline{\mathbb{Q}}\_l$. In that case there is no hope of having a well-defined weight function with values in $\mathbb{Z}$, indeed, some power of $q$ will be equal to $1$ modulo $l$. However I can assign to each power of $q$ a weight which is an element of $\mathbb{Z}/2n$ where $n$ is the order of $q$ in $\mathbb{F}\_l$. I would like to know if I can extend this kind of weight function to the whole monoid of Weil numbers. Precisely, if I denote by $W$ the monoid of Weil numbers and by $\overline{W}$ the image of $W$ in the residue field of $\overline{\mathbb{Q}}\_l$, I would like to know if I can construct a monoid map $\overline{W}\to\mathbb{Z}/m$ that is compatible with the weight function (in the sense that it makes the obvious square commute). Obviously this can be done with $m=1$, on the other hand, the order of $q$ in $\mathbb{F}\_l$ gives an upper bound for what $m$ can be. Is it known if I can take $m$ to be $2n$ ? and if not is there a way to evaluate what the optimal m is ?



Weight of Weil numbers in the residue field of $\overline{\mathbb{Q}}_l$
Presumably by a map $\phi: \overline{W} \rightarrow \mathbb{Z}/m$ that is "compatible with the weight function" you would require that Weil numbers $W^0$ of weight zero lie in the kernel of $\phi$. But every non-zero element the residue field of $\overline{\mathbb{Q}}\_l$ lifts to a root of unity of $\overline{\mathbb{Q}}\_l$, and all roots of unity are Weil numbers of weight zero. So $W^0$ surjects onto $\overline{W}$, and so $\phi$ is automatically trivial. 


