2
$\begingroup$

I have a question about following argument used in Szamuely's "Galois Groups and Fundamental Groups" in the excerpt below (or look up at page 159):

enter image description here

Let $X,Y$ schemes which are finite and locally free over base scheme $S$. Let $U \subset S$ be dense in $S$.

If we have a morphism $X \times_S U \to Y \times_S U$ over $U$ why does it extend to a morphism $X \to Y$ over $S$?

My considerations: The problem is local so wlog $S= Spec(R), X= Spec(R^n), Y= Spec(R^m)$(by locally freeness) and $U= D(f)$ for a $f \in R$.

Therefore the problem is reduced to the following affine case:

If $R^m \otimes_R R_f = R_f^m \to R_f^n$ a ring morphism why does it uniquely induce a ring morphism $R^m \to R^n$ uniquely?

Remark: Denote by $R_f$ the localization of $R$ at $f$.

Sure I can compose it with $R^m \to R^m_f$ but does the resulting map $R^m \to R_f^n$ factorize (uniquely) thought $R^n$?

Or did the author mean another approach?

$\endgroup$
  • 1
    $\begingroup$ An open set of the form $D(f)$ has a complement of a codimension $1$ (by the Hauptidealsatz). Here $U$ is assumed to have a complement of codimension $\geq 2$ so is contained in no $D(f)$ and sections on $U$ should be the same as on $S$. It's too late and I'm a bit too tired to fill in the dots, but this should at least get you started. $\endgroup$ – Gro-Tsen May 4 at 22:00
  • $\begingroup$ @Gro-Tsen: If I understood your argument correctly then the point is that the given condition for $U$ having complement of codim $\ge2$ garantees that every global section on $S$ localized/restricted to $U$ doesn't arise from a localization on a $f \in H^0(S,O_S)$. So by restructing $S$ to $U$ is "nothing localized" (I know sounds a bit sloppy)? Or did I misunderstood your explanation? The only aspect that puzzles me a bit is that in your argumentation you didn't used the assumption that $U$ is dense,right? Or is this indeed not necassary? $\endgroup$ – Karl_Peter May 5 at 20:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.