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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?

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It seems you have mostly figured this out by yourself. I can happily confirm you are on the right track.

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

They are indeed. So they are in fact algebraic numbers.

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?

Yes, this is 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.

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