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> Let $C$ be a proper smooth curve over a perfect field $K$ of positive characteristic $p$, $u: U \hookrightarrow C$ strictly open and $\mathfrak{F}$ a lisse (lcc) $\mathbb{F}\_l $-sheaf $(l \neq p)$ on $U$. Let $K'/K$ be **any** field extension and $x'$ a geometric point of $C'= C \times\_K K'$ that maps to $x \in C$. Is it true that $Sw\_{x'}(u'\_!\mathfrak{F}') = Sw\_x(u\_!\mathfrak{F})$?
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Here $Sw\_x(u\_!\mathfrak{F})$ denotes the Swan conductor of $u\_!\mathfrak{F}$ at $x$ and the primes indicate the canonical pullbacks. 


If $K'/K$ is algebraic, then by (Kindler, Rülling- Intro to l-adic sheaves and ramification theory, Lemma 10.2) this holds in fact true -but it is not obvious. The Swan conductor depends on representation étale fundamental groups which do not behave very well under base change. Hence I would not really expect this to hold true.


My problem is: in the paper "Semi-continuité du conducteur de Swan" Laumon seems to use the result above several times for non algebraic basechange $K'/K$.



Swan-conductor and base change
Yes.


We have some finite etale Galois cover $D \to C$ with automorphism group $G$. We can base change this cover, obtaining a finite etale Galois cover $D' \to C'$ with the same automorphism group. Because the Swan conductor is determined by the lower numbering filtration on this Galois group, it suffices to show that the lower numbering filtrations on $G$ arising from these two covers are the same.


But this is obvious - the $i$th subgroup consist of automorphisms $f$ of $D$ such that for a uniformizer $\pi$ at $x$, $f(\pi)-\pi$ is divisible by $\pi^i$. This condition is manifestly preserved (and reflected) by base change.


The moral is that while the etale fundamental group does not behave well under non-algebraic base change, representations of the etale fundamental group do. The problems on the group side come from new representations that appear.


