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