Mixing
if the base chage of all fibres of a morphism is faithfully flat, do we have flatness of the morphism?
$begingroup$
Let $f:Xrightarrow Y$ be a morphism of schemes over $S$ (with possibly Noetherian conditions all over the place).
For every point $sin S$, the morphism base changed to the fiber $f_s:X_srightarrow Y_s$ is faithfully flat.
Do we have that $f$ is flat? or even faithfully flat?
Edit: what if moreover both $X$ and $Y$ are flat over $S$?
algebraic-geometry commutative-algebra flatness
edited Feb 17 at 17:29
user26857
39.4k124183
asked Jan 26 at 11:53
Lance WuLance Wu
1016
$endgroup$
add a comment |
$begingroup$
Let $f:Xrightarrow Y$ be a morphism of schemes over $S$ (with possibly Noetherian conditions all over the place).
For every point $sin S$, the morphism base changed to the fiber $f_s:X_srightarrow Y_s$ is faithfully flat.
Do we have that $f$ is flat? or even faithfully flat?
Edit: what if moreover both $X$ and $Y$ are flat over $S$?
algebraic-geometry commutative-algebra flatness
edited Feb 17 at 17:29
user26857
39.4k124183
asked Jan 26 at 11:53
Lance WuLance Wu
1016
$endgroup$
$begingroup$
If $Y=S$ and $f:Xto S$ is onto, then for every $sin S, f_s:X_sto s$ is faithfully flat since $s$ is the spectrum of the field. But $f$ is obviously not necessarily flat.
$endgroup$
– Roland
Jan 26 at 12:41
$begingroup$
@Roland what if both of $X$ and $Y$ are flat over $S$? Then your example doesn't apply.
$endgroup$
– Lance Wu
Jan 26 at 13:11
add a comment |
$begingroup$
Let $f:Xrightarrow Y$ be a morphism of schemes over $S$ (with possibly Noetherian conditions all over the place).
For every point $sin S$, the morphism base changed to the fiber $f_s:X_srightarrow Y_s$ is faithfully flat.
Do we have that $f$ is flat? or even faithfully flat?
Edit: what if moreover both $X$ and $Y$ are flat over $S$?
algebraic-geometry commutative-algebra flatness
edited Feb 17 at 17:29
user26857
39.4k124183
asked Jan 26 at 11:53
Lance WuLance Wu
1016
$endgroup$
Let $f:Xrightarrow Y$ be a morphism of schemes over $S$ (with possibly Noetherian conditions all over the place).
For every point $sin S$, the morphism base changed to the fiber $f_s:X_srightarrow Y_s$ is faithfully flat.
Do we have that $f$ is flat? or even faithfully flat?
Edit: what if moreover both $X$ and $Y$ are flat over $S$?
algebraic-geometry commutative-algebra flatness
algebraic-geometry commutative-algebra flatness
edited Feb 17 at 17:29
user26857
39.4k124183
asked Jan 26 at 11:53
Lance WuLance Wu
1016
edited Feb 17 at 17:29
user26857
39.4k124183
asked Jan 26 at 11:53
Lance WuLance Wu
1016
edited Feb 17 at 17:29
user26857
39.4k124183
edited Feb 17 at 17:29
user26857
39.4k124183
edited Feb 17 at 17:29
user26857
39.4k124183
39.4k124183
asked Jan 26 at 11:53
Lance WuLance Wu
1016
asked Jan 26 at 11:53
Lance WuLance Wu
1016
asked Jan 26 at 11:53
Lance WuLance Wu
1016
1016
$begingroup$
If $Y=S$ and $f:Xto S$ is onto, then for every $sin S, f_s:X_sto s$ is faithfully flat since $s$ is the spectrum of the field. But $f$ is obviously not necessarily flat.
$endgroup$
– Roland
Jan 26 at 12:41
$begingroup$
@Roland what if both of $X$ and $Y$ are flat over $S$? Then your example doesn't apply.
$endgroup$
– Lance Wu
Jan 26 at 13:11
add a comment |
$begingroup$
If $Y=S$ and $f:Xto S$ is onto, then for every $sin S, f_s:X_sto s$ is faithfully flat since $s$ is the spectrum of the field. But $f$ is obviously not necessarily flat.
$endgroup$
– Roland
Jan 26 at 12:41
$begingroup$
@Roland what if both of $X$ and $Y$ are flat over $S$? Then your example doesn't apply.
$endgroup$
– Lance Wu
Jan 26 at 13:11
$begingroup$
If $Y=S$ and $f:Xto S$ is onto, then for every $sin S, f_s:X_sto s$ is faithfully flat since $s$ is the spectrum of the field. But $f$ is obviously not necessarily flat.
$endgroup$
– Roland
Jan 26 at 12:41
$begingroup$
If $Y=S$ and $f:Xto S$ is onto, then for every $sin S, f_s:X_sto s$ is faithfully flat since $s$ is the spectrum of the field. But $f$ is obviously not necessarily flat.
$endgroup$
– Roland
Jan 26 at 12:41
$begingroup$
@Roland what if both of $X$ and $Y$ are flat over $S$? Then your example doesn't apply.
$endgroup$
– Lance Wu
Jan 26 at 13:11
$begingroup$
@Roland what if both of $X$ and $Y$ are flat over $S$? Then your example doesn't apply.
$endgroup$
– Lance Wu
Jan 26 at 13:11
add a comment |
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$begingroup$
If $Y=S$ and $f:Xto S$ is onto, then for every $sin S, f_s:X_sto s$ is faithfully flat since $s$ is the spectrum of the field. But $f$ is obviously not necessarily flat.
$endgroup$
– Roland
Jan 26 at 12:41
$begingroup$
@Roland what if both of $X$ and $Y$ are flat over $S$? Then your example doesn't apply.
$endgroup$
– Lance Wu
Jan 26 at 13:11