1
vote
0answers
299 views

Submodule embeddable in a finitely generated module

I have a terminology question. For a commutative ring $A$ (not necessarily Noetherian), $A$-modules that are isomorphic to an $A$-submodule of a finitely generated $A$-module form a fairly good class ...
4
votes
1answer
131 views

$ mult(R/I) = d_1 \cdots d_r \quad \Rightarrow \quad f_1,\dots,f_r \quad \text{is a $R$-regular sequence?}$

We define multiplicity of a module M of dimension $d>0$ as $$mult(M) := lc (P_M) (d-1)!$$ where $P_M$ denotes the Hilbert polynomial of M. Equivalently, we have $mult(M) = Q_M(1)$, where $HP_M (z) ...
-3
votes
1answer
84 views

Isomorphic quotient of a Module over Noetherian commutative algebra [closed]

I have a nice solution to the following problem and I thought of writing a paper about it but beforehand, I wanted to ask the problem here to see if this is an easy problem and if you people can solve ...
11
votes
7answers
693 views

Properties of rings that have an elegant description in terms of the associated category of modules

Suppose $A$ is a ring. Then $A$ happens to be a division ring iff every left $A$-module is free. (See here for proofs). I think this is very beautiful; what other properties of rings have an elegant ...
0
votes
1answer
220 views

The equivalence of Artinian and Noetherinan for the modules of a semisimple ring

We know in a semisimple ring R, for every R-module, Noetherian is equivalent to Artinian, my question is: If for every R-module M Noetherian is equivalent to Artinian, can we prove R is a semisimple ...
1
vote
2answers
323 views

A Question About Free Resolutions

I would warmly appreciate it if someone could tell me whether the following question has an affirmative answer. I am new to the field of commutative algebra, so I am simply trying to fill in some ...
6
votes
1answer
408 views

Are there non-reflexive modules isomorphic to their bi-dual?

Let $M$ be an $R$-module. We say that $M$ is reflexive if the natural map $M\rightarrow M^{**}$ is an isomorphism. I'd like to know if there exists a module isomorphic to its bi-dual but not ...
7
votes
2answers
669 views

Modules over Laurent series rings

Let $k[x]$ be the ring of polynomials over a field k in one variable x. A $k[x]$-module is a k-vector space together with a linear endomorphism (the action of x). The field $k(x)$ of rational ...
1
vote
1answer
344 views

Unimodular column property

Hi, I know that if $R$ is a ring such that every projective $R$-module finitely generated is free then $R$ has the unimodular column property. I would like to know if there is a ring $R$ that doesn't ...
7
votes
1answer
664 views

Direct sum of injective modules over non-Noetherian rings

Hi. I know, by the Bass-Papp theorem, that if every direct sum of injective $R$-modules is injective then $R$ is Noetherian. I would like to know if there exists a direct sum of injective $R$-modules ...
1
vote
3answers
493 views

Stably free module not finitely generated is free

Hi. I have read that stably free modules not finitely generated are free; this is proved in M.R. Gabel, stably free projectives over commutative rings, Thesis, Brandeis Univ., Waltham, MA 1972. But ...
2
votes
2answers
352 views

Related to fractional ideals

$K$ a field, $A\subset K$ a subring, $M\subset K$ an $A$-submodule. Define $$(A:_{K}M):= \lbrace s\in K|sM\subset A\rbrace$$ Then it is easy to see that $$M\subset A\Longleftrightarrow A\subset ...
2
votes
2answers
336 views

Homological dimensions of module

$(A,\mathfrak{m})$ a Noetherian local ring, $M\neq 0$ a finitely generated $A$-module. As I understand, $\mbox{Ext }^{j}(A/\mathfrak{m}, M) = 0$ for $j<\mbox{depth }(M)$ and for $j>\mbox{inj. ...
1
vote
1answer
564 views

Existence of a minimal generating set of a module

Does a module (over a commutative ring) always possess a minimal generating set? When the module is not finitely generated, the typical Zorn's lemma type argument doesn't seem to work. More precisely, ...
17
votes
4answers
2k views

Flatness and local freeness

The following statement is well-known: $A$ a commutative Noetherian ring, $M$ a finitely generated $A$-module. Than $M$ is flat if and only if $M_{\mathfrak{p}}$ is free for all $\mathfrak{p}$. My ...
1
vote
2answers
599 views

Non-finite version of Nakayama's lemma?

Let $A$ be a local ring with nilpotent maximal ideal $\mathfrak{m}$ (i.e., some power of $\mathfrak{m}$ vanishes), and $M$ an $A$-module (not necessarily finitely generated). Let $\bar{S}\subset ...