Codimension of non-flat locusFlat locus of $S_1$-morphismProjectivity in flat familiesAre irreducible components of a flat family flat?When is the determinant of the push-forward of an ample line bundle ampleon flat morphismsflat and finite type morphismsWhen is the flatness locus non-emptyIs the zero locus of a global section flat?Do arithmetic schemes have non-singular alterations?Connected components in flat families

Codimension of non-flat locus


Flat locus of $S_1$-morphismProjectivity in flat familiesAre irreducible components of a flat family flat?When is the determinant of the push-forward of an ample line bundle ampleon flat morphismsflat and finite type morphismsWhen is the flatness locus non-emptyIs the zero locus of a global section flat?Do arithmetic schemes have non-singular alterations?Connected components in flat families













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$begingroup$


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







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      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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      asked 3 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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          4 Answers
          4






          active

          oldest

          votes


















          2












          $begingroup$

          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
            1 hour ago


















          2












          $begingroup$

          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









          $endgroup$




















            1












            $begingroup$

            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











            $endgroup$




















              0












              $begingroup$

              Try $Y=mathrmSpec,mathbbC[x^4, x^3y, xy^3, y^4]$ 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
                1 hour ago










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









              2












              $begingroup$

              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
                1 hour ago















              2












              $begingroup$

              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
                1 hour ago













              2












              2








              2





              $begingroup$

              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 1 hour ago









              Sándor KovácsSándor Kovács

              36.9k284127




              36.9k284127











              • $begingroup$
                what happens if $Y$ is smooth?
                $endgroup$
                – Stepan Banach
                1 hour ago
















              • $begingroup$
                what happens if $Y$ is smooth?
                $endgroup$
                – Stepan Banach
                1 hour ago















              $begingroup$
              what happens if $Y$ is smooth?
              $endgroup$
              – Stepan Banach
              1 hour ago




              $begingroup$
              what happens if $Y$ is smooth?
              $endgroup$
              – Stepan Banach
              1 hour ago











              2












              $begingroup$

              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









              $endgroup$

















                2












                $begingroup$

                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









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  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 21 mins ago









                  SashaSasha

                  21.3k22756




                  21.3k22756





















                      1












                      $begingroup$

                      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












                        $begingroup$

                        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












                          1








                          1





                          $begingroup$

                          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 1 hour ago

























                          answered 1 hour ago









                          David E SpeyerDavid E Speyer

                          108k9282540




                          108k9282540





















                              0












                              $begingroup$

                              Try $Y=mathrmSpec,mathbbC[x^4, x^3y, xy^3, y^4]$ 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
                                1 hour ago










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















                              0












                              $begingroup$

                              Try $Y=mathrmSpec,mathbbC[x^4, x^3y, xy^3, y^4]$ 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
                                1 hour ago










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













                              0












                              0








                              0





                              $begingroup$

                              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 1 hour 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
                                1 hour ago










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












                              • 1




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










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







                              1




                              1




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




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












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




                              $begingroup$
                              @StepanBanach You did not mention normal in your question.
                              $endgroup$
                              – Mohan
                              43 mins 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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                              When We Were Young (canção de Adele) Índice Antecedentes e lançamento | Faixas e formatos | Performances e covers | Desempenho nas tabelas musicais | Créditos | Histórico de lançamento | Referências Menu de navegação«Best albums of 2015»«Adele - When We Were Young (Radio Date: 22-01-2016)»«When We Were Young - Single by Adele»«Adele: Inside Her Private Life and Triumphant Return»«adele interview: world exclusive first interview in three years»«Tobias Jesso Jr: since Adele tweeted his song, he's even bigger than his dad»«Adele interviews Tobias Jesso Jr: 'I think that a couple of the ideas we had could be rap songs'»«Adele on Her Return: 'I Was So Frightened That No One Cared'»«Tobias Jesso Jr. on working with Adele: 'I was as nervous as shit'»«How Ariel Rechtshaid Pushed Adele To Her Limit»«Adele Previews 'When We Were Young' on '60 Minutes' Teaser, Tops Trending 140»«Adele Performs New '25' Ballad, "When We Were Young" Live: Watch»«Which '25' Song Should Be Adele's Next Single?»«Adele's 'When We Were Young' Confirmed As Second Single From '25'»«Adele's new single artwork for 'When We Were Young' is perfectly adorable»«Adele at the BBC review: honest, funny and spectacular – Celebrity News News»«'Saturday Night Live': Adele Sings 'Hello' and 'When We Were Young'»«'Adele: Live in New York City' NBC Special – Set List Revealed!»«Adele Closes Out the 2016 Brit Awards With 'When We Were Young'»«What Is The Adele Live Tour Set List? There's No Way She'd Leave Out These 8 Songs»«See Demi Lovato's Soaring Cover of Adele's 'When We Were Young'»«'The Voice': 5 Best Moments From Week 1 Blind Auditions»«Adele – When We Were Young (Media Control Charts)»«Adele – When We Were Young (Entertainment Monitoring Africa)»«Adele – When We Were Young (ARIA Charts)»«Adele – When We Were Young (Ö3 Austria Top 40)»«Adele – When We Were Young (Ultratop 50)»«Adele – When We Were Young (Ultratop 40)»«Adele – When We Were Young (Canadian Hot 100)»«Adele – When We Were Young (Tracklisten)»«Adele – When We Were Young (The Official Charts Company)»«Adele – Hello (IFPI Slovenská Republika)»«Adele – When We Were Young (Productores de Música de España)»«Adele – When We Were Young (Billboard Hot 100)»«Adele – When We Were Young (Pop Songs)»«Adele – When We Were Young (Adult Pop Songs)»«Adele – When We Were Young (Hot Adult Contemporary Charts)»«Adele – When We Were Young (Hot Dance Club Songs)»«Adele – When We Were Young (Rock Airplay)»«Adele – When We Were Young (IFPI Finlândia)»«Adele – When We Were Young (Syndicat National de l'Éditon Phonographique)»«Adele – When We Were Young (Magyar Hanglemezkiadók Szövetsége)»«Adele – When We Were Young (Irish Recorded Music Association)»«Adele – When We Were Young (Mexico Airplay)»«Adele – When We Were Young (VG-lista)»«Adele – When We Were Young (NZ Top 40 Singles)»«Adele – When We Were Young (MegaCharts)»«Adele – When We Were Young (Związek Producentów Audio Video)»«Adele – When We Were Young (Portugal Digital Songs)»«Adele – When We Were Young (UK Indie Singles Chart)»«Adele – When We Were Young (UK Singles Chart)»«Adele – When We Were Young (Sverigetopplistan)»«Adele – When We Were Young (Schweizer Hitparade)»«Adele – When We Were Young (Portugal Digital Songs)»«Music Canada – Gold/Platinum – When We Were Young»«NZ Top 40 Singles Chart»«Certified Awards»e