I am reading the book Weil Conjectures, perverse sheaves and l'adic Fourier transform by Kiehl and Weissauer. In I.6, this book claims that for a smooth affine curve $U_0$ over finite field $\mathbf{F}_q$ and a lisse sheaf $F$ on $U_0$, by Noether normalization, we can find a finite morphism $f:U_0 \rightarrow \mathbf{A}_{\mathbf{F}_q}^1$. Then the direct image of $F$ from $U_0$ onto the affine curve is lisse on an open nonempty subset of the affine curve.
I'm very confused about this statement. Can anyone explain the reason?
I think this can be argued as follows: $f$ is surjective by the going up theorem, and as $f$ is proper, by the Proper Base Change theorem we have an isomorphism of sheaves on $V_0$ $$ (f_\ast F)_{\mid V_0} \cong (f_{\mid f ^{-1}V_0})_\ast F_{\mid f^{-1}V_0}$$ for any open subset $V_0 \subset \mathbf{A}^1_{\mathbf{F}_q}$. Thus, via base-change to an open subset $V_0 \subset \mathbf{A}^1_{\mathbf{F}_q}$ we may assume $f$ is a finite étale cover since $f$ is such outside of its ramification divisor $D \subset \mathbf{A}^1_{\mathbf{F}_q}$***. Then $f_\ast F$ is indeed (an inverse system of) finite locally constant sheaves - if we base change $f$ along $f$ itself then we obtain the projection $U_0':=U_0\times_{V_0} U_0 \to U_0$ which admits the diagonal $\Delta: U_0 \to U_0'$ as a section, mapping isomorphically onto a connected component of $U_0'$ since $f$ is unramified. Since $f' : U_0' \to U_0$ is again a finite étale cover, we can base change once more along itself and so forth, eventually getting a map $U\to V$ which is the base change of $f:U_0 \to \mathbf{A}^1_{\mathbf{F}_q}$ along an étale cover $V\to \mathbf{A}^1_{\mathbf{F}_q}$, so that $U$ is isomorphic to a finite number of copies of the base $V$ and $f'$ is the identity on each of these components. Then here we evidently have that $(f_\ast F)_{\mid U} = f'_\ast F_{\mid U'}$ (again via pbc) is finite locally constant since $F_{\mid U'}$ is.
Hope this helps and I didn't mess something up :P I guess I could've referenced the Stacks Project for some similar fact but I've seen this trick used quite often in étale cohomology and thought it would've been nice to explain it here:)
***Edit: thanks to a correction by Aphelli in the comments, the differential of the section $s\in \Gamma(U_0,\mathcal{O}_{U_0})$ corresponding to the map $U_0 \to \mathbf{A}^1_{\mathbf{F}_q}$ might be a zero divisor and thus the zero section, since $U_0$ is integral. However, if this occurs then we may factor $f$ as a composition $$ U_0 \xrightarrow{f'} \mathbf{A}_{\mathbf{F_q}}^1 \xrightarrow{\text{Frob}_q^n} \mathbf{A}^1_{\mathbf{F}_q}, $$ $\text{Frob}_q$ being the Frobenius, where $f'$ has a non-zero differential $df'$ and thus its vanishing locus defines a divisor $D$. Then $f_\ast F$ is finite locally free on an open subset of $\mathbf{A}^1_{\mathbf{F}_q}$ if $f_\ast' F$ is, since the Frobenius preserves finite locally constant sheaves via pushforwards (because it preserves constant sheaves and is proper).