Revisions - MathOverflow

Return to Answer

typo

view source

edited Feb 20 2011 at 16:29

Qing Liu
8,585●17●26

A partial answer: if $X$ is normal and $Y$ is smooth of dimension $1$, then $X_y$ only have constant regular functions ($X\to Y$ is cohomologically flat in relative dimension $0$). This is proved in Raynaud: Spcialisation Spécialisation du foncteur de Picard, Prop. 6.4.2 (use the characteristic 0 hypothesis here, otherwise it is false even when $X$ is also smooth.) It is also proved in the begining of ''Surfaces fibrées en courbes de genre deux'', Lecture Notes in Math. 1137 (1985) by Gang Xiao.

Add: and of course in this situation $X_y$ has no embedded point as $X$ is (S$_2$).

X = S_2

view source

edited Feb 20 2011 at 16:20

Qing Liu
8,585●17●26

A partial answer: if $X$ is normal and $Y$ is smooth of dimension $1$, then $X_y$ only have constant regular functions ($X\to Y$ is cohomologically flat in relative dimension $0$). This is proved in Raynaud: Spcialisation du foncteur de Picard, Prop. 6.4.2 (use the characteristic 0 hypothesis here, otherwise it is false even when $X$ is also smooth.) It is also proved in the begining of ''Surfaces fibrées en courbes de genre deux'', Lecture Notes in Math. 1137 (1985) by Gang Xiao.

Add: and of course in this situation $X_y$ has no embedded point as $X$ is (S$_2$).

view source

answered Feb 20 2011 at 16:14

Qing Liu
8,585●17●26

A partial answer: if $X$ is normal and $Y$ is smooth of dimension $1$, then $X_y$ only have constant regular functions ($X\to Y$ is cohomologically flat in relative dimension $0$). This is proved in Raynaud: Spcialisation du foncteur de Picard, Prop. 6.4.2 (use the characteristic 0 hypothesis here, otherwise it is false even when $X$ is also smooth.) It is also proved in the begining of ''Surfaces fibrées en courbes de genre deux'', Lecture Notes in Math. 1137 (1985) by Gang Xiao.