Let $X$ be a proper, geometrically connected, geometrically integral variety over $\mathbf{F}\_q$. There exists a finite field extension $k/\mathbf{F}\_q$ of degree $d$ and an alteration $X'\to X\_k$ defined over $k$, where $X'$ is smooth and projective over $k$.


For $\ell$ a prime not dividing $q$, we have an injective map
$$H^i\_{ét}(X\_{\overline{k}},\overline{\mathbf{Q}}\_{\ell})\subset H^i\_{ét}(X'\_{\overline{k}},\overline{\mathbf{Q}}\_{\ell})$$


such that geometric Frobenius $F$ on $H^i\_{ét}(X'\_{\overline{k}},\overline{\mathbf{Q}}\_{\ell})$ restricts to the $d$-th power of geometric Frobenius on $H^i\_{ét}(X\_{\overline{k}},\overline{\mathbf{Q}}\_{\ell})\simeq H^i\_{ét}(X\_{\overline{\mathbf{F}}\_q},\overline{\mathbf{Q}}\_{\ell})$.


Fix a field isomorphism $\iota: \overline{\mathbf{Q}}\_{\ell}\simeq\mathbf{C}$.


By the Weil conjectures applied to $X'$, $F$ has eigenvalues $\lambda$ with complex absolute value $$|\iota(\lambda)|=q^{di/2}$$ It looks like this implies that this is also satisfied by geometric Frobenius (to the power $d$) on $H^i\_{ét}(X\_{\overline{\mathbf{F}}\_q},\overline{\mathbf{Q}}\_{\ell})$.


This also seems to imply that geometric Frobenius on $H^i\_{ét}(X\_{\overline{\mathbf{F}}\_q},\overline{\mathbf{Q}}\_{\ell})$ has eigenvalues with complex absolute value $q^{i/2}$.


This feels wrong: I would expect the eigenvalues to have complex absolute value **bounded above** by $q^{i/2}$ (resp. $q^{di/2}$). If $X$ was itself smooth then Poincaré duality would give the reverse inequality, and so purity, but without smoothness I would expect this to not be true.



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(maybe an incorrect use/statement of de Jong's theorem? I guess $X'$ needs not be projective, but only open in smooth projective)



Purity for proper varieties
The claim that alteration induces an injection on cohomology is wrong, as the example of the resolution of a nodal cubic curve by $\mathbb P^1$ shows (resolutions being a special case of alterations).


The induced map from $H^1$ of the nodal cubic to $H^1$ of $\mathbb P^1$ is not an injection, because $H^1$ of $\mathbb P^1$ vanishes but $H^1$ of the nodal cubic does not.


