# Frobenius eigenvalues algebraic numbers

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?

Are they not just the eigenvalues of geometric Frobenius acting on $$H^{2j}(\overline{X},{\mathbf{Q}}_{\ell})$$, renormalized by $$q^{-j}$$?
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?
For a simple example you could work with a surface that is the product of two elliptic curves. $$H^2( \overline{E_1 \times E_2},\mathbb Q_\ell (-1))$$ has dimension $$6$$ and it is possible for $$2$$, $$4$$, or $$6$$ of the eigenvalues to be algebraic integers (and therefore roots of unity ), with the remainder algebraic numbers, but not algebraic integers.