Codimension of non-flat locusFlat locus of $S_1$-morphismProjectivity in flat familiesAre irreducible components of a flat family flat?When is the determinant of the push-forward of an ample line bundle ampleon flat morphismsflat and finite type morphismsWhen is the flatness locus non-emptyIs the zero locus of a global section flat?Do arithmetic schemes have non-singular alterations?Connected components in flat families
Codimension of non-flat locus
Flat locus of $S_1$-morphismProjectivity in flat familiesAre irreducible components of a flat family flat?When is the determinant of the push-forward of an ample line bundle ampleon flat morphismsflat and finite type morphismsWhen is the flatness locus non-emptyIs the zero locus of a global section flat?Do arithmetic schemes have non-singular alterations?Connected components in flat families
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Let $X$, $Y$ be integral separated schemes of finite type over $mathbbC$, $Y$ be normal, $f:Xrightarrow Y$ be a surjective morphism of schemes. Can the non-flat locus of $f$ be non-empty and have codimension $geq 2$ in $X$?
ag.algebraic-geometry
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Let $X$, $Y$ be integral separated schemes of finite type over $mathbbC$, $Y$ be normal, $f:Xrightarrow Y$ be a surjective morphism of schemes. Can the non-flat locus of $f$ be non-empty and have codimension $geq 2$ in $X$?
ag.algebraic-geometry
New contributor
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add a comment |
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Let $X$, $Y$ be integral separated schemes of finite type over $mathbbC$, $Y$ be normal, $f:Xrightarrow Y$ be a surjective morphism of schemes. Can the non-flat locus of $f$ be non-empty and have codimension $geq 2$ in $X$?
ag.algebraic-geometry
New contributor
$endgroup$
Let $X$, $Y$ be integral separated schemes of finite type over $mathbbC$, $Y$ be normal, $f:Xrightarrow Y$ be a surjective morphism of schemes. Can the non-flat locus of $f$ be non-empty and have codimension $geq 2$ in $X$?
ag.algebraic-geometry
ag.algebraic-geometry
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asked 11 hours ago
Stepan BanachStepan Banach
1236
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4 Answers
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Let $Y$ be a cone over a smooth projective variety which is a product and projectively normal (so $Y$ is normal). For example let $Y$ be the cone over $mathbb P^1times mathbb P^1subseteq mathbb P^3$. In general, say $Y$ is a cone over $Vtimes W$.
Next let $Hsubseteq W$ be an effective Cartier divisor (In the $mathbb P^1times mathbb P^1$ example, $H$ is simply a point) and let $H'=Vtimes H$. Finally, let
$f:Xto Y$ be the blow up of $Y$ along its subscheme $Z$ which is the cone over $H'$. Then $f$ is an isomorphism (and hence flat) outside $Z$. Since $Z$ is a Weil divisor, which is Cartier except at the vertex, $f$ is a small morphism, so it is an isom outside a codimension $2$ subset.
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$begingroup$
what happens if $Y$ is smooth?
$endgroup$
– Stepan Banach
9 hours ago
add a comment |
$begingroup$
Let $n ge 2$, $X = mathbbA^n$, $Y = mathbbA^n/pm 1$ and $f colon X to Y$ the quotient morphism. The non-flat locus of $f$ is the point $f(0) in Y$.
$endgroup$
add a comment |
$begingroup$
Yes. Let $Y$ be $mathrmSpec k[w,x,y,z]/(wz-xy)$. Let $X$ be $mathrmProj k[w,x,y,z, s,t]/(wz-xy, wt-xs, yt-zs)$ where $w$, $x$, $y$ and $z$ are in degree $0$ and $s$ and $t$ are in degree $1$. Then $Y$ is three dimensional with a singularity at $w=x=y=z=0$. The map $X to Y$ is a resolution of this singularity; it is an ismorphism (and hence flat) away from the singularity and the fiber over the singularity is $mathbbP^1$, so the nonflat locus is $1$-dimensional inside the $3$-fold $X$.
This is a special case Sándor Kovács example, namely, it is what happens when you blow up the cone on $mathbbP^1 times (mathrmpoint)$ inside the cone on $mathbbP^1 times mathbbP^1$.
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add a comment |
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Try $Y=mathrmSpec,mathbbC[x^4, x^3y, xy^3, y^4]$ and $X$ its normalization.
$endgroup$
2
$begingroup$
isn't the normalization of a normal scheme the identity morphism (whose non-flat locus is empty)?
$endgroup$
– Stepan Banach
10 hours ago
1
$begingroup$
@StepanBanach You did not mention normal in your question.
$endgroup$
– Mohan
8 hours ago
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $Y$ be a cone over a smooth projective variety which is a product and projectively normal (so $Y$ is normal). For example let $Y$ be the cone over $mathbb P^1times mathbb P^1subseteq mathbb P^3$. In general, say $Y$ is a cone over $Vtimes W$.
Next let $Hsubseteq W$ be an effective Cartier divisor (In the $mathbb P^1times mathbb P^1$ example, $H$ is simply a point) and let $H'=Vtimes H$. Finally, let
$f:Xto Y$ be the blow up of $Y$ along its subscheme $Z$ which is the cone over $H'$. Then $f$ is an isomorphism (and hence flat) outside $Z$. Since $Z$ is a Weil divisor, which is Cartier except at the vertex, $f$ is a small morphism, so it is an isom outside a codimension $2$ subset.
$endgroup$
$begingroup$
what happens if $Y$ is smooth?
$endgroup$
– Stepan Banach
9 hours ago
add a comment |
$begingroup$
Let $Y$ be a cone over a smooth projective variety which is a product and projectively normal (so $Y$ is normal). For example let $Y$ be the cone over $mathbb P^1times mathbb P^1subseteq mathbb P^3$. In general, say $Y$ is a cone over $Vtimes W$.
Next let $Hsubseteq W$ be an effective Cartier divisor (In the $mathbb P^1times mathbb P^1$ example, $H$ is simply a point) and let $H'=Vtimes H$. Finally, let
$f:Xto Y$ be the blow up of $Y$ along its subscheme $Z$ which is the cone over $H'$. Then $f$ is an isomorphism (and hence flat) outside $Z$. Since $Z$ is a Weil divisor, which is Cartier except at the vertex, $f$ is a small morphism, so it is an isom outside a codimension $2$ subset.
$endgroup$
$begingroup$
what happens if $Y$ is smooth?
$endgroup$
– Stepan Banach
9 hours ago
add a comment |
$begingroup$
Let $Y$ be a cone over a smooth projective variety which is a product and projectively normal (so $Y$ is normal). For example let $Y$ be the cone over $mathbb P^1times mathbb P^1subseteq mathbb P^3$. In general, say $Y$ is a cone over $Vtimes W$.
Next let $Hsubseteq W$ be an effective Cartier divisor (In the $mathbb P^1times mathbb P^1$ example, $H$ is simply a point) and let $H'=Vtimes H$. Finally, let
$f:Xto Y$ be the blow up of $Y$ along its subscheme $Z$ which is the cone over $H'$. Then $f$ is an isomorphism (and hence flat) outside $Z$. Since $Z$ is a Weil divisor, which is Cartier except at the vertex, $f$ is a small morphism, so it is an isom outside a codimension $2$ subset.
$endgroup$
Let $Y$ be a cone over a smooth projective variety which is a product and projectively normal (so $Y$ is normal). For example let $Y$ be the cone over $mathbb P^1times mathbb P^1subseteq mathbb P^3$. In general, say $Y$ is a cone over $Vtimes W$.
Next let $Hsubseteq W$ be an effective Cartier divisor (In the $mathbb P^1times mathbb P^1$ example, $H$ is simply a point) and let $H'=Vtimes H$. Finally, let
$f:Xto Y$ be the blow up of $Y$ along its subscheme $Z$ which is the cone over $H'$. Then $f$ is an isomorphism (and hence flat) outside $Z$. Since $Z$ is a Weil divisor, which is Cartier except at the vertex, $f$ is a small morphism, so it is an isom outside a codimension $2$ subset.
answered 9 hours ago
Sándor KovácsSándor Kovács
36.9k284127
36.9k284127
$begingroup$
what happens if $Y$ is smooth?
$endgroup$
– Stepan Banach
9 hours ago
add a comment |
$begingroup$
what happens if $Y$ is smooth?
$endgroup$
– Stepan Banach
9 hours ago
$begingroup$
what happens if $Y$ is smooth?
$endgroup$
– Stepan Banach
9 hours ago
$begingroup$
what happens if $Y$ is smooth?
$endgroup$
– Stepan Banach
9 hours ago
add a comment |
$begingroup$
Let $n ge 2$, $X = mathbbA^n$, $Y = mathbbA^n/pm 1$ and $f colon X to Y$ the quotient morphism. The non-flat locus of $f$ is the point $f(0) in Y$.
$endgroup$
add a comment |
$begingroup$
Let $n ge 2$, $X = mathbbA^n$, $Y = mathbbA^n/pm 1$ and $f colon X to Y$ the quotient morphism. The non-flat locus of $f$ is the point $f(0) in Y$.
$endgroup$
add a comment |
$begingroup$
Let $n ge 2$, $X = mathbbA^n$, $Y = mathbbA^n/pm 1$ and $f colon X to Y$ the quotient morphism. The non-flat locus of $f$ is the point $f(0) in Y$.
$endgroup$
Let $n ge 2$, $X = mathbbA^n$, $Y = mathbbA^n/pm 1$ and $f colon X to Y$ the quotient morphism. The non-flat locus of $f$ is the point $f(0) in Y$.
answered 8 hours ago
SashaSasha
21.3k22756
21.3k22756
add a comment |
add a comment |
$begingroup$
Yes. Let $Y$ be $mathrmSpec k[w,x,y,z]/(wz-xy)$. Let $X$ be $mathrmProj k[w,x,y,z, s,t]/(wz-xy, wt-xs, yt-zs)$ where $w$, $x$, $y$ and $z$ are in degree $0$ and $s$ and $t$ are in degree $1$. Then $Y$ is three dimensional with a singularity at $w=x=y=z=0$. The map $X to Y$ is a resolution of this singularity; it is an ismorphism (and hence flat) away from the singularity and the fiber over the singularity is $mathbbP^1$, so the nonflat locus is $1$-dimensional inside the $3$-fold $X$.
This is a special case Sándor Kovács example, namely, it is what happens when you blow up the cone on $mathbbP^1 times (mathrmpoint)$ inside the cone on $mathbbP^1 times mathbbP^1$.
$endgroup$
add a comment |
$begingroup$
Yes. Let $Y$ be $mathrmSpec k[w,x,y,z]/(wz-xy)$. Let $X$ be $mathrmProj k[w,x,y,z, s,t]/(wz-xy, wt-xs, yt-zs)$ where $w$, $x$, $y$ and $z$ are in degree $0$ and $s$ and $t$ are in degree $1$. Then $Y$ is three dimensional with a singularity at $w=x=y=z=0$. The map $X to Y$ is a resolution of this singularity; it is an ismorphism (and hence flat) away from the singularity and the fiber over the singularity is $mathbbP^1$, so the nonflat locus is $1$-dimensional inside the $3$-fold $X$.
This is a special case Sándor Kovács example, namely, it is what happens when you blow up the cone on $mathbbP^1 times (mathrmpoint)$ inside the cone on $mathbbP^1 times mathbbP^1$.
$endgroup$
add a comment |
$begingroup$
Yes. Let $Y$ be $mathrmSpec k[w,x,y,z]/(wz-xy)$. Let $X$ be $mathrmProj k[w,x,y,z, s,t]/(wz-xy, wt-xs, yt-zs)$ where $w$, $x$, $y$ and $z$ are in degree $0$ and $s$ and $t$ are in degree $1$. Then $Y$ is three dimensional with a singularity at $w=x=y=z=0$. The map $X to Y$ is a resolution of this singularity; it is an ismorphism (and hence flat) away from the singularity and the fiber over the singularity is $mathbbP^1$, so the nonflat locus is $1$-dimensional inside the $3$-fold $X$.
This is a special case Sándor Kovács example, namely, it is what happens when you blow up the cone on $mathbbP^1 times (mathrmpoint)$ inside the cone on $mathbbP^1 times mathbbP^1$.
$endgroup$
Yes. Let $Y$ be $mathrmSpec k[w,x,y,z]/(wz-xy)$. Let $X$ be $mathrmProj k[w,x,y,z, s,t]/(wz-xy, wt-xs, yt-zs)$ where $w$, $x$, $y$ and $z$ are in degree $0$ and $s$ and $t$ are in degree $1$. Then $Y$ is three dimensional with a singularity at $w=x=y=z=0$. The map $X to Y$ is a resolution of this singularity; it is an ismorphism (and hence flat) away from the singularity and the fiber over the singularity is $mathbbP^1$, so the nonflat locus is $1$-dimensional inside the $3$-fold $X$.
This is a special case Sándor Kovács example, namely, it is what happens when you blow up the cone on $mathbbP^1 times (mathrmpoint)$ inside the cone on $mathbbP^1 times mathbbP^1$.
edited 9 hours ago
answered 9 hours ago
David E SpeyerDavid E Speyer
108k9282540
108k9282540
add a comment |
add a comment |
$begingroup$
Try $Y=mathrmSpec,mathbbC[x^4, x^3y, xy^3, y^4]$ and $X$ its normalization.
$endgroup$
2
$begingroup$
isn't the normalization of a normal scheme the identity morphism (whose non-flat locus is empty)?
$endgroup$
– Stepan Banach
10 hours ago
1
$begingroup$
@StepanBanach You did not mention normal in your question.
$endgroup$
– Mohan
8 hours ago
add a comment |
$begingroup$
Try $Y=mathrmSpec,mathbbC[x^4, x^3y, xy^3, y^4]$ and $X$ its normalization.
$endgroup$
2
$begingroup$
isn't the normalization of a normal scheme the identity morphism (whose non-flat locus is empty)?
$endgroup$
– Stepan Banach
10 hours ago
1
$begingroup$
@StepanBanach You did not mention normal in your question.
$endgroup$
– Mohan
8 hours ago
add a comment |
$begingroup$
Try $Y=mathrmSpec,mathbbC[x^4, x^3y, xy^3, y^4]$ and $X$ its normalization.
$endgroup$
Try $Y=mathrmSpec,mathbbC[x^4, x^3y, xy^3, y^4]$ and $X$ its normalization.
answered 10 hours ago
MohanMohan
3,45411312
3,45411312
2
$begingroup$
isn't the normalization of a normal scheme the identity morphism (whose non-flat locus is empty)?
$endgroup$
– Stepan Banach
10 hours ago
1
$begingroup$
@StepanBanach You did not mention normal in your question.
$endgroup$
– Mohan
8 hours ago
add a comment |
2
$begingroup$
isn't the normalization of a normal scheme the identity morphism (whose non-flat locus is empty)?
$endgroup$
– Stepan Banach
10 hours ago
1
$begingroup$
@StepanBanach You did not mention normal in your question.
$endgroup$
– Mohan
8 hours ago
2
2
$begingroup$
isn't the normalization of a normal scheme the identity morphism (whose non-flat locus is empty)?
$endgroup$
– Stepan Banach
10 hours ago
$begingroup$
isn't the normalization of a normal scheme the identity morphism (whose non-flat locus is empty)?
$endgroup$
– Stepan Banach
10 hours ago
1
1
$begingroup$
@StepanBanach You did not mention normal in your question.
$endgroup$
– Mohan
8 hours ago
$begingroup$
@StepanBanach You did not mention normal in your question.
$endgroup$
– Mohan
8 hours ago
add a comment |
Stepan Banach is a new contributor. Be nice, and check out our Code of Conduct.
Stepan Banach is a new contributor. Be nice, and check out our Code of Conduct.
Stepan Banach is a new contributor. Be nice, and check out our Code of Conduct.
Stepan Banach is a new contributor. Be nice, and check out our Code of Conduct.
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