Filtration in crystalline Poincaré Lemma
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
add a comment |
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
add a comment |
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
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 at 16:38
slin0
1288
1288
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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 :
$$
begin{CD}
F^2(Omega^*_{P/A}):@.{}@.{}@.Pdx_1wedge dx_2@>>> Pdx_1wedge dx_2wedge dx_3\
@.@.@.@VVV@VVV\
F^1(Omega^*_{P/A}):@.{}@.Pdx_1oplus Pdx_2@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3\
@.@.@VVV@VVV@VVV\
F^0(Omega^*_{P/A}):@.P@>>>Pdx_1oplus Pdx_2oplus Pdx_3@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3
end{CD}
$$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3028735%2ffiltration-in-crystalline-poincar%25c3%25a9-lemma%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 :
$$
begin{CD}
F^2(Omega^*_{P/A}):@.{}@.{}@.Pdx_1wedge dx_2@>>> Pdx_1wedge dx_2wedge dx_3\
@.@.@.@VVV@VVV\
F^1(Omega^*_{P/A}):@.{}@.Pdx_1oplus Pdx_2@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3\
@.@.@VVV@VVV@VVV\
F^0(Omega^*_{P/A}):@.P@>>>Pdx_1oplus Pdx_2oplus Pdx_3@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3
end{CD}
$$
add a comment |
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 :
$$
begin{CD}
F^2(Omega^*_{P/A}):@.{}@.{}@.Pdx_1wedge dx_2@>>> Pdx_1wedge dx_2wedge dx_3\
@.@.@.@VVV@VVV\
F^1(Omega^*_{P/A}):@.{}@.Pdx_1oplus Pdx_2@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3\
@.@.@VVV@VVV@VVV\
F^0(Omega^*_{P/A}):@.P@>>>Pdx_1oplus Pdx_2oplus Pdx_3@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3
end{CD}
$$
add a comment |
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 :
$$
begin{CD}
F^2(Omega^*_{P/A}):@.{}@.{}@.Pdx_1wedge dx_2@>>> Pdx_1wedge dx_2wedge dx_3\
@.@.@.@VVV@VVV\
F^1(Omega^*_{P/A}):@.{}@.Pdx_1oplus Pdx_2@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3\
@.@.@VVV@VVV@VVV\
F^0(Omega^*_{P/A}):@.P@>>>Pdx_1oplus Pdx_2oplus Pdx_3@>>>Pdx_1wedge dx_2oplus Pdx_1wedge dx_3oplus Pdx_2wedge dx_3@>>> Pdx_1wedge dx_2wedge dx_3
end{CD}
$$
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 at 9:43
Roland
6,8891913
6,8891913
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3028735%2ffiltration-in-crystalline-poincar%25c3%25a9-lemma%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown