Filtration in crystalline Poincaré Lemma
$begingroup$
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
asked Dec 6 '18 at 16:38
slin0slin0
1288
$endgroup$
add a comment |
$begingroup$
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
asked Dec 6 '18 at 16:38
slin0slin0
1288
$endgroup$
add a comment |
$begingroup$
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
asked Dec 6 '18 at 16:38
slin0slin0
1288
$endgroup$
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
asked Dec 6 '18 at 16:38
slin0slin0
1288
asked Dec 6 '18 at 16:38
slin0slin0
1288
asked Dec 6 '18 at 16:38
slin0slin0
1288
asked Dec 6 '18 at 16:38
slin0slin0
1288
asked Dec 6 '18 at 16:38
slin0slin0
1288
1288
add a comment |
add a comment |
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 :
$$
begin{CD}
F^2(Omega^*_{P/A}):@.{}@.{}@.Pdx_1wedge dx_2@>>> Pdx_1wedge dx_2wedge dx_3\
@.@.@.@VVV@VVV\
F^1(Omega^*_{P/A}):@.{}@.Pdx_1oplus Pdx_2@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3\
@.@.@VVV@VVV@VVV\
F^0(Omega^*_{P/A}):@.P@>>>Pdx_1oplus Pdx_2oplus Pdx_3@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3
end{CD}
$$
answered Dec 7 '18 at 9:43
RolandRoland
7,1891913
$endgroup$
add a comment |
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1 Answer
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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 :
$$
begin{CD}
F^2(Omega^*_{P/A}):@.{}@.{}@.Pdx_1wedge dx_2@>>> Pdx_1wedge dx_2wedge dx_3\
@.@.@.@VVV@VVV\
F^1(Omega^*_{P/A}):@.{}@.Pdx_1oplus Pdx_2@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3\
@.@.@VVV@VVV@VVV\
F^0(Omega^*_{P/A}):@.P@>>>Pdx_1oplus Pdx_2oplus Pdx_3@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3
end{CD}
$$
answered Dec 7 '18 at 9:43
RolandRoland
7,1891913
$endgroup$
add a comment |
$begingroup$
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 :
$$
begin{CD}
F^2(Omega^*_{P/A}):@.{}@.{}@.Pdx_1wedge dx_2@>>> Pdx_1wedge dx_2wedge dx_3\
@.@.@.@VVV@VVV\
F^1(Omega^*_{P/A}):@.{}@.Pdx_1oplus Pdx_2@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3\
@.@.@VVV@VVV@VVV\
F^0(Omega^*_{P/A}):@.P@>>>Pdx_1oplus Pdx_2oplus Pdx_3@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3
end{CD}
$$
answered Dec 7 '18 at 9:43
RolandRoland
7,1891913
$endgroup$
add a comment |
$begingroup$
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 :
$$
begin{CD}
F^2(Omega^*_{P/A}):@.{}@.{}@.Pdx_1wedge dx_2@>>> Pdx_1wedge dx_2wedge dx_3\
@.@.@.@VVV@VVV\
F^1(Omega^*_{P/A}):@.{}@.Pdx_1oplus Pdx_2@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3\
@.@.@VVV@VVV@VVV\
F^0(Omega^*_{P/A}):@.P@>>>Pdx_1oplus Pdx_2oplus Pdx_3@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3
end{CD}
$$
answered Dec 7 '18 at 9:43
RolandRoland
7,1891913
$endgroup$
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 :
$$
begin{CD}
F^2(Omega^*_{P/A}):@.{}@.{}@.Pdx_1wedge dx_2@>>> Pdx_1wedge dx_2wedge dx_3\
@.@.@.@VVV@VVV\
F^1(Omega^*_{P/A}):@.{}@.Pdx_1oplus Pdx_2@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3\
@.@.@VVV@VVV@VVV\
F^0(Omega^*_{P/A}):@.P@>>>Pdx_1oplus Pdx_2oplus Pdx_3@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3
end{CD}
$$
answered Dec 7 '18 at 9:43
RolandRoland
7,1891913
answered Dec 7 '18 at 9:43
RolandRoland
7,1891913
answered Dec 7 '18 at 9:43
RolandRoland
7,1891913
answered Dec 7 '18 at 9:43
RolandRoland
7,1891913
7,1891913
add a comment |
add a comment |
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