Let $X$ be a smooth projective variety over $\mathbf{F}_q$ and $\overline{X}$ its base change to $\overline{\mathbf{F}_q}$.

By Deligne’s Weil I, the eigenvalues of the geometric Frobenius acting on $H^{2j}(\overline{X},{\mathbf{Q}}_{\ell})$ are all algebraic numbers.

Is it true that the eigenvalues of the geometric Frobenius acting on $H^{2j}(\overline{X},{\mathbf{Q}}_{\ell}(j))$ are also algebraic numbers? Are they not just the eigenvalues of geometric Frobenius acting on $H^{2j}(\overline{X},{\mathbf{Q}}_{\ell})$, renormalized by $q^{-j}$?

This feels wrong because otherwise geometric Frobenius would act on $H^{2j}(\overline{X},\mathbf{Q}_{\ell}(j))$ in a unipotent way, since its eigenvalues would all be algebraic and of complex absolute value $1$, then roots of unity. This cannot be the case, and my question is “why?”:

why do Tate twists mess up algebraicity of geometric Frobenius eigenvalues?

**Edit:** it’s possible that my confusion is about “all absolute value one algebraic numbers are roots of unity”, and Tate twists are not guilty.
The sentence in quotation marks is false: only the absolute value one roots of a monic polynomial with **integer** coefficients are roots of unity, and the characteristic polynomial of geometric Frobenius on $H^{2j}(\overline{X},\mathbf{Q}_{\ell}(j))$ may well not be with integer coefficients but only rational coefficients (or in $\mathbf{Z}[q^{-j}]$). Is this the problem?