Questions tagged [graded-rings-modules]
The graded-rings-modules tag has no usage guidance.
62
questions
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Projectivization of graded vector spaces
Let $E=E^0\oplus E^1$ be a $\mathbb{Z}_2$-graded vector space. Is there a somehow graded notion of projectivization
$$\mathbb{P}(E)=\mathbb{P}(E^0\oplus E^1)$$
maybe in terms of $\mathbb{P}(E^0)$ and $...
9
votes
2
answers
288
views
Over which (graded) rings are all modules decomposable into indecomposables?
A module is decomposable if it is the direct sum of two modules. The process of splitting summands off of a decomposable module does not need to terminate, so infinitely generated modules do not ...
2
votes
0
answers
90
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Hilbert-function-like results for weighted projective spaces
Let $R = k[X_1,\ldots,X_N]/I$ be a finitely generated graded generated $k$-algebra, where the $X_i$'s have nonequal degrees, say $\deg(X_i)=a_i$, and $I$ is a (weighted-)homogeneous ideal.
We can ...
2
votes
1
answer
140
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Proving the graded structure of the tensor algebra from only the universal property
When the tensor algebra is presented, it is usually constructed as the direct sum of all tensor powers of the space. By this construction, the graded structure of the tensor algebra is easy to prove. ...
1
vote
1
answer
307
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Is there an operad homotopifying the Koszul rule?
In homotopy theory one has the idea of a homotopy-commutative multiplication, in which one replaces the relation $$ab=ba$$ in a commutative monoid/group/ring/etc. for an unspecified homotopy. One ...
1
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0
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91
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Commutative monoid gradings via group scheme actions
$\newcommand{\Spec}{\mathrm{Spec}}$Recall the following result, proved in Section 2.9 of Neil Strickland's Formal schemes and formal groups, and in Lemma 1.3.2 of Eric Peterson's Formal Geometry and ...
6
votes
1
answer
209
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Connection between the classifications of group extensions and group-graded algebras in terms of non-abelian cohomology
First, consider group extensions with non-abelian kernel
$$1\to K\to G \to Q \to 1$$
It is well-known that these are classified by certain cohomological objects, specifically: Any such extension ...
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181
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Dimension of the associated graded module at an ideal
Let $I$ be an ideal of a Noetherian local ring $(R, \mathfrak m)$. Define the associated graded ring $G_I(R):=\bigoplus_{n=0}^\infty I^n/I^{n+1}$. Then $G_I(R)$ is a Noetherian ring of the same ...
3
votes
1
answer
123
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On the degree of the Hilbert polynomial of a graded module over the Rees algebra
If $A=\oplus_{n=0}^\infty A_n$ is a Noetherian graded ring of finite dimension such that $A_0$ is local and $A=A_0[A_1]$, and if $M=\oplus_{n=0}^\infty M_n$ is a finitely generated graded $A$-module ...
1
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0
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28
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Making a generating set of a section of a graded polynomial $R$-module coming from a quotient into a basis of a quotient by higher degree polynomial
Denote the graded rings $R:=\mathbb{R}[x_{1},\dots x_{n}]$ and $S:=R[x_{0}]$ adding the homogenizing variable $x_{0}.$ Consider $h\in S$ a homogenous polynomial of degree $d$ with leading coefficient $...
1
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112
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Surjection from finite rank free $R$-module to finitely generated $R$-module and basis associated to generator set
Suppose the we have an epimorphism $s\colon M\to N,$ where $M$ is a free $R$-module of rank $r$ and $N$ is a finitely generated $R$-module, such that there exists a basis $B:=\{m_{1},\dots, m_{r}\}$ ...
1
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0
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95
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Strongly graded rings
In Theorem 3.1 of Graded rings over arithmetical orders, the authors prove that for a strongly $\mathbb{Z}$-graded ring $R$, if $R_0$ is left and right Goldie and a maximal order in its (classical) ...
2
votes
1
answer
59
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Bigraded operadic suspension
I know from this paper by Ward that one can obtain the (signs of) the Gerstenhaber bracket using operadic suspension on any operad $\mathcal{O}$. More precisely, the insertion $\tilde{\circ}$ of the ...
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0
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87
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Quotient of $\text{Proj}(A)$ by the action of a finite group
Let $X$ be $ \operatorname{Proj}(A)$ for some graded ring A, and let $G$ be a finite group acting on $A$ with morphisms of graded rings; consequently $G$ acts on $X$.
I know I can write $X = \bigcup_{...
7
votes
1
answer
355
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$\mathbb{Z}$-graded algebras and tensor products
Let $A = \bigoplus_{k \in \mathbb{Z}} A_k$ be a not necessarily commutative $\mathbb{Z}$-graded unital algebra over a field $\mathbb{K}$, and assume that it is strongly graded:
$$
A_kA_l = A_{k+l}.
$$
...
1
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0
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48
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Homomorphisms and indecomposable decompositions of finite modules over polynomial rings [closed]
I am studying $\mathbb{N}^n$-graded, finitely generated modules $M$ over the $\mathbb{N}^n$-graded polynomial ring $K[X_1,X_2,X_3...,X_n]$. I know every such $M$ has an indecomposable decomposition $M\...
4
votes
1
answer
141
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Weighted blowup of a Cohen-Macaulay ring along a regular sequence is Cohen-Macaulay?
Let $(R, \mathfrak{m})$ be a Cohen-Macaulay ring, let $f_1, \dotsc, f_d \in \mathfrak{m}$ be a regular sequence, and let $n_1, \dotsc, n_d > 0$ be weights (feel free to assume that $n_1 = 1$ if it ...
5
votes
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75
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Ring with different graded and ungraded global dimensions
Let $A$ be a $\mathbb N$-graded ring. One can consider the two categories $M_A^g$ and $M_A^u$ of graded and ungraded modules over $A$. Both have, say, enough projectives, hence one can compute various ...
8
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2
answers
319
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Conceptual explanation for the sign in front of some binary operations
In several situations, I've seen that given a binary operation on a graded module $m:A\otimes A\to A$, a new operation $M(x,y)=(-1)^{|x|}m(x,y)$ is defined so that it satisfies some properties.
One ...
5
votes
2
answers
459
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Representation theory in braided monoidal categories
The crux of what I wish to know is what results from representation theory, a subject usually framed within the category $\text{Vect}_\mathbb{k}$, follow in more general braided monoidal categories? I ...
2
votes
1
answer
166
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Detailed proof of $\mathfrak{s}^{-1}\mathrm{End}_V\cong \mathrm{End}_{\Sigma V}$
I asked this question on MSE but I want to ask it again here with some more context sine it received no answers. In Chapter 3 (Algebra) of the book Operads in Algebra, Topology and Physics by Markl, ...
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100
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Is the free algebra over an operad an algebra over that operad?
I'm asking here this question I asked on MSE that got no answers.
Let $V$ be a dg-module and $P$ an operad. The free $P$-algebra on $V$ is defined by $P(V)=\bigoplus_{r=0}^\infty (P(r)\otimes V^{\...
5
votes
1
answer
214
views
Is operadic desuspension inverse to operadic suspension?
Given a graded vector space $V$ over a field $k$, consider it's suspension $\Sigma V$ such that $(\Sigma V)^i=V^{i-1}$. For an operad of graded vector spaces over a field $\mathcal{O}$, the operadic ...
3
votes
2
answers
176
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Polynomial identities of supercommutative-gradable algebras
All algebras below are associative, and not assumed unital, and, to fix ideas, over the complex numbers.
An algebra $A$ is supercommutative-gradable if it admits a grading $A=A_0\oplus A_1$ in $\...
0
votes
0
answers
92
views
Certain morphism between graded modules
Say we have a morphism $f=(f_{i,j}) : \oplus_{i}M_{i} \rightarrow \oplus_{j}N_{j}$ (The direct sum of graded modules is finite) and $f_{i,j}$ are morphisms of graded modules but not necessary with the ...
1
vote
1
answer
79
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non-archimedean valuations on graded rings
Let $R$ be a commutative (non-trivially) graded ring. By a non-archimedean valuation I mean a map $v: R \to \Gamma \cup {0}$ such that for all $x,y \in R$, we have $v(x+y) \leq \max\{v(x),v(y)\}$, $v(...
5
votes
2
answers
1k
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Reason to apply the Koszul sign rule everywhere in graded contexts
The Koszul sign rule is a sign rule that arises from graded-commutative algebras. For instance, let $\bigwedge(x_1,\dots, x_n)$ be the free graded-commutative algebra generated by $n$ elements of ...
5
votes
2
answers
388
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Are open $\mathbb{G}_m$-invariant subschemes of an affine scheme precisely the homogeneous radical ideals of its coordinate ring?
This question is certainly not research level and in fact quite elementary which is why I asked it on math.stackexchange before: math.stackexchange. However it doesn't seem to get much attention there ...
1
vote
1
answer
171
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Strongly graded algebras with no zero divisors
Let $A = \bigoplus_{i \in \mathbb{Z}} A_i$ be a strongly graded unital algebra over $\mathbb{C}$, with no zero divisors. Is it always true that
$$
m: A_i \otimes_{A_0} A_j \to A_{i+j}
$$
is an ...
3
votes
0
answers
44
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Asymptotic stability of associated primes of graded local cohomology modules
This question concerns the asymptotic behaviour of associated primes of graded components of local cohomology modules. A survey of this can be found in M. Brodmann, Asymptotic behaviour of cohomology: ...
1
vote
1
answer
499
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Affine cone example
Consider the complete intersection ideal ${\displaystyle (f,g_{1},g_{2},g_{3})\subset \mathbb {C} [x_{0},\ldots ,x_{n}]}$ and let $X$ be a projective variety defined by the (product) ideal sheaf ${\...
5
votes
0
answers
251
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Homogeneous regular sequence
Consider a $\mathbb{Z}$-graded polynomial ring $R = k[x_1,\cdots,x_n]$ over a field $k$, where the elements $x_i$ are homogeneous (they may have negative degree). Let $I$ be a homogeneous ideal that ...
2
votes
1
answer
131
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infinite fold tensor product of universal enveloping algebra
Let $\mathfrak a$ be a Lie algebra graded by the abelian semigroup $S$, then the universal enveloping algebra $U(\mathfrak a)$ of $\mathfrak a$ is $S \sqcup \{0\}$ graded. I have the following ...
0
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0
answers
47
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Induced grading on free non-associative algebra
Let $A_X$ be a free non-associative algebra over a field $k$ on a set $X$ of free generators, where $X = X_0 \cup X_1$ and $X_0 \cap X_1 = \phi$. We will think of the elements of $X_0$ as even, and ...
2
votes
1
answer
597
views
Relation between the Spec and the Proj of a ring
I am reading Thaddeus' paper on GIT and flips (https://arxiv.org/pdf/alg-geom/9405004.pdf), and I am confused with a claim in the begining.
Let $R$ be a finitely generated integral algebra over an ...
3
votes
0
answers
164
views
Derivations of algebras graded by a group
Let $A$ be an algebra. A derivation of $A$ is a linear map $d:A \rightarrow A$ such that $d(ab)=d(a)b + a d(b)$ for $a, b \in A$.
If $A$ is a $\mathbb{Z}$-graded algebra, where $\mathbb{Z}$ is the set ...
5
votes
0
answers
295
views
Does the associated graded functor take products of filtered k-coalgebras to graded k-coalgebras?
Let's suppose we have two noncommutative graded k-coalgebras $C_1$ and $C_2$ with respective admissible filtrations (i.e $F_{0}C_i=0$ and $\mathrm{colim}_k F_kC_i=C_i$), I would like to know if there ...
0
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0
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229
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Defining a graded ring structure on $\bigoplus_{i} \text{Ext}^i (A,A)$
I read that, using the fact that $\text{Tot}^{\prod}(\text{Hom}(P_{\bullet} ,Q_{\bullet}))$ can by used to compute $\text{Ext}^* (A,B)$ (which I understand), we can give $\bigoplus_{i} \text{Ext}^i (A,...
5
votes
1
answer
285
views
If a commutative graded algebra is free over a graded subalgebra, then must it have a graded basis?
Fix a field $\mathbf{k}$ and an $\mathbb{N}$-graded commutative $\mathbf{k}$-algebra $A = \bigoplus\limits_{n = 0}^{\infty} A_n$ of finite type. ("Finite type" means that each $A_n$ is a ...
6
votes
1
answer
155
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Is $[Im:(x)][Im:(y,z)]\subseteq Im$ in $k[x,y,z]$?
Let $k$ be a field and $S=k[x,y,z]$. Let $m=(x,y,z)$ and $I\subseteq m$ a proper homogeneous ideal in $S$. Is this true that we always have:
$$[Im:(x)][Im:(y,z)]\subseteq Im \ ?$$
In a paper we ...
22
votes
3
answers
997
views
Graded analogues of theorems in commutative algebra
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 ...
4
votes
1
answer
282
views
Global dimension of a graded algebra
Let $A= \bigoplus\limits_{n=0}^{\infty}{A_n}$ be an $\mathbb{N}$-graded algebra with semisimple $A_0$.
Question: Do we have that the global dimension of $A$ is equal to $\sup \{i \geq 0 | Ext_A^i(...
5
votes
1
answer
306
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Projective dimension of graded modules
Short version:
Why is the projective dimension of a graded module the same as the projective dimension of its underlying ungraded module?
Longer version:
Let $G$ be a commutative group, let $R$ ...
9
votes
2
answers
455
views
Divided power algebra is artinian as a module over the polynomial ring
I already asked this on math.stackexchange.com, but did not receive much responses. I hope this is also appropriate for mathoverflow.
In the paper Homological algebra on a complete intersection, with ...
8
votes
1
answer
259
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The different gradings of a graded ring, and their schemes
Let $(A,g)$ be a graded commutative ring, where $A$ denotes the commutative ring, and $g$ its grading. What can be said about the set $\mathcal{G}_A := \{ \mathrm{Proj}\Big((A,g)\Big)\ \vert\ g \}$? ...
6
votes
1
answer
275
views
Is every (left) graded-Noetherian graded ring (left) Noetherian?
I call a $\mathbb{Z}$-graded (non-commutative, associative, unital) ring $A$ (left) graded-Noetherian if every homogeneous (left) ideal is finitely generated, and (left) Noetherian if it is (left) ...
1
vote
0
answers
32
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Hilbert functions of graded modules generated by mapped generators
I need to prove the following claim for my work. Intuitively, the claim should hold as it is analogous to the concept of an initial module, but a rigorous approach for the same I cannot find. Any help ...
5
votes
0
answers
300
views
Can the Artin-Rees lemma be derived from Krull Intersection theorem?
The Krull Intersection theorem states that : For a finitely generated module $M$ over a Noetherian ring $R$ and any ideal $I$ of $R$, we have $I(\cap_{k=1}^{\infty}I^k M)=\cap_{k=1}^{\infty}I^k M$ . ...
0
votes
0
answers
93
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Hochster-Roberts theorem
I am trying to understand this theorem, and as an example I try this case: Let $R=\mathbb{C}[z_1^{\pm},\ldots,z_n^{\pm}]^{S_n}$ be the algebra of Laurent polynomials. Now $R$ should be somehow ...
1
vote
1
answer
179
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If $A\subset B$, what to say about their $\operatorname{Proj}$?
Let $k$ be a field and $A$ and $B$ be two graded $k$-algebras satisfying $A\subset B$. The $\operatorname{Proj}$ construction is not functorial but is there nothing to say about $\operatorname{Proj}(A)...