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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    2












    $begingroup$


    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.







    $endgroup$















      2












      2








      2





      $begingroup$


      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




      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 8 hours ago









      Stepan BanachStepan Banach

      1236




      1236




      New contributor




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



















          4












          $begingroup$

          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









          $endgroup$





















            3












            $begingroup$

            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











            $endgroup$





















              0












              $begingroup$

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






              share|cite|improve this answer









              $endgroup$









              • 1




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










              • $begingroup$
                @StepanBanach You did not mention normal in your question.
                $endgroup$
                – Mohan
                5 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
                6 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
                6 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 6 hours ago









              Sándor KovácsSándor Kovács

              36.9k284127




              36.9k284127












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


















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
















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




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











              4












              $begingroup$

              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












                $begingroup$

                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












                  4








                  4





                  $begingroup$

                  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 5 hours ago









                  SashaSasha

                  21.3k22756




                  21.3k22756























                      3












                      $begingroup$

                      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











                      $endgroup$


















                        3












                        $begingroup$

                        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











                        $endgroup$
















                          3












                          3








                          3





                          $begingroup$

                          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 6 hours ago

























                          answered 6 hours ago









                          David E SpeyerDavid E Speyer

                          108k9282540




                          108k9282540























                              0












                              $begingroup$

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






                              share|cite|improve this answer









                              $endgroup$









                              • 1




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










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
















                              0












                              $begingroup$

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






                              share|cite|improve this answer









                              $endgroup$









                              • 1




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










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














                              0












                              0








                              0





                              $begingroup$

                              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 7 hours ago









                              MohanMohan

                              3,44411312




                              3,44411312








                              • 1




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










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














                              • 1




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










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








                              1




                              1




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




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












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




                              $begingroup$
                              @StepanBanach You did not mention normal in your question.
                              $endgroup$
                              – Mohan
                              5 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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                              變成蝙蝠會怎樣？ 參考資料 外部連結 导航菜单Thomas Nagel, "What is it like to be a...

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                              心靈哲學思想實驗 湯瑪斯·內格爾感質物理心靈解釋鴻溝物理信息蝙蝠物理主義科學還原經驗現象學聲吶超音波神經衝動 成為一隻蝙蝠可能是什麼樣子？ 〈 成為一隻蝙蝠可能是什麼樣子 〉（ What is it like to be a bat? ）是湯瑪斯·內格爾在1974年發表的論文，旨在說明感質在物理與心靈之間的解釋鴻溝（ Explanatory gap （ 英语 ： Explanatory gap ） ）。 根據內格爾的論證，物理信息（ Physical information （ 英语 ： Physical information ） ）無法讓我們知道身為一隻蝙蝠的感覺是什麼，也藉此推知，我們無法知道他人對於顏色、聲音、氣味、疼痛等等的感覺「是什麼感覺」。首先，根據物理主義，世界上所有事物似乎都能由科學來解釋，每個人個別的、主觀的感質，是否也能還原為客觀的物理信息？內格爾認為，「從還原中排除經驗的現象學特徵是不可能的」。 [1] 譬如蝙蝠是由聲吶或回聲來感知世界，藉此得知物體的距離、大小、形狀等等。由於蝙蝠的感受方式與我們有很大的不同，我們無法以自身的內在生活去推知蝙蝠的內在生活，所以我們無法得知身為一隻蝙蝠真正的感覺是什麼。就算以科學方法得知蝙蝠在接收到超音波時，腦中會出現特定的神經衝動，但人類不能知道腦中產生那種神經衝動「是什麼感覺」。只有從蝙蝠的主觀觀點出發，才會知道是蝙蝠是什麼感覺，牠的感覺無法用物理的術語去理解，因為物理術語必定是以許多觀點都能理解為基礎的。 參考資料 ^ 內格爾：〈成為一隻蝙蝠可能是什麼樣子？〉，高新民、儲昭華主編：《心靈哲學》（北京：商務印書館，2002），頁107。 外部連結 Thomas Nagel, "What is it like to be a bat?" 这是一篇與哲學相關的小作品。你可以通过编辑或修订扩充其内容。 查 论 编 This page is only for reference, If you need detailed information, please check here
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