Codimension of non-flat locusFlat locus of $S_{1}$-morphismProjectivity in flat familiesAre irreducible...



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 $mathbb{C}$, $Y$ 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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    3












    $begingroup$


    Let $X$, $Y$ be integral separated schemes of finite type over $mathbb{C}$, $Y$ 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







    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.
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      3












      3








      3





      $begingroup$


      Let $X$, $Y$ be integral separated schemes of finite type over $mathbb{C}$, $Y$ 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







      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 $mathbb{C}$, $Y$ 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






      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.











      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.









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      asked 13 hours ago









      Stepan BanachStepan Banach

      1286




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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


















          5












          $begingroup$

          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
            11 hours ago



















          5












          $begingroup$

          Let $n ge 2$, $X = mathbb{A}^n$, $Y = mathbb{A}^n/{pm 1}$ 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












            $begingroup$

            Yes. Let $Y$ be $mathrm{Spec} k[w,x,y,z]/(wz-xy)$. Let $X$ be $mathrm{Proj} k[w,x,y,z, s,t]/(wz-xy, wt-xs, yt-zs)$ 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 $mathbb{P}^1$, 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 $mathbb{P}^1 times (mathrm{point})$ inside the cone on $mathbb{P}^1 times mathbb{P}^1$.






            share|cite|improve this answer











            $endgroup$





















              1












              $begingroup$

              Try $Y=mathrm{Spec},mathbb{C}[x^4, x^3y, xy^3, y^4]$ and $X$ its normalization.






              share|cite|improve this answer









              $endgroup$









              • 3




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






              • 1




                $begingroup$
                @StepanBanach You did not mention normal in your question.
                $endgroup$
                – Mohan
                10 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









              5












              $begingroup$

              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
                11 hours ago
















              5












              $begingroup$

              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
                11 hours ago














              5












              5








              5





              $begingroup$

              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 11 hours ago









              Sándor KovácsSándor Kovács

              36.9k284127




              36.9k284127












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


















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
















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




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











              5












              $begingroup$

              Let $n ge 2$, $X = mathbb{A}^n$, $Y = mathbb{A}^n/{pm 1}$ 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









              $endgroup$


















                5












                $begingroup$

                Let $n ge 2$, $X = mathbb{A}^n$, $Y = mathbb{A}^n/{pm 1}$ 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









                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  Let $n ge 2$, $X = mathbb{A}^n$, $Y = mathbb{A}^n/{pm 1}$ 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 $n ge 2$, $X = mathbb{A}^n$, $Y = mathbb{A}^n/{pm 1}$ 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 10 hours ago









                  SashaSasha

                  21.3k22756




                  21.3k22756























                      4












                      $begingroup$

                      Yes. Let $Y$ be $mathrm{Spec} k[w,x,y,z]/(wz-xy)$. Let $X$ be $mathrm{Proj} k[w,x,y,z, s,t]/(wz-xy, wt-xs, yt-zs)$ 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 $mathbb{P}^1$, 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 $mathbb{P}^1 times (mathrm{point})$ inside the cone on $mathbb{P}^1 times mathbb{P}^1$.






                      share|cite|improve this answer











                      $endgroup$


















                        4












                        $begingroup$

                        Yes. Let $Y$ be $mathrm{Spec} k[w,x,y,z]/(wz-xy)$. Let $X$ be $mathrm{Proj} k[w,x,y,z, s,t]/(wz-xy, wt-xs, yt-zs)$ 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 $mathbb{P}^1$, 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 $mathbb{P}^1 times (mathrm{point})$ inside the cone on $mathbb{P}^1 times mathbb{P}^1$.






                        share|cite|improve this answer











                        $endgroup$
















                          4












                          4








                          4





                          $begingroup$

                          Yes. Let $Y$ be $mathrm{Spec} k[w,x,y,z]/(wz-xy)$. Let $X$ be $mathrm{Proj} k[w,x,y,z, s,t]/(wz-xy, wt-xs, yt-zs)$ 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 $mathbb{P}^1$, 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 $mathbb{P}^1 times (mathrm{point})$ inside the cone on $mathbb{P}^1 times mathbb{P}^1$.






                          share|cite|improve this answer











                          $endgroup$



                          Yes. Let $Y$ be $mathrm{Spec} k[w,x,y,z]/(wz-xy)$. Let $X$ be $mathrm{Proj} k[w,x,y,z, s,t]/(wz-xy, wt-xs, yt-zs)$ 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 $mathbb{P}^1$, 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 $mathbb{P}^1 times (mathrm{point})$ inside the cone on $mathbb{P}^1 times mathbb{P}^1$.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 11 hours ago

























                          answered 11 hours ago









                          David E SpeyerDavid E Speyer

                          108k9282540




                          108k9282540























                              1












                              $begingroup$

                              Try $Y=mathrm{Spec},mathbb{C}[x^4, x^3y, xy^3, y^4]$ and $X$ its normalization.






                              share|cite|improve this answer









                              $endgroup$









                              • 3




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






                              • 1




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
















                              1












                              $begingroup$

                              Try $Y=mathrm{Spec},mathbb{C}[x^4, x^3y, xy^3, y^4]$ and $X$ its normalization.






                              share|cite|improve this answer









                              $endgroup$









                              • 3




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






                              • 1




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














                              1












                              1








                              1





                              $begingroup$

                              Try $Y=mathrm{Spec},mathbb{C}[x^4, x^3y, xy^3, y^4]$ and $X$ its normalization.






                              share|cite|improve this answer









                              $endgroup$



                              Try $Y=mathrm{Spec},mathbb{C}[x^4, x^3y, xy^3, y^4]$ and $X$ its normalization.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 11 hours ago









                              MohanMohan

                              3,45411312




                              3,45411312








                              • 3




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






                              • 1




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














                              • 3




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






                              • 1




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








                              3




                              3




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




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




                              1




                              1




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




                              $begingroup$
                              @StepanBanach You did not mention normal in your question.
                              $endgroup$
                              – Mohan
                              10 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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