If $X,Y$ and $S$ are varieties (over an algebraically closed field) and $X\to S$ and $Y\to S$ are maps with geometrically irreducible generic fibres; why it is true that there is a unique irreducible component of $X\times_{S}Y$ dominating over $S$?
Asked
Active
Viewed 491 times
3
-
This is related, since it covers the case $S=\operatorname{spec} k$: http://math.stackexchange.com/questions/152056/is-fibre-product-of-varieties-irreducible-integral – MooS Feb 16 '17 at 07:51
1 Answers
2
Let $\eta$ be the generic point of $S$.
Note that the fibres $X_\eta$ and $Y_\eta$ are by assumption geometrically irreducible over $k(\eta)$, i.e. irreducible after base change to the algebraic closure. By Is fibre product of varieties irreducible (integral)? this implies that $X_\eta \times_{k(\eta)} Y_\eta$ is also geometrically irreducible. The closure of its image in $X \times_S Y$ is the unique irreducible component that dominates $S$.
