Mixing
Graded analogues of theorems in commutative algebra
$begingroup$
Many theorems in commutative algebra hold true in a ($mathbb{Z}$-)graded context. More precisely, we can take any theorem in commutative algebra and replace every occurrence of the word
- commutative ring by commutative graded ring (without the sign for commutativity)
- module by graded module
- element by homogeneous element
- ideal by homogeneous ideal (i.e. ideal generated by homogeneous elements)
This results in further substitutions, e.g. a $ast$local ring is a graded ring with a unique maximal homogeneous ideal, we get a notion of graded depth etc. After all these substitutions we can ask whether the theorem is still true.
One book that does some steps in this direction is Cohen-Macaulay rings by Bruns and Herzog, especially Section 1.5. For example, they have as Exercise 1.5.24 the following graded analogue of the Nakayama lemma:
Let $(R,mathfrak{m})$ be a $ast$local ring, $M$ be a finitely generated graded $R$-module and $N$ a graded submodule. Assume $M = N + mathfrak{m}M$. Then $M = N$.
Moreover, a student of mine recently showed that the graded analogue of Lazard's theorem (a module is flat if and only if it is a filtered colimit of free modules) is also true.
Usually one proves this kind of theorems essentially by a combination of two techniques:
- Copy the ungraded proof and just substitute ungraded for graded concepts in the manner sketched above.
- If one is annoyed by the length of the resulting argument, use some shortcuts by some translations between ungraded and graded. (E.g. a noetherian graded ring that is graded Cohen-Macaulay is also ungraded Cohen-Macaulay.)
Sometimes one can also be lucky and the statement is suitably algebro-geometry that one can argue geometrically with the stack $[Spec R/mathbb{G}_m]$ for a graded ring $R$, using that a $mathbb{Z}$-grading corresponds to a $mathbb{G}_m$-action.
In any case, my question is the following:
Is there any class of statements, where one knows automatically that the graded analogue is true if the original statement in ungraded commutative algebra is true, without going through the whole proof?
I am not sure whether one can hope here for a model-theoretic approach as I know almost nothing about model theory, but any such statement could save a lot of work in proving graded analogues of known theorems.
ac.commutative-algebra graded-rings-modules
asked Jan 7 at 8:31
Lennart MeierLennart Meier
6,47924573
$endgroup$
add a comment |
$begingroup$
Many theorems in commutative algebra hold true in a ($mathbb{Z}$-)graded context. More precisely, we can take any theorem in commutative algebra and replace every occurrence of the word
- commutative ring by commutative graded ring (without the sign for commutativity)
- module by graded module
- element by homogeneous element
- ideal by homogeneous ideal (i.e. ideal generated by homogeneous elements)
This results in further substitutions, e.g. a $ast$local ring is a graded ring with a unique maximal homogeneous ideal, we get a notion of graded depth etc. After all these substitutions we can ask whether the theorem is still true.
One book that does some steps in this direction is Cohen-Macaulay rings by Bruns and Herzog, especially Section 1.5. For example, they have as Exercise 1.5.24 the following graded analogue of the Nakayama lemma:
Let $(R,mathfrak{m})$ be a $ast$local ring, $M$ be a finitely generated graded $R$-module and $N$ a graded submodule. Assume $M = N + mathfrak{m}M$. Then $M = N$.
Moreover, a student of mine recently showed that the graded analogue of Lazard's theorem (a module is flat if and only if it is a filtered colimit of free modules) is also true.
Usually one proves this kind of theorems essentially by a combination of two techniques:
- Copy the ungraded proof and just substitute ungraded for graded concepts in the manner sketched above.
- If one is annoyed by the length of the resulting argument, use some shortcuts by some translations between ungraded and graded. (E.g. a noetherian graded ring that is graded Cohen-Macaulay is also ungraded Cohen-Macaulay.)
Sometimes one can also be lucky and the statement is suitably algebro-geometry that one can argue geometrically with the stack $[Spec R/mathbb{G}_m]$ for a graded ring $R$, using that a $mathbb{Z}$-grading corresponds to a $mathbb{G}_m$-action.
In any case, my question is the following:
Is there any class of statements, where one knows automatically that the graded analogue is true if the original statement in ungraded commutative algebra is true, without going through the whole proof?
I am not sure whether one can hope here for a model-theoretic approach as I know almost nothing about model theory, but any such statement could save a lot of work in proving graded analogues of known theorems.
ac.commutative-algebra graded-rings-modules
asked Jan 7 at 8:31
Lennart MeierLennart Meier
6,47924573
$endgroup$
$begingroup$
Do you have a reference for how to argue geometrically using stacks as you allude to?
$endgroup$
– Avi Steiner
Jan 13 at 4:14
1
$begingroup$
Call a graded module over a graded ring $R$ graded flat if its tensor product with every exact sequence of graded modules is graded again. I claim that a module is graded flat if and only if its flat. Indeed, the category of graded modules is monoidally equivalent to quasi-coherent sheaves on $[Spec R/mathbb{G}_m]$. As $Spec R to [Spec R/mathbb{G}_m]$ is fpqc, exactness of a sequence can be tested on $Spec R$, which implies the result. Thus every theorem about flat modules also applies to graded flat modules.
$endgroup$
– Lennart Meier
Jan 14 at 7:16
add a comment |
$begingroup$
Many theorems in commutative algebra hold true in a ($mathbb{Z}$-)graded context. More precisely, we can take any theorem in commutative algebra and replace every occurrence of the word
- commutative ring by commutative graded ring (without the sign for commutativity)
- module by graded module
- element by homogeneous element
- ideal by homogeneous ideal (i.e. ideal generated by homogeneous elements)
This results in further substitutions, e.g. a $ast$local ring is a graded ring with a unique maximal homogeneous ideal, we get a notion of graded depth etc. After all these substitutions we can ask whether the theorem is still true.
One book that does some steps in this direction is Cohen-Macaulay rings by Bruns and Herzog, especially Section 1.5. For example, they have as Exercise 1.5.24 the following graded analogue of the Nakayama lemma:
Let $(R,mathfrak{m})$ be a $ast$local ring, $M$ be a finitely generated graded $R$-module and $N$ a graded submodule. Assume $M = N + mathfrak{m}M$. Then $M = N$.
Moreover, a student of mine recently showed that the graded analogue of Lazard's theorem (a module is flat if and only if it is a filtered colimit of free modules) is also true.
Usually one proves this kind of theorems essentially by a combination of two techniques:
- Copy the ungraded proof and just substitute ungraded for graded concepts in the manner sketched above.
- If one is annoyed by the length of the resulting argument, use some shortcuts by some translations between ungraded and graded. (E.g. a noetherian graded ring that is graded Cohen-Macaulay is also ungraded Cohen-Macaulay.)
Sometimes one can also be lucky and the statement is suitably algebro-geometry that one can argue geometrically with the stack $[Spec R/mathbb{G}_m]$ for a graded ring $R$, using that a $mathbb{Z}$-grading corresponds to a $mathbb{G}_m$-action.
In any case, my question is the following:
Is there any class of statements, where one knows automatically that the graded analogue is true if the original statement in ungraded commutative algebra is true, without going through the whole proof?
I am not sure whether one can hope here for a model-theoretic approach as I know almost nothing about model theory, but any such statement could save a lot of work in proving graded analogues of known theorems.
ac.commutative-algebra graded-rings-modules
asked Jan 7 at 8:31
Lennart MeierLennart Meier
6,47924573
$endgroup$
Many theorems in commutative algebra hold true in a ($mathbb{Z}$-)graded context. More precisely, we can take any theorem in commutative algebra and replace every occurrence of the word
- commutative ring by commutative graded ring (without the sign for commutativity)
- module by graded module
- element by homogeneous element
- ideal by homogeneous ideal (i.e. ideal generated by homogeneous elements)
This results in further substitutions, e.g. a $ast$local ring is a graded ring with a unique maximal homogeneous ideal, we get a notion of graded depth etc. After all these substitutions we can ask whether the theorem is still true.
One book that does some steps in this direction is Cohen-Macaulay rings by Bruns and Herzog, especially Section 1.5. For example, they have as Exercise 1.5.24 the following graded analogue of the Nakayama lemma:
Let $(R,mathfrak{m})$ be a $ast$local ring, $M$ be a finitely generated graded $R$-module and $N$ a graded submodule. Assume $M = N + mathfrak{m}M$. Then $M = N$.
Moreover, a student of mine recently showed that the graded analogue of Lazard's theorem (a module is flat if and only if it is a filtered colimit of free modules) is also true.
Usually one proves this kind of theorems essentially by a combination of two techniques:
- Copy the ungraded proof and just substitute ungraded for graded concepts in the manner sketched above.
- If one is annoyed by the length of the resulting argument, use some shortcuts by some translations between ungraded and graded. (E.g. a noetherian graded ring that is graded Cohen-Macaulay is also ungraded Cohen-Macaulay.)
Sometimes one can also be lucky and the statement is suitably algebro-geometry that one can argue geometrically with the stack $[Spec R/mathbb{G}_m]$ for a graded ring $R$, using that a $mathbb{Z}$-grading corresponds to a $mathbb{G}_m$-action.
In any case, my question is the following:
Is there any class of statements, where one knows automatically that the graded analogue is true if the original statement in ungraded commutative algebra is true, without going through the whole proof?
I am not sure whether one can hope here for a model-theoretic approach as I know almost nothing about model theory, but any such statement could save a lot of work in proving graded analogues of known theorems.
ac.commutative-algebra graded-rings-modules
ac.commutative-algebra graded-rings-modules
asked Jan 7 at 8:31
Lennart MeierLennart Meier
6,47924573
asked Jan 7 at 8:31
Lennart MeierLennart Meier
6,47924573
asked Jan 7 at 8:31
Lennart MeierLennart Meier
6,47924573
asked Jan 7 at 8:31
Lennart MeierLennart Meier
6,47924573
asked Jan 7 at 8:31
Lennart MeierLennart Meier
6,47924573
6,47924573
$begingroup$
Do you have a reference for how to argue geometrically using stacks as you allude to?
$endgroup$
– Avi Steiner
Jan 13 at 4:14
1
$begingroup$
Call a graded module over a graded ring $R$ graded flat if its tensor product with every exact sequence of graded modules is graded again. I claim that a module is graded flat if and only if its flat. Indeed, the category of graded modules is monoidally equivalent to quasi-coherent sheaves on $[Spec R/mathbb{G}_m]$. As $Spec R to [Spec R/mathbb{G}_m]$ is fpqc, exactness of a sequence can be tested on $Spec R$, which implies the result. Thus every theorem about flat modules also applies to graded flat modules.
$endgroup$
– Lennart Meier
Jan 14 at 7:16
add a comment |
$begingroup$
Do you have a reference for how to argue geometrically using stacks as you allude to?
$endgroup$
– Avi Steiner
Jan 13 at 4:14
1
$begingroup$
Call a graded module over a graded ring $R$ graded flat if its tensor product with every exact sequence of graded modules is graded again. I claim that a module is graded flat if and only if its flat. Indeed, the category of graded modules is monoidally equivalent to quasi-coherent sheaves on $[Spec R/mathbb{G}_m]$. As $Spec R to [Spec R/mathbb{G}_m]$ is fpqc, exactness of a sequence can be tested on $Spec R$, which implies the result. Thus every theorem about flat modules also applies to graded flat modules.
$endgroup$
– Lennart Meier
Jan 14 at 7:16
$begingroup$
Do you have a reference for how to argue geometrically using stacks as you allude to?
$endgroup$
– Avi Steiner
Jan 13 at 4:14
$begingroup$
Do you have a reference for how to argue geometrically using stacks as you allude to?
$endgroup$
– Avi Steiner
Jan 13 at 4:14
1
1
$begingroup$
Call a graded module over a graded ring $R$ graded flat if its tensor product with every exact sequence of graded modules is graded again. I claim that a module is graded flat if and only if its flat. Indeed, the category of graded modules is monoidally equivalent to quasi-coherent sheaves on $[Spec R/mathbb{G}_m]$. As $Spec R to [Spec R/mathbb{G}_m]$ is fpqc, exactness of a sequence can be tested on $Spec R$, which implies the result. Thus every theorem about flat modules also applies to graded flat modules.
$endgroup$
– Lennart Meier
Jan 14 at 7:16
$begingroup$
Call a graded module over a graded ring $R$ graded flat if its tensor product with every exact sequence of graded modules is graded again. I claim that a module is graded flat if and only if its flat. Indeed, the category of graded modules is monoidally equivalent to quasi-coherent sheaves on $[Spec R/mathbb{G}_m]$. As $Spec R to [Spec R/mathbb{G}_m]$ is fpqc, exactness of a sequence can be tested on $Spec R$, which implies the result. Thus every theorem about flat modules also applies to graded flat modules.
$endgroup$
– Lennart Meier
Jan 14 at 7:16
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $mathbb P(A,M)$ be a property of a finitely generated module $M$ over a
Noetherian local ring $A$.
Let $R$ be a $mathbb Z^n$-graded ring, $N$ a finitely generated graded $R$-module, and
$P$ a prime ideal of $R$.
Let $P^*$ be the largest homogeneous ideal contained in $P$ (it is a prime ideal).
For many important $mathbb P$, we have that $mathbb P(R_P,N_P)$ is equivalent to
$mathbb P(R_{P^*},N_{P^*})$.
For example, the property $M=0$ satisfies the equivalence, and the graded Nakayama
follows from this.
Indeed, if $mathcal C$ and $mathcal D$ are classes of Noetherian local rings and
$mathbb P(A,M)$ is a property of a finitely generated module over a Noetherian
local ring, and assume that
(1) If $Ainmathcal C$, $M$ a finite $A$-module, $mathbb P(A,M)$, and $Arightarrow B$ is a regular local
homomorphism essentially of finite type, then $Binmathcal D$ and $mathbb P(B,B
otimes_AM)$ holds.
(2) If $A$ is a Noetherian local ring and $M$ a finite $A$-module, $Arightarrow B$ is a regular local homomorphism essentially of finite type, $Binmathcal D$ and $mathbb P(B,Botimes_AM)$ holds, then $Ainmathcal D$ and $mathbb P(A,M)$ holds.
Then for any $mathbb Z^n$-graded Noetherian ring $R$ and a finitely generated graded $R$-module
$N$, if $R_{P^*}inmathcal C$ and $mathbb P(R_{P^*},N_{P^*})$ holds, then $R_Pinmathcal D$ and $mathbb P(R_P,N_P)$.
For example, letting $mathcal C=mathcal D=text{Cohen-Macaulay}$ and $mathbb P$ be
anything, we have that $R_{P^*}$ CM implies $R_P$ CM.
So if $(R,mathfrak m)$ is a ${}^*$local ring and $R$ is graded CM, then
$R_{mathfrak m}$ is CM, and so $R_Q$ is CM for any graded prime ideal $Q$.
So $R_P$ is CM for any prime, since $R_{P^*}$ is CM.
We have proved that $R$ is CM.
Similar argument is applicable to Gorenstein, local complete intersection, and regular.
It is also applicable to some $F$-singularities.
On the other hand, letting $mathcal C=mathcal D=text{Noetherian local}$, and
letting $mathbb P(A,M)$ to be finite projective dimension,'finite injective
dimension,' torsionless,'reflexive' and so on, the condition is satisfied.
Maybe this does not produce any graded "analogue," but I hope this is useful.
Please see section 7 of M. Hashimoto and M. Miyazaki, Comm. Algebra 2013 for details.
answered Jan 7 at 14:37
P. GrapeP. Grape
814
$endgroup$
add a comment |
$begingroup$
Maybe not really what you are after, but anyway: This holds for all those statements that depend only on the structure of abelian category on the category of graded modules. Namely, if $G$ is any commutative group, then the category of $G$-graded $R$-modules, where $R$ is a $G$-graded ring, is abelian and fulfills AB5 and AB4, hence has a projective generator and an injective cogenerator.
However, one has to be careful with this approach, since there are still many differences between these categories for different groups $G$. For example, suppose that $R$ is a noetherian $G$-graded ring. If $G$ is finite, then the category of $G$-graded $R$-modules has a noetherian generator, but if $G$ is infinite then it does not.
answered Jan 7 at 12:24
Fred RohrerFred Rohrer
4,39611734
$endgroup$
add a comment |
$begingroup$
Since the OP is asking for some "class of statements" which can be readily transferred from the ungraded to the graded case, let me outline a couple of thoughts, which are not really tied to neither the commutative case, nor the $mathbb{Z}$-gradation itself:
- On the one hand, if $R$ is a $k$-algebra and $G$ is a group, then a $G$-gradation on $R$ can be equivalently viewed either as a $(kG)^*$-action or as a $kG$-coaction (where $kG$ is the group hopf algebra and $(kG)^*$ its dual hopf algebra). So, the graded algebra $R$ is alternatively viewed as an algebra in the category of $(kG)^*$-modules or an algebra in the category of $kG$-comodules. In this sense, the statements and the properties which are "naturally" transferred between the graded and the ungraded case are those which are in some sense "compatible" or "invariant" with respect to the $kG$-coaction (or the $(kG)^*$-action). I do not know if it sounds like an abuse of terminology to speak about
properties in the category of $(kG)^*$-modules or properties in the category of $kG$-comodules
on the other hand, the Category $R_{gr}$-$mod$ of graded modules over the $G$-graded ring $R$, is equivalent (actually i think it is isomorphic in most cases) to a closed subcategory of modules (in the ungraded sense) over the smash product algebra $Rsharp kG$. (Here the smash product is understood in the sense of the one associated to a Hopf module algebra).
In this sense, we could say that the statements and properties which one might expect to "naturally transfer" over between graded and ungraded, are
the ones which are preserved under the smash product between $R$ and the group hopf algebra $kG$.
(In some sense, this implies that it might be meaningful to consider properties which "transfer" between the $R$-modules and the $Rsharp kG$-modules).
One might reasonably expect that
all those statements that depend only on the structure of abelian category on the category of graded modules
mentioned in Fred Rohrer's answer fall into the classes mentioned above, however i can't tell for sure on the exact overlap between these classes.
answered Jan 8 at 2:44
Konstantinos KanakoglouKonstantinos Kanakoglou
3,30021234
$endgroup$
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: "504"
};
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%2fmathoverflow.net%2fquestions%2f320269%2fgraded-analogues-of-theorems-in-commutative-algebra%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $mathbb P(A,M)$ be a property of a finitely generated module $M$ over a
Noetherian local ring $A$.
Let $R$ be a $mathbb Z^n$-graded ring, $N$ a finitely generated graded $R$-module, and
$P$ a prime ideal of $R$.
Let $P^*$ be the largest homogeneous ideal contained in $P$ (it is a prime ideal).
For many important $mathbb P$, we have that $mathbb P(R_P,N_P)$ is equivalent to
$mathbb P(R_{P^*},N_{P^*})$.
For example, the property $M=0$ satisfies the equivalence, and the graded Nakayama
follows from this.
Indeed, if $mathcal C$ and $mathcal D$ are classes of Noetherian local rings and
$mathbb P(A,M)$ is a property of a finitely generated module over a Noetherian
local ring, and assume that
(1) If $Ainmathcal C$, $M$ a finite $A$-module, $mathbb P(A,M)$, and $Arightarrow B$ is a regular local
homomorphism essentially of finite type, then $Binmathcal D$ and $mathbb P(B,B
otimes_AM)$ holds.
(2) If $A$ is a Noetherian local ring and $M$ a finite $A$-module, $Arightarrow B$ is a regular local homomorphism essentially of finite type, $Binmathcal D$ and $mathbb P(B,Botimes_AM)$ holds, then $Ainmathcal D$ and $mathbb P(A,M)$ holds.
Then for any $mathbb Z^n$-graded Noetherian ring $R$ and a finitely generated graded $R$-module
$N$, if $R_{P^*}inmathcal C$ and $mathbb P(R_{P^*},N_{P^*})$ holds, then $R_Pinmathcal D$ and $mathbb P(R_P,N_P)$.
For example, letting $mathcal C=mathcal D=text{Cohen-Macaulay}$ and $mathbb P$ be
anything, we have that $R_{P^*}$ CM implies $R_P$ CM.
So if $(R,mathfrak m)$ is a ${}^*$local ring and $R$ is graded CM, then
$R_{mathfrak m}$ is CM, and so $R_Q$ is CM for any graded prime ideal $Q$.
So $R_P$ is CM for any prime, since $R_{P^*}$ is CM.
We have proved that $R$ is CM.
Similar argument is applicable to Gorenstein, local complete intersection, and regular.
It is also applicable to some $F$-singularities.
On the other hand, letting $mathcal C=mathcal D=text{Noetherian local}$, and
letting $mathbb P(A,M)$ to be finite projective dimension,'finite injective
dimension,' torsionless,'reflexive' and so on, the condition is satisfied.
Maybe this does not produce any graded "analogue," but I hope this is useful.
Please see section 7 of M. Hashimoto and M. Miyazaki, Comm. Algebra 2013 for details.
answered Jan 7 at 14:37
P. GrapeP. Grape
814
$endgroup$
add a comment |
$begingroup$
Let $mathbb P(A,M)$ be a property of a finitely generated module $M$ over a
Noetherian local ring $A$.
Let $R$ be a $mathbb Z^n$-graded ring, $N$ a finitely generated graded $R$-module, and
$P$ a prime ideal of $R$.
Let $P^*$ be the largest homogeneous ideal contained in $P$ (it is a prime ideal).
For many important $mathbb P$, we have that $mathbb P(R_P,N_P)$ is equivalent to
$mathbb P(R_{P^*},N_{P^*})$.
For example, the property $M=0$ satisfies the equivalence, and the graded Nakayama
follows from this.
Indeed, if $mathcal C$ and $mathcal D$ are classes of Noetherian local rings and
$mathbb P(A,M)$ is a property of a finitely generated module over a Noetherian
local ring, and assume that
(1) If $Ainmathcal C$, $M$ a finite $A$-module, $mathbb P(A,M)$, and $Arightarrow B$ is a regular local
homomorphism essentially of finite type, then $Binmathcal D$ and $mathbb P(B,B
otimes_AM)$ holds.
(2) If $A$ is a Noetherian local ring and $M$ a finite $A$-module, $Arightarrow B$ is a regular local homomorphism essentially of finite type, $Binmathcal D$ and $mathbb P(B,Botimes_AM)$ holds, then $Ainmathcal D$ and $mathbb P(A,M)$ holds.
Then for any $mathbb Z^n$-graded Noetherian ring $R$ and a finitely generated graded $R$-module
$N$, if $R_{P^*}inmathcal C$ and $mathbb P(R_{P^*},N_{P^*})$ holds, then $R_Pinmathcal D$ and $mathbb P(R_P,N_P)$.
For example, letting $mathcal C=mathcal D=text{Cohen-Macaulay}$ and $mathbb P$ be
anything, we have that $R_{P^*}$ CM implies $R_P$ CM.
So if $(R,mathfrak m)$ is a ${}^*$local ring and $R$ is graded CM, then
$R_{mathfrak m}$ is CM, and so $R_Q$ is CM for any graded prime ideal $Q$.
So $R_P$ is CM for any prime, since $R_{P^*}$ is CM.
We have proved that $R$ is CM.
Similar argument is applicable to Gorenstein, local complete intersection, and regular.
It is also applicable to some $F$-singularities.
On the other hand, letting $mathcal C=mathcal D=text{Noetherian local}$, and
letting $mathbb P(A,M)$ to be finite projective dimension,'finite injective
dimension,' torsionless,'reflexive' and so on, the condition is satisfied.
Maybe this does not produce any graded "analogue," but I hope this is useful.
Please see section 7 of M. Hashimoto and M. Miyazaki, Comm. Algebra 2013 for details.
answered Jan 7 at 14:37
P. GrapeP. Grape
814
$endgroup$
add a comment |
$begingroup$
Let $mathbb P(A,M)$ be a property of a finitely generated module $M$ over a
Noetherian local ring $A$.
Let $R$ be a $mathbb Z^n$-graded ring, $N$ a finitely generated graded $R$-module, and
$P$ a prime ideal of $R$.
Let $P^*$ be the largest homogeneous ideal contained in $P$ (it is a prime ideal).
For many important $mathbb P$, we have that $mathbb P(R_P,N_P)$ is equivalent to
$mathbb P(R_{P^*},N_{P^*})$.
For example, the property $M=0$ satisfies the equivalence, and the graded Nakayama
follows from this.
Indeed, if $mathcal C$ and $mathcal D$ are classes of Noetherian local rings and
$mathbb P(A,M)$ is a property of a finitely generated module over a Noetherian
local ring, and assume that
(1) If $Ainmathcal C$, $M$ a finite $A$-module, $mathbb P(A,M)$, and $Arightarrow B$ is a regular local
homomorphism essentially of finite type, then $Binmathcal D$ and $mathbb P(B,B
otimes_AM)$ holds.
(2) If $A$ is a Noetherian local ring and $M$ a finite $A$-module, $Arightarrow B$ is a regular local homomorphism essentially of finite type, $Binmathcal D$ and $mathbb P(B,Botimes_AM)$ holds, then $Ainmathcal D$ and $mathbb P(A,M)$ holds.
Then for any $mathbb Z^n$-graded Noetherian ring $R$ and a finitely generated graded $R$-module
$N$, if $R_{P^*}inmathcal C$ and $mathbb P(R_{P^*},N_{P^*})$ holds, then $R_Pinmathcal D$ and $mathbb P(R_P,N_P)$.
For example, letting $mathcal C=mathcal D=text{Cohen-Macaulay}$ and $mathbb P$ be
anything, we have that $R_{P^*}$ CM implies $R_P$ CM.
So if $(R,mathfrak m)$ is a ${}^*$local ring and $R$ is graded CM, then
$R_{mathfrak m}$ is CM, and so $R_Q$ is CM for any graded prime ideal $Q$.
So $R_P$ is CM for any prime, since $R_{P^*}$ is CM.
We have proved that $R$ is CM.
Similar argument is applicable to Gorenstein, local complete intersection, and regular.
It is also applicable to some $F$-singularities.
On the other hand, letting $mathcal C=mathcal D=text{Noetherian local}$, and
letting $mathbb P(A,M)$ to be finite projective dimension,'finite injective
dimension,' torsionless,'reflexive' and so on, the condition is satisfied.
Maybe this does not produce any graded "analogue," but I hope this is useful.
Please see section 7 of M. Hashimoto and M. Miyazaki, Comm. Algebra 2013 for details.
answered Jan 7 at 14:37
P. GrapeP. Grape
814
$endgroup$
Let $mathbb P(A,M)$ be a property of a finitely generated module $M$ over a
Noetherian local ring $A$.
Let $R$ be a $mathbb Z^n$-graded ring, $N$ a finitely generated graded $R$-module, and
$P$ a prime ideal of $R$.
Let $P^*$ be the largest homogeneous ideal contained in $P$ (it is a prime ideal).
For many important $mathbb P$, we have that $mathbb P(R_P,N_P)$ is equivalent to
$mathbb P(R_{P^*},N_{P^*})$.
For example, the property $M=0$ satisfies the equivalence, and the graded Nakayama
follows from this.
Indeed, if $mathcal C$ and $mathcal D$ are classes of Noetherian local rings and
$mathbb P(A,M)$ is a property of a finitely generated module over a Noetherian
local ring, and assume that
(1) If $Ainmathcal C$, $M$ a finite $A$-module, $mathbb P(A,M)$, and $Arightarrow B$ is a regular local
homomorphism essentially of finite type, then $Binmathcal D$ and $mathbb P(B,B
otimes_AM)$ holds.
(2) If $A$ is a Noetherian local ring and $M$ a finite $A$-module, $Arightarrow B$ is a regular local homomorphism essentially of finite type, $Binmathcal D$ and $mathbb P(B,Botimes_AM)$ holds, then $Ainmathcal D$ and $mathbb P(A,M)$ holds.
Then for any $mathbb Z^n$-graded Noetherian ring $R$ and a finitely generated graded $R$-module
$N$, if $R_{P^*}inmathcal C$ and $mathbb P(R_{P^*},N_{P^*})$ holds, then $R_Pinmathcal D$ and $mathbb P(R_P,N_P)$.
For example, letting $mathcal C=mathcal D=text{Cohen-Macaulay}$ and $mathbb P$ be
anything, we have that $R_{P^*}$ CM implies $R_P$ CM.
So if $(R,mathfrak m)$ is a ${}^*$local ring and $R$ is graded CM, then
$R_{mathfrak m}$ is CM, and so $R_Q$ is CM for any graded prime ideal $Q$.
So $R_P$ is CM for any prime, since $R_{P^*}$ is CM.
We have proved that $R$ is CM.
Similar argument is applicable to Gorenstein, local complete intersection, and regular.
It is also applicable to some $F$-singularities.
On the other hand, letting $mathcal C=mathcal D=text{Noetherian local}$, and
letting $mathbb P(A,M)$ to be finite projective dimension,'finite injective
dimension,' torsionless,'reflexive' and so on, the condition is satisfied.
Maybe this does not produce any graded "analogue," but I hope this is useful.
Please see section 7 of M. Hashimoto and M. Miyazaki, Comm. Algebra 2013 for details.
answered Jan 7 at 14:37
P. GrapeP. Grape
814
answered Jan 7 at 14:37
P. GrapeP. Grape
814
answered Jan 7 at 14:37
P. GrapeP. Grape
814
answered Jan 7 at 14:37
P. GrapeP. Grape
814
814
add a comment |
add a comment |
$begingroup$
Maybe not really what you are after, but anyway: This holds for all those statements that depend only on the structure of abelian category on the category of graded modules. Namely, if $G$ is any commutative group, then the category of $G$-graded $R$-modules, where $R$ is a $G$-graded ring, is abelian and fulfills AB5 and AB4, hence has a projective generator and an injective cogenerator.
However, one has to be careful with this approach, since there are still many differences between these categories for different groups $G$. For example, suppose that $R$ is a noetherian $G$-graded ring. If $G$ is finite, then the category of $G$-graded $R$-modules has a noetherian generator, but if $G$ is infinite then it does not.
answered Jan 7 at 12:24
Fred RohrerFred Rohrer
4,39611734
$endgroup$
add a comment |
$begingroup$
Maybe not really what you are after, but anyway: This holds for all those statements that depend only on the structure of abelian category on the category of graded modules. Namely, if $G$ is any commutative group, then the category of $G$-graded $R$-modules, where $R$ is a $G$-graded ring, is abelian and fulfills AB5 and AB4, hence has a projective generator and an injective cogenerator.
However, one has to be careful with this approach, since there are still many differences between these categories for different groups $G$. For example, suppose that $R$ is a noetherian $G$-graded ring. If $G$ is finite, then the category of $G$-graded $R$-modules has a noetherian generator, but if $G$ is infinite then it does not.
answered Jan 7 at 12:24
Fred RohrerFred Rohrer
4,39611734
$endgroup$
add a comment |
$begingroup$
Maybe not really what you are after, but anyway: This holds for all those statements that depend only on the structure of abelian category on the category of graded modules. Namely, if $G$ is any commutative group, then the category of $G$-graded $R$-modules, where $R$ is a $G$-graded ring, is abelian and fulfills AB5 and AB4, hence has a projective generator and an injective cogenerator.
However, one has to be careful with this approach, since there are still many differences between these categories for different groups $G$. For example, suppose that $R$ is a noetherian $G$-graded ring. If $G$ is finite, then the category of $G$-graded $R$-modules has a noetherian generator, but if $G$ is infinite then it does not.
answered Jan 7 at 12:24
Fred RohrerFred Rohrer
4,39611734
$endgroup$
Maybe not really what you are after, but anyway: This holds for all those statements that depend only on the structure of abelian category on the category of graded modules. Namely, if $G$ is any commutative group, then the category of $G$-graded $R$-modules, where $R$ is a $G$-graded ring, is abelian and fulfills AB5 and AB4, hence has a projective generator and an injective cogenerator.
However, one has to be careful with this approach, since there are still many differences between these categories for different groups $G$. For example, suppose that $R$ is a noetherian $G$-graded ring. If $G$ is finite, then the category of $G$-graded $R$-modules has a noetherian generator, but if $G$ is infinite then it does not.
answered Jan 7 at 12:24
Fred RohrerFred Rohrer
4,39611734
answered Jan 7 at 12:24
Fred RohrerFred Rohrer
4,39611734
answered Jan 7 at 12:24
Fred RohrerFred Rohrer
4,39611734
answered Jan 7 at 12:24
Fred RohrerFred Rohrer
4,39611734
4,39611734
add a comment |
add a comment |
$begingroup$
Since the OP is asking for some "class of statements" which can be readily transferred from the ungraded to the graded case, let me outline a couple of thoughts, which are not really tied to neither the commutative case, nor the $mathbb{Z}$-gradation itself:
- On the one hand, if $R$ is a $k$-algebra and $G$ is a group, then a $G$-gradation on $R$ can be equivalently viewed either as a $(kG)^*$-action or as a $kG$-coaction (where $kG$ is the group hopf algebra and $(kG)^*$ its dual hopf algebra). So, the graded algebra $R$ is alternatively viewed as an algebra in the category of $(kG)^*$-modules or an algebra in the category of $kG$-comodules. In this sense, the statements and the properties which are "naturally" transferred between the graded and the ungraded case are those which are in some sense "compatible" or "invariant" with respect to the $kG$-coaction (or the $(kG)^*$-action). I do not know if it sounds like an abuse of terminology to speak about
properties in the category of $(kG)^*$-modules or properties in the category of $kG$-comodules
on the other hand, the Category $R_{gr}$-$mod$ of graded modules over the $G$-graded ring $R$, is equivalent (actually i think it is isomorphic in most cases) to a closed subcategory of modules (in the ungraded sense) over the smash product algebra $Rsharp kG$. (Here the smash product is understood in the sense of the one associated to a Hopf module algebra).
In this sense, we could say that the statements and properties which one might expect to "naturally transfer" over between graded and ungraded, are
the ones which are preserved under the smash product between $R$ and the group hopf algebra $kG$.
(In some sense, this implies that it might be meaningful to consider properties which "transfer" between the $R$-modules and the $Rsharp kG$-modules).
One might reasonably expect that
all those statements that depend only on the structure of abelian category on the category of graded modules
mentioned in Fred Rohrer's answer fall into the classes mentioned above, however i can't tell for sure on the exact overlap between these classes.
answered Jan 8 at 2:44
Konstantinos KanakoglouKonstantinos Kanakoglou
3,30021234
$endgroup$
add a comment |
$begingroup$
Since the OP is asking for some "class of statements" which can be readily transferred from the ungraded to the graded case, let me outline a couple of thoughts, which are not really tied to neither the commutative case, nor the $mathbb{Z}$-gradation itself:
- On the one hand, if $R$ is a $k$-algebra and $G$ is a group, then a $G$-gradation on $R$ can be equivalently viewed either as a $(kG)^*$-action or as a $kG$-coaction (where $kG$ is the group hopf algebra and $(kG)^*$ its dual hopf algebra). So, the graded algebra $R$ is alternatively viewed as an algebra in the category of $(kG)^*$-modules or an algebra in the category of $kG$-comodules. In this sense, the statements and the properties which are "naturally" transferred between the graded and the ungraded case are those which are in some sense "compatible" or "invariant" with respect to the $kG$-coaction (or the $(kG)^*$-action). I do not know if it sounds like an abuse of terminology to speak about
properties in the category of $(kG)^*$-modules or properties in the category of $kG$-comodules
on the other hand, the Category $R_{gr}$-$mod$ of graded modules over the $G$-graded ring $R$, is equivalent (actually i think it is isomorphic in most cases) to a closed subcategory of modules (in the ungraded sense) over the smash product algebra $Rsharp kG$. (Here the smash product is understood in the sense of the one associated to a Hopf module algebra).
In this sense, we could say that the statements and properties which one might expect to "naturally transfer" over between graded and ungraded, are
the ones which are preserved under the smash product between $R$ and the group hopf algebra $kG$.
(In some sense, this implies that it might be meaningful to consider properties which "transfer" between the $R$-modules and the $Rsharp kG$-modules).
One might reasonably expect that
all those statements that depend only on the structure of abelian category on the category of graded modules
mentioned in Fred Rohrer's answer fall into the classes mentioned above, however i can't tell for sure on the exact overlap between these classes.
answered Jan 8 at 2:44
Konstantinos KanakoglouKonstantinos Kanakoglou
3,30021234
$endgroup$
add a comment |
$begingroup$
Since the OP is asking for some "class of statements" which can be readily transferred from the ungraded to the graded case, let me outline a couple of thoughts, which are not really tied to neither the commutative case, nor the $mathbb{Z}$-gradation itself:
- On the one hand, if $R$ is a $k$-algebra and $G$ is a group, then a $G$-gradation on $R$ can be equivalently viewed either as a $(kG)^*$-action or as a $kG$-coaction (where $kG$ is the group hopf algebra and $(kG)^*$ its dual hopf algebra). So, the graded algebra $R$ is alternatively viewed as an algebra in the category of $(kG)^*$-modules or an algebra in the category of $kG$-comodules. In this sense, the statements and the properties which are "naturally" transferred between the graded and the ungraded case are those which are in some sense "compatible" or "invariant" with respect to the $kG$-coaction (or the $(kG)^*$-action). I do not know if it sounds like an abuse of terminology to speak about
properties in the category of $(kG)^*$-modules or properties in the category of $kG$-comodules
on the other hand, the Category $R_{gr}$-$mod$ of graded modules over the $G$-graded ring $R$, is equivalent (actually i think it is isomorphic in most cases) to a closed subcategory of modules (in the ungraded sense) over the smash product algebra $Rsharp kG$. (Here the smash product is understood in the sense of the one associated to a Hopf module algebra).
In this sense, we could say that the statements and properties which one might expect to "naturally transfer" over between graded and ungraded, are
the ones which are preserved under the smash product between $R$ and the group hopf algebra $kG$.
(In some sense, this implies that it might be meaningful to consider properties which "transfer" between the $R$-modules and the $Rsharp kG$-modules).
One might reasonably expect that
all those statements that depend only on the structure of abelian category on the category of graded modules
mentioned in Fred Rohrer's answer fall into the classes mentioned above, however i can't tell for sure on the exact overlap between these classes.
answered Jan 8 at 2:44
Konstantinos KanakoglouKonstantinos Kanakoglou
3,30021234
$endgroup$
Since the OP is asking for some "class of statements" which can be readily transferred from the ungraded to the graded case, let me outline a couple of thoughts, which are not really tied to neither the commutative case, nor the $mathbb{Z}$-gradation itself:
- On the one hand, if $R$ is a $k$-algebra and $G$ is a group, then a $G$-gradation on $R$ can be equivalently viewed either as a $(kG)^*$-action or as a $kG$-coaction (where $kG$ is the group hopf algebra and $(kG)^*$ its dual hopf algebra). So, the graded algebra $R$ is alternatively viewed as an algebra in the category of $(kG)^*$-modules or an algebra in the category of $kG$-comodules. In this sense, the statements and the properties which are "naturally" transferred between the graded and the ungraded case are those which are in some sense "compatible" or "invariant" with respect to the $kG$-coaction (or the $(kG)^*$-action). I do not know if it sounds like an abuse of terminology to speak about
properties in the category of $(kG)^*$-modules or properties in the category of $kG$-comodules
on the other hand, the Category $R_{gr}$-$mod$ of graded modules over the $G$-graded ring $R$, is equivalent (actually i think it is isomorphic in most cases) to a closed subcategory of modules (in the ungraded sense) over the smash product algebra $Rsharp kG$. (Here the smash product is understood in the sense of the one associated to a Hopf module algebra).
In this sense, we could say that the statements and properties which one might expect to "naturally transfer" over between graded and ungraded, are
the ones which are preserved under the smash product between $R$ and the group hopf algebra $kG$.
(In some sense, this implies that it might be meaningful to consider properties which "transfer" between the $R$-modules and the $Rsharp kG$-modules).
One might reasonably expect that
all those statements that depend only on the structure of abelian category on the category of graded modules
mentioned in Fred Rohrer's answer fall into the classes mentioned above, however i can't tell for sure on the exact overlap between these classes.
answered Jan 8 at 2:44
Konstantinos KanakoglouKonstantinos Kanakoglou
3,30021234
edited Jan 8 at 21:47
answered Jan 8 at 2:44
Konstantinos KanakoglouKonstantinos Kanakoglou
3,30021234
answered Jan 8 at 2:44
Konstantinos KanakoglouKonstantinos Kanakoglou
3,30021234
answered Jan 8 at 2:44
Konstantinos KanakoglouKonstantinos Kanakoglou
3,30021234
3,30021234
add a comment |
add a comment |
Thanks for contributing an answer to MathOverflow!
- 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.
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%2fmathoverflow.net%2fquestions%2f320269%2fgraded-analogues-of-theorems-in-commutative-algebra%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
$begingroup$
Do you have a reference for how to argue geometrically using stacks as you allude to?
$endgroup$
– Avi Steiner
Jan 13 at 4:14
1
$begingroup$
Call a graded module over a graded ring $R$ graded flat if its tensor product with every exact sequence of graded modules is graded again. I claim that a module is graded flat if and only if its flat. Indeed, the category of graded modules is monoidally equivalent to quasi-coherent sheaves on $[Spec R/mathbb{G}_m]$. As $Spec R to [Spec R/mathbb{G}_m]$ is fpqc, exactness of a sequence can be tested on $Spec R$, which implies the result. Thus every theorem about flat modules also applies to graded flat modules.
$endgroup$
– Lennart Meier
Jan 14 at 7:16