Let $E \in D^{b}\_{c}(X,\overline{\mathbb{Q}}\_{l})$ where $X$ is a $k$ scheme of finite type for a field $k$.


Let $Y\rightarrow X$ a finite flat surjective morphism such that $f^{\*}E$ is perverse and irreducible.


Do we have that $E$ is perverse and irreducible?



flat descent for perverse sheaves
$f\_\*$ for a finite morphism preserves perversity, so $f\_\* f^\* E$ is going to be perverse. $f\_\* f^\* E= E \otimes f\_( \overline{\mathbb Q\_l}$. $f\_\* (\overline{\mathbb Q\_l})$ has the constant sheaf as a summand, (by averaging), so $E$ is a summand of a perverse sheaf, hence is perverse.


Since inseperable morphisms don't change the etale site, we may assume that $f$ is separable, hence finite etale over an open set. By etale descent, over that set $E$ is irreducible. Thus if $E$ is reducible, it has a subobject or quotient object with lower-dimensional support. Call this $F$. It can't be a subobject, because then we would have a nontrivial element in $Hom(F, f\_\* f^\* E)$ giving a nontrivial $Hom(f^\* F, f^\* E)$ which is impossible because the second is irreducible and their supports are different dimensions. Since $f$ is finite, $f\_\*= f\_!$, and we can apply the same argument to quotient objects: $Hom(f\_! f^\* F,E)$ is nontrivial, so $Hom(f^\* F, f^! E)$ is nontrivial, but again one is irreducible and their supports are different dimensions.


