Filtration in crystalline Poincaré Lemma












1












$begingroup$


I am trying to understand section 20 in https://stacks.math.columbia.edu/download/crystalline.pdf, especially the proof of Lemma 20.2.



If $Arightarrow B$ is a map of rings and $P=B[x_i]$ is some polynomial algebra, then we can look at the map of De-Rham complexes $Omega^*_{B/A}rightarrow Omega^*_{P/A}$. We have a filtration $F^i$ on $Omega^*_{B/A}$ given by truncating $Omega^*_{B/A}$, i.e. by setting $F^i(Omega^*_{B/A})$ to be the complex which is 0 in degree $j<i$ and $Omega^j_{B/A}$ for $jgeq 0$.



Now I want to use this filtration to define a filtration on $Omega^*_{P/A}$. In the pdf above they do it by setting $F^i(Omega^*_{P/A})=F^i(Omega^*_{B/A})wedge Omega^*_{P/A}$ but I fail to see how this gives me a filtration. How is the wedge product defined (as $F^i(Omega^*_{B/A})$ and $Omega^*_{P/A}$ live over different rings)?






algebraic-geometry commutative-algebra homological-algebra filtrations







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I am trying to understand section 20 in https://stacks.math.columbia.edu/download/crystalline.pdf, especially the proof of Lemma 20.2.



    If $Arightarrow B$ is a map of rings and $P=B[x_i]$ is some polynomial algebra, then we can look at the map of De-Rham complexes $Omega^*_{B/A}rightarrow Omega^*_{P/A}$. We have a filtration $F^i$ on $Omega^*_{B/A}$ given by truncating $Omega^*_{B/A}$, i.e. by setting $F^i(Omega^*_{B/A})$ to be the complex which is 0 in degree $j<i$ and $Omega^j_{B/A}$ for $jgeq 0$.



    Now I want to use this filtration to define a filtration on $Omega^*_{P/A}$. In the pdf above they do it by setting $F^i(Omega^*_{P/A})=F^i(Omega^*_{B/A})wedge Omega^*_{P/A}$ but I fail to see how this gives me a filtration. How is the wedge product defined (as $F^i(Omega^*_{B/A})$ and $Omega^*_{P/A}$ live over different rings)?






    algebraic-geometry commutative-algebra homological-algebra filtrations







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am trying to understand section 20 in https://stacks.math.columbia.edu/download/crystalline.pdf, especially the proof of Lemma 20.2.



      If $Arightarrow B$ is a map of rings and $P=B[x_i]$ is some polynomial algebra, then we can look at the map of De-Rham complexes $Omega^*_{B/A}rightarrow Omega^*_{P/A}$. We have a filtration $F^i$ on $Omega^*_{B/A}$ given by truncating $Omega^*_{B/A}$, i.e. by setting $F^i(Omega^*_{B/A})$ to be the complex which is 0 in degree $j<i$ and $Omega^j_{B/A}$ for $jgeq 0$.



      Now I want to use this filtration to define a filtration on $Omega^*_{P/A}$. In the pdf above they do it by setting $F^i(Omega^*_{P/A})=F^i(Omega^*_{B/A})wedge Omega^*_{P/A}$ but I fail to see how this gives me a filtration. How is the wedge product defined (as $F^i(Omega^*_{B/A})$ and $Omega^*_{P/A}$ live over different rings)?






      algebraic-geometry commutative-algebra homological-algebra filtrations







      share|cite|improve this question









      $endgroup$




      I am trying to understand section 20 in https://stacks.math.columbia.edu/download/crystalline.pdf, especially the proof of Lemma 20.2.



      If $Arightarrow B$ is a map of rings and $P=B[x_i]$ is some polynomial algebra, then we can look at the map of De-Rham complexes $Omega^*_{B/A}rightarrow Omega^*_{P/A}$. We have a filtration $F^i$ on $Omega^*_{B/A}$ given by truncating $Omega^*_{B/A}$, i.e. by setting $F^i(Omega^*_{B/A})$ to be the complex which is 0 in degree $j<i$ and $Omega^j_{B/A}$ for $jgeq 0$.



      Now I want to use this filtration to define a filtration on $Omega^*_{P/A}$. In the pdf above they do it by setting $F^i(Omega^*_{P/A})=F^i(Omega^*_{B/A})wedge Omega^*_{P/A}$ but I fail to see how this gives me a filtration. How is the wedge product defined (as $F^i(Omega^*_{B/A})$ and $Omega^*_{P/A}$ live over different rings)?






      algebraic-geometry commutative-algebra homological-algebra filtrations




      algebraic-geometry commutative-algebra homological-algebra filtrations






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 6 '18 at 16:38









      slin0slin0

      1288




      1288






















          1 Answer
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          $begingroup$

          As you said, there is a map $Omega^*_{B/A}toOmega^*_{P/A}$. This induces $F^i(Omega^*_{B/A})to Omega^*_{P/A}$. Then $F^i(Omega^*_{P/A})$ are the classes that can be written as a product of an element of the image of this map and an element of $Omega^*_{P/A}$. Note that this is not all of $Omega^*_{P/A}$ since $F^i(Omega^*_{B/A})$ contains only elements of degree $geq i$. Intuitively $F^i(Omega^*_{P/A})$ is then spanned by the classes $a_1wedge ...wedge a_n$ such that at least $i$ of the $a_1,...,a_n$ comes from $Omega^*_{B/A}$.



          As an example, let $B=A[x_1,x_2]$ and $P=B[x_3]=A[x_1,x_2,x_3]$. Then $Omega_{B/A}$ is the following complex with the following filtration :



          $$
          require{AMScd}
          begin{CD}
          F^2(Omega^*_{B/A}):@.{}@.{}@.Bdx_1wedge dx_2\
          @.@.@.@VVV\
          F^1(Omega^*_{B/A}):@.{}@.Bdx_1oplus Bdx_2@>>>Bdx_1wedge dx_2\
          @.@.@VVV@VVV\
          F^0(Omega^*_{B/A}):@.B@>>>Bdx_1oplus Bdx_2@>>>Bdx_1wedge dx_2
          end{CD}
          $$
          Then $Omega_{P/A}$ is the following complex with filtration :
          $$
          begin{CD}
          F^2(Omega^*_{P/A}):@.{}@.{}@.Pdx_1wedge dx_2@>>> Pdx_1wedge dx_2wedge dx_3\
          @.@.@.@VVV@VVV\
          F^1(Omega^*_{P/A}):@.{}@.Pdx_1oplus Pdx_2@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3\
          @.@.@VVV@VVV@VVV\
          F^0(Omega^*_{P/A}):@.P@>>>Pdx_1oplus Pdx_2oplus Pdx_3@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3
          end{CD}
          $$






          share|cite|improve this answer









          $endgroup$













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            3












            $begingroup$

            As you said, there is a map $Omega^*_{B/A}toOmega^*_{P/A}$. This induces $F^i(Omega^*_{B/A})to Omega^*_{P/A}$. Then $F^i(Omega^*_{P/A})$ are the classes that can be written as a product of an element of the image of this map and an element of $Omega^*_{P/A}$. Note that this is not all of $Omega^*_{P/A}$ since $F^i(Omega^*_{B/A})$ contains only elements of degree $geq i$. Intuitively $F^i(Omega^*_{P/A})$ is then spanned by the classes $a_1wedge ...wedge a_n$ such that at least $i$ of the $a_1,...,a_n$ comes from $Omega^*_{B/A}$.



            As an example, let $B=A[x_1,x_2]$ and $P=B[x_3]=A[x_1,x_2,x_3]$. Then $Omega_{B/A}$ is the following complex with the following filtration :



            $$
            require{AMScd}
            begin{CD}
            F^2(Omega^*_{B/A}):@.{}@.{}@.Bdx_1wedge dx_2\
            @.@.@.@VVV\
            F^1(Omega^*_{B/A}):@.{}@.Bdx_1oplus Bdx_2@>>>Bdx_1wedge dx_2\
            @.@.@VVV@VVV\
            F^0(Omega^*_{B/A}):@.B@>>>Bdx_1oplus Bdx_2@>>>Bdx_1wedge dx_2
            end{CD}
            $$
            Then $Omega_{P/A}$ is the following complex with filtration :
            $$
            begin{CD}
            F^2(Omega^*_{P/A}):@.{}@.{}@.Pdx_1wedge dx_2@>>> Pdx_1wedge dx_2wedge dx_3\
            @.@.@.@VVV@VVV\
            F^1(Omega^*_{P/A}):@.{}@.Pdx_1oplus Pdx_2@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3\
            @.@.@VVV@VVV@VVV\
            F^0(Omega^*_{P/A}):@.P@>>>Pdx_1oplus Pdx_2oplus Pdx_3@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3
            end{CD}
            $$






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              As you said, there is a map $Omega^*_{B/A}toOmega^*_{P/A}$. This induces $F^i(Omega^*_{B/A})to Omega^*_{P/A}$. Then $F^i(Omega^*_{P/A})$ are the classes that can be written as a product of an element of the image of this map and an element of $Omega^*_{P/A}$. Note that this is not all of $Omega^*_{P/A}$ since $F^i(Omega^*_{B/A})$ contains only elements of degree $geq i$. Intuitively $F^i(Omega^*_{P/A})$ is then spanned by the classes $a_1wedge ...wedge a_n$ such that at least $i$ of the $a_1,...,a_n$ comes from $Omega^*_{B/A}$.



              As an example, let $B=A[x_1,x_2]$ and $P=B[x_3]=A[x_1,x_2,x_3]$. Then $Omega_{B/A}$ is the following complex with the following filtration :



              $$
              require{AMScd}
              begin{CD}
              F^2(Omega^*_{B/A}):@.{}@.{}@.Bdx_1wedge dx_2\
              @.@.@.@VVV\
              F^1(Omega^*_{B/A}):@.{}@.Bdx_1oplus Bdx_2@>>>Bdx_1wedge dx_2\
              @.@.@VVV@VVV\
              F^0(Omega^*_{B/A}):@.B@>>>Bdx_1oplus Bdx_2@>>>Bdx_1wedge dx_2
              end{CD}
              $$
              Then $Omega_{P/A}$ is the following complex with filtration :
              $$
              begin{CD}
              F^2(Omega^*_{P/A}):@.{}@.{}@.Pdx_1wedge dx_2@>>> Pdx_1wedge dx_2wedge dx_3\
              @.@.@.@VVV@VVV\
              F^1(Omega^*_{P/A}):@.{}@.Pdx_1oplus Pdx_2@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3\
              @.@.@VVV@VVV@VVV\
              F^0(Omega^*_{P/A}):@.P@>>>Pdx_1oplus Pdx_2oplus Pdx_3@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3
              end{CD}
              $$






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                As you said, there is a map $Omega^*_{B/A}toOmega^*_{P/A}$. This induces $F^i(Omega^*_{B/A})to Omega^*_{P/A}$. Then $F^i(Omega^*_{P/A})$ are the classes that can be written as a product of an element of the image of this map and an element of $Omega^*_{P/A}$. Note that this is not all of $Omega^*_{P/A}$ since $F^i(Omega^*_{B/A})$ contains only elements of degree $geq i$. Intuitively $F^i(Omega^*_{P/A})$ is then spanned by the classes $a_1wedge ...wedge a_n$ such that at least $i$ of the $a_1,...,a_n$ comes from $Omega^*_{B/A}$.



                As an example, let $B=A[x_1,x_2]$ and $P=B[x_3]=A[x_1,x_2,x_3]$. Then $Omega_{B/A}$ is the following complex with the following filtration :



                $$
                require{AMScd}
                begin{CD}
                F^2(Omega^*_{B/A}):@.{}@.{}@.Bdx_1wedge dx_2\
                @.@.@.@VVV\
                F^1(Omega^*_{B/A}):@.{}@.Bdx_1oplus Bdx_2@>>>Bdx_1wedge dx_2\
                @.@.@VVV@VVV\
                F^0(Omega^*_{B/A}):@.B@>>>Bdx_1oplus Bdx_2@>>>Bdx_1wedge dx_2
                end{CD}
                $$
                Then $Omega_{P/A}$ is the following complex with filtration :
                $$
                begin{CD}
                F^2(Omega^*_{P/A}):@.{}@.{}@.Pdx_1wedge dx_2@>>> Pdx_1wedge dx_2wedge dx_3\
                @.@.@.@VVV@VVV\
                F^1(Omega^*_{P/A}):@.{}@.Pdx_1oplus Pdx_2@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3\
                @.@.@VVV@VVV@VVV\
                F^0(Omega^*_{P/A}):@.P@>>>Pdx_1oplus Pdx_2oplus Pdx_3@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3
                end{CD}
                $$






                share|cite|improve this answer









                $endgroup$



                As you said, there is a map $Omega^*_{B/A}toOmega^*_{P/A}$. This induces $F^i(Omega^*_{B/A})to Omega^*_{P/A}$. Then $F^i(Omega^*_{P/A})$ are the classes that can be written as a product of an element of the image of this map and an element of $Omega^*_{P/A}$. Note that this is not all of $Omega^*_{P/A}$ since $F^i(Omega^*_{B/A})$ contains only elements of degree $geq i$. Intuitively $F^i(Omega^*_{P/A})$ is then spanned by the classes $a_1wedge ...wedge a_n$ such that at least $i$ of the $a_1,...,a_n$ comes from $Omega^*_{B/A}$.



                As an example, let $B=A[x_1,x_2]$ and $P=B[x_3]=A[x_1,x_2,x_3]$. Then $Omega_{B/A}$ is the following complex with the following filtration :



                $$
                require{AMScd}
                begin{CD}
                F^2(Omega^*_{B/A}):@.{}@.{}@.Bdx_1wedge dx_2\
                @.@.@.@VVV\
                F^1(Omega^*_{B/A}):@.{}@.Bdx_1oplus Bdx_2@>>>Bdx_1wedge dx_2\
                @.@.@VVV@VVV\
                F^0(Omega^*_{B/A}):@.B@>>>Bdx_1oplus Bdx_2@>>>Bdx_1wedge dx_2
                end{CD}
                $$
                Then $Omega_{P/A}$ is the following complex with filtration :
                $$
                begin{CD}
                F^2(Omega^*_{P/A}):@.{}@.{}@.Pdx_1wedge dx_2@>>> Pdx_1wedge dx_2wedge dx_3\
                @.@.@.@VVV@VVV\
                F^1(Omega^*_{P/A}):@.{}@.Pdx_1oplus Pdx_2@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3\
                @.@.@VVV@VVV@VVV\
                F^0(Omega^*_{P/A}):@.P@>>>Pdx_1oplus Pdx_2oplus Pdx_3@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3
                end{CD}
                $$







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                share|cite|improve this answer










                answered Dec 7 '18 at 9:43









                RolandRoland

                7,1891913




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