-3
votes
1answer
195 views

Doubt in this proof of Horrocks theorem

I'm beginning to study some research papers and I need right now to understand the solution of Vaseršteĭn of Serre's theorem (simplest proof of this theorem), to do so, I'm beginning to understand ...
3
votes
1answer
239 views

Splitting as $\mathbb{F}_p[[X]]$-modules

Let $A$ be a finitely generated torsion $\mathbb{Z}_p[[X]]$-module, $B$ = { $x \in A$ such that $px=0$ } and $C=A/B$ where $\mathbb{Z}_p$ denotes the $p$-adic integers. Given $ 0 \rightarrow B/pB ...
13
votes
5answers
686 views

is the category of coherent sheaves some kind of abelian envelope of the category of vector bundles?

This might be obvious to experts, but I'm not sure where to look for the answer. On a reasonably nice, at least noetherian, scheme (or variety, algebraic space, stack), can the category of coherent ...
4
votes
0answers
205 views

Epimorphisms between external tensor products

Let $k$ be a field, $R,S$ commutative $k$-algebras. If $\mathsf{Mod}_{fp}$ denotes the category of finitely presented modules, the external tensor product $\mathsf{Mod}_{fp}(R) \otimes_k ...
0
votes
1answer
169 views

cofree modules and dual

1, Why do people pay special attention to Q/Z in the definition of cofree modules instead of ordinary abelian groups? 2, Over a PID, is every injective module cofree? Just like the relationship ...
3
votes
1answer
130 views

is every finitely n-presented (S^{-1})R-module a localization of a finitely n-presented R-module?

Let S be a multiplicative set in a ring R. We can see that every finitely generated $(S^{-1})R$-module is a localization of a finitely generated R-module. Then, more generally, is every finitely ...
1
vote
1answer
159 views

Classification of pairs of commuting endomorphisms

Let $K$ be an algebraically closed field. I'm interested in isomorphism classes of triples $(V,f,g)$ where $V$ is a finite dimensional $K$-vector space and $f,g$ are commuting endomorphisms of $V$. ...
2
votes
1answer
80 views

Homocyclic primary module over PID

I posed the question here, but get no answers yet. Let $R$ be a PID, $M$ be an $R$-module. If $M$ is isomorphic to $r$ copies of cyclic primary module $R/\langle p^s\rangle$ where $p$ is a prime ...
2
votes
1answer
142 views

Is there an $A$ such that $B$ injective iff 1st Ext functor vanishes?

In the category of $\mathbb{Z}$-modules, there exists a module $A$---for instance $\bigoplus_{k=2}^\infty \mathbb{Z}/k\mathbb{Z}$---such that a $\mathbb{Z}$-module $B$ is injective iff ...
2
votes
1answer
269 views

An example of a tensor product consisting of only simple tensors?

Hy guys. I'm doing some independent analysis which makes use of the tensor product of modules (over commutative rings with unit 1, and ring homomorphisms map $1 \mapsto 1$). Let $\pi: A' \to A$ be a ...
0
votes
0answers
80 views

Invariants of a module, 'readily' computable from the presentation matrix

Suppose we know the presentation matrix of a module. (For simplicity: the module is over a local Noetherian ring, the ring is over a field.) For which invariants of the module some explicit(!) ...
3
votes
3answers
304 views

Support of a module over a polynomial algebra

In Atiyah and Bott's paper "The Moment Map and Equivariant Cohomology", they say that for any exact sequence of modules over $\mathbb{C}[u_1,...,u_l]$ $$D \to E \to F,$$ we have that Supp $E \subset$ ...
8
votes
2answers
224 views

A criterion for freeness over a local ring

Let $A=K[[X_1,\dots,X_n]]$ where $K$ is a field. Let $M$ be a finitely generated torsion-free $A$-module, such that for all $k$, the $A[1/X_k]$-module $M[1/X_k]$ is free of rank $d$; for every $i ...
5
votes
1answer
370 views

Baer's criterion for projective modules

Let $R$ be a commutative ring. If necessary, assume that $R$ has any convenient properties you like. Is there some $R$-module $Q$ such that an $R$-module $P$ is projective if and only if ...
2
votes
1answer
123 views

Open idempotents in modules over a local ring

Let $R$ be a local ring. By an open idempotent I mean an $R$-module $F$ equipped with a homomorphism $e : F \to R$ such that $e \otimes F = F \otimes e$ is an isomorphism $F \otimes F \cong F$ (this ...
7
votes
1answer
268 views

flatness of power series rings

It is known that $A[[X]]$ is flat if $A$ is noetherian (see for example Bourbaki, Algèbre commutative, Ch. III, §3, Cor. 3 p. 146). What happens if A is not noetherian? Is there an easy ...
9
votes
1answer
445 views

Classification of symtrivial modules over a PID

Let us call a module $M$ over a commutative ring $R$ symtrivial if the symmetry $M \otimes M \to M \otimes M, a \otimes b \mapsto b \otimes a$ equals the identity (the same notion applies to arbitrary ...
0
votes
1answer
199 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
305 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 ...
3
votes
1answer
168 views

Liftability of a submodule from an associated graded module

Let $k$ be a field, $A$ a $k$-algebra (probably noncommutative), and $M$ an $A$-module that's finite-dimensional as a vector space over $k$. Let $Gr(M;k)$ denote the set of all $k$-subspaces of $M$, ...
3
votes
1answer
645 views

Are there two non-isomorphic modules such that all the Hom-sets are isomorphic? [closed]

Prove or disprove: If $M, N$ are R-module and for all $P$ R-module $Hom(M,P) \simeq Hom(N,P)$ then $M\simeq N$
1
vote
1answer
245 views

When fitting ideals determine the module?

Let $M$ be a module over a local ring $(R,m)$, everything is finitely generated/presented. The fitting ideals, $I_j(M)$ carry a lot of information about the module. When do they actually determine the ...
3
votes
0answers
252 views

Inductive proof of a version of Nakayama's lemma

I have already asked this at math.stackexchange, but since no one answered there after my edit, I decided to try here, although it might be a non-research level question. The following version of ...
0
votes
0answers
133 views

Length of $\mathfrak{m}$-torsion module

Let $(R,\mathfrak{m})$ be a commutative Noetherian local ring and $M$ is an $R$-module such that $\mathfrak{m}^tM=0$ for some non-negative integer $t$. Then the length of $M$ is finite. Is that ...
1
vote
2answers
331 views

Is a reduced, torsion-free module of finite rank over an Henselian ring free?

Let $R$ be an Henselian discrete valuation ring with field of fractions $K$. Let $M$ be a torsion-free $R$-module of finite rank (i.e. $dim_K(M\otimes_RK)<+\infty$). Let $D$ be the maximal ...
3
votes
1answer
440 views

Alternative module-theoretic characterization of flatness

Let $A \to B$ be a homomorphism of commutative rings. I would like to find a criterion for the flatness of $A \to B$ which does not involve the notion of kernels; it should rather involve cokernels. ...
3
votes
2answers
321 views

Modules of finite support

I'm reading Dwyer and Fried's paper "Homology of free abelian covers, I". In it, they make the following claim, which I'm having trouble verifying. Let $F$ be a field and $A = F[x_1^{\pm ...
6
votes
1answer
393 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 ...
0
votes
1answer
211 views

Equivalent functors

Let $R$ be a commutative Noetherian ring, $M$ is a finitely generated $R$-module. If $F: Mod \to Mod$ is a left exact functor and $R^iF(E)=0$ where $E$ is injective module. Assume that $F(-) \cong ...
7
votes
2answers
635 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 ...
3
votes
1answer
293 views

spurious torsion under compositions of linear maps

Say we have a PID $R$, integers $1 \leq a \leq b$, and $R$-homomorphisms $R^a \stackrel f\to R^b \stackrel g\to R^a$ with $g \circ f$ of full rank. For $h = f, g, g \circ f$, let $c(h)$ be the ...
1
vote
1answer
419 views

graded noetherian module

Let M be a R graded module with $M= \oplus M_i$. If M is noetherian then $M_i=0 $ for i << 0. My question is this, isn't $M_i = 0$ for all i >> 0 as well? If $(M_{n_i})_{i} \neq 0, n_i > 0$ ...
3
votes
1answer
720 views

structure theorem for modules

Can structure theorem for modules be extended to modules over UFDS , to modules over Neotherian rings ? if yes then can one get the statement and reference? Since operations on matrices with ...
3
votes
1answer
327 views

Modules with flat duals

Let $R$ be a commutative ring, $M$ an $R$-module, $M^*=Hom_R(M,R)$ its dual. What are sufficient (and possibly necessary) conditions on $M$ that ensure that $M^*$ is flat? Is there a name for such ...
1
vote
1answer
331 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
655 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
475 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
349 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 ...
13
votes
4answers
4k views

Flat Module and Torsion-Free Module

All rings in this question are integral. It is known that flat modules are torsion-free. Conversely, torsion-free modules over Prüfer domain (in particular, Dedekind domain) are flat, please see ...
1
vote
1answer
393 views

Cardinality of a linear independent subset of a free module over a commutative ring which is not an integral domain

If R is a commutative ring with unity and not an integral domain and F is a free R-module with rank k,is there a linear independent set with cardinality > k? I prooved that this is not true if R is an ...
40
votes
11answers
3k views

How to introduce notions of flat, projective and free modules?

In the coming spring semester I will be teaching for the first time an introductory (graduate) course in Commutative Algebra. As many people know, I have been plugging away for a while at this ...
11
votes
2answers
2k views

Wikipedia's definition of 'locally free sheaf'

Let $R$ be a, say, noetherian ring and $M$ an $R$-module. The Wikipedia article on 'locally free sheaf' tells me that the following two statements are equivalent: The module $M$ is locally free ...
5
votes
1answer
594 views

Torsion submodule

$A$ a commutative Noetherian domain, $M$ a finitely generated $A$-module. How can I show that the kernel of the natural map $M\rightarrow M^{**}$, where $ M^{ * *}$ is the double dual (with respect to ...
1
vote
2answers
813 views

Extension problem

As I understand, if $0\rightarrow A\rightarrow X\rightarrow B\rightarrow 0$ is a short exact sequence of abelian groups, $\mbox{Ext }_{\mathbb{Z}}^{1}(B,A)$ gives all the isomorphism classes of what ...
2
votes
2answers
332 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
552 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, ...
16
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
575 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 ...
10
votes
4answers
1k views

Linearly independent subsets of a free module

Is it true that the cardinality of every maximal linearly independent subset of a finitely generated free module $A^{n}$ is equal to $n$ (not just at most $n$, but in fact $n$)? Here $A$ is a nonzero ...
0
votes
1answer
569 views

Useless question on rank

What is the rank of $A^{n}$ if A is the zero ring? It's clearly not $n$ as many careless authors claim, since it's not even invariant. I don't think it's 0 either because it does have a linearly ...