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













2












$begingroup$


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 $X$?










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    2












    $begingroup$


    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 $X$?










    share|cite|improve this question







    New contributor




    Stepan Banach is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$














      2












      2








      2





      $begingroup$


      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 $X$?










      share|cite|improve this question







      New contributor




      Stepan Banach is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      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






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      Stepan Banach is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 9 hours ago









      Stepan BanachStepan Banach

      1236




      1236




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      New contributor





      Stepan Banach is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      Check out our Code of Conduct.




















          4 Answers
          4






          active

          oldest

          votes


















          4












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            what happens if $Y$ is smooth?
            $endgroup$
            – Stepan Banach
            7 hours ago


















          4












          $begingroup$

          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 $f(0) in Y$.






          share|cite|improve this answer









          $endgroup$




















            3












            $begingroup$

            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 $mathbbP^1 times mathbbP^1$.






            share|cite|improve this answer











            $endgroup$




















              1












              $begingroup$

              Try $Y=mathrmSpec,mathbbC[x^4, x^3y, xy^3, y^4]$ and $X$ its normalization.






              share|cite|improve this answer









              $endgroup$








              • 2




                $begingroup$
                isn't the normalization of a normal scheme the identity morphism (whose non-flat locus is empty)?
                $endgroup$
                – Stepan Banach
                7 hours ago






              • 1




                $begingroup$
                @StepanBanach You did not mention normal in your question.
                $endgroup$
                – Mohan
                6 hours ago











              Your Answer





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              4 Answers
              4






              active

              oldest

              votes








              4 Answers
              4






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              4












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                what happens if $Y$ is smooth?
                $endgroup$
                – Stepan Banach
                7 hours ago















              4












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                what happens if $Y$ is smooth?
                $endgroup$
                – Stepan Banach
                7 hours ago













              4












              4








              4





              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$



              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.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 7 hours ago









              Sándor KovácsSándor Kovács

              36.9k284127




              36.9k284127











              • $begingroup$
                what happens if $Y$ is smooth?
                $endgroup$
                – Stepan Banach
                7 hours ago
















              • $begingroup$
                what happens if $Y$ is smooth?
                $endgroup$
                – Stepan Banach
                7 hours ago















              $begingroup$
              what happens if $Y$ is smooth?
              $endgroup$
              – Stepan Banach
              7 hours ago




              $begingroup$
              what happens if $Y$ is smooth?
              $endgroup$
              – Stepan Banach
              7 hours ago











              4












              $begingroup$

              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 $f(0) in Y$.






              share|cite|improve this answer









              $endgroup$

















                4












                $begingroup$

                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 $f(0) in Y$.






                share|cite|improve this answer









                $endgroup$















                  4












                  4








                  4





                  $begingroup$

                  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 $f(0) in Y$.






                  share|cite|improve this answer









                  $endgroup$



                  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 $f(0) in Y$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 6 hours ago









                  SashaSasha

                  21.3k22756




                  21.3k22756





















                      3












                      $begingroup$

                      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 $mathbbP^1 times mathbbP^1$.






                      share|cite|improve this answer











                      $endgroup$

















                        3












                        $begingroup$

                        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 $mathbbP^1 times mathbbP^1$.






                        share|cite|improve this answer











                        $endgroup$















                          3












                          3








                          3





                          $begingroup$

                          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 $mathbbP^1 times mathbbP^1$.






                          share|cite|improve this answer











                          $endgroup$



                          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 $mathbbP^1 times mathbbP^1$.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 7 hours ago

























                          answered 7 hours ago









                          David E SpeyerDavid E Speyer

                          108k9282540




                          108k9282540





















                              1












                              $begingroup$

                              Try $Y=mathrmSpec,mathbbC[x^4, x^3y, xy^3, y^4]$ and $X$ its normalization.






                              share|cite|improve this answer









                              $endgroup$








                              • 2




                                $begingroup$
                                isn't the normalization of a normal scheme the identity morphism (whose non-flat locus is empty)?
                                $endgroup$
                                – Stepan Banach
                                7 hours ago






                              • 1




                                $begingroup$
                                @StepanBanach You did not mention normal in your question.
                                $endgroup$
                                – Mohan
                                6 hours ago















                              1












                              $begingroup$

                              Try $Y=mathrmSpec,mathbbC[x^4, x^3y, xy^3, y^4]$ and $X$ its normalization.






                              share|cite|improve this answer









                              $endgroup$








                              • 2




                                $begingroup$
                                isn't the normalization of a normal scheme the identity morphism (whose non-flat locus is empty)?
                                $endgroup$
                                – Stepan Banach
                                7 hours ago






                              • 1




                                $begingroup$
                                @StepanBanach You did not mention normal in your question.
                                $endgroup$
                                – Mohan
                                6 hours ago













                              1












                              1








                              1





                              $begingroup$

                              Try $Y=mathrmSpec,mathbbC[x^4, x^3y, xy^3, y^4]$ and $X$ its normalization.






                              share|cite|improve this answer









                              $endgroup$



                              Try $Y=mathrmSpec,mathbbC[x^4, x^3y, xy^3, y^4]$ and $X$ its normalization.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 7 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
                                7 hours ago






                              • 1




                                $begingroup$
                                @StepanBanach You did not mention normal in your question.
                                $endgroup$
                                – Mohan
                                6 hours ago












                              • 2




                                $begingroup$
                                isn't the normalization of a normal scheme the identity morphism (whose non-flat locus is empty)?
                                $endgroup$
                                – Stepan Banach
                                7 hours ago






                              • 1




                                $begingroup$
                                @StepanBanach You did not mention normal in your question.
                                $endgroup$
                                – Mohan
                                6 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
                              7 hours ago




                              $begingroup$
                              isn't the normalization of a normal scheme the identity morphism (whose non-flat locus is empty)?
                              $endgroup$
                              – Stepan Banach
                              7 hours ago




                              1




                              1




                              $begingroup$
                              @StepanBanach You did not mention normal in your question.
                              $endgroup$
                              – Mohan
                              6 hours ago




                              $begingroup$
                              @StepanBanach You did not mention normal in your question.
                              $endgroup$
                              – Mohan
                              6 hours ago










                              Stepan Banach is a new contributor. Be nice, and check out our Code of Conduct.









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                              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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