All Questions
8 questions
8
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Do graded-commutative rings satisfy the strong rank condition?
Let $R$ be a ring. Recall that $R$ is said to satisfy the strong rank condition if, whenever $R^m \to R^n$ is a monomorphism of right $R$-modules (with $m,n \in \mathbb N$), we have $m \leq n$.
It is ...
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1
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0
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45
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Lifting module homomorphisms imposing conditions on characteristic polynomials
Suppose that we are in the setting described in the first two paragraphs of this MSE post. My question wants to deal with an instance of the study of the amount of freedom that the choice of the ...
- 573
1
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0
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33
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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 $...
- 573
1
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0
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212
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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 $B:=\{m_{1},\dots, m_{r}\}$ ...
- 573
9
votes
2
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513
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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 ...
- 1,286
1
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0
answers
40
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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 ...
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5
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0
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337
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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 $I(\cap_{k=1}^{\infty}I^k M)=\cap_{k=1}^{\infty}I^k M$ . ...
CommunityBot
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3
votes
1
answer
701
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On graded projective modules
If $R$ is a PID, then we know that any projective module over $R$ is free. Is there any graded version of this? That is, let us call a graded commutative ring $R=\oplus _{g \in G} R_g$ a graded PID if ...
CommunityBot
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