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?”: 



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



Frobenius eigenvalues algebraic numbers
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 $q^{-j}$?
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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.


