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.
    Check out our Code of Conduct.







    $endgroup$















      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.









      share|cite|improve this question




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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 13 hours ago









      Stepan BanachStepan Banach

      1286




      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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                              Castillo d'Acher Características Menú de navegación

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                              Montañas de los Pirineos de AragónMontañas de la provincia de Huesca msnmSelva de OzaPirineo aragonésEspañabosque Este artículo o sección necesita referencias que aparezcan en una publicación acreditada. Este aviso fue puesto el 11 de agosto de 2011. El pico desde Selva de Oza. El Castillo d' Acher es una cima de 2.384msnm en la zona de la Selva de Oza en la parte occidental del Pirineo aragonés, en España. Debe su nombre a su aspecto de fortaleza que presenta hacia todas sus vertientes. Características En orden ascendente, la cima pasa de un bosque frondoso de tonalidades verde-amarillenta a laderas rojizas. Su cima, que suele presentarse mucha nieve, es un pequeño valle rectangular rodeado en sus cuatro caras por una muralla que le da un aspecto de fortaleza. This page is only for reference, If you need detailed information, please check here
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