Questions tagged [free-modules]

Use this tag for questions about free modules and related notions as projective modules or free abelian groups. This tag should be used together with the tags of abstract algebra and modules.

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Finitness conditions under change of rings?

Let $R \to S$ be a homomorphism of commutative rings, such that $S$ is finitely generated and projective $R$-module. Proof that if $A$ is a finitely presented $S$-module then $A$ is a finitely ...
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If $R$ is a right hereditary ring, then any submodule of a right projective $R$-module is again projective.

Recall that a ring $R$ is called right hereditary if every right right ideal $I\subset R$ is projective as a right $R$-module. I need to prove that if $M_R$ is projective module and $N\leq M$ is any ...
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Free objects in the category of all R-modules without linearly independent generating set

We can define a free module $F$ over a commutative ring with identity $R$ as a free object (with universal property) in the category of unitary $R$-modules. This is equivalent to the existence of a ...
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Equality of two free modules of the same finite rank under strong hypothesis.

So basically the question is the one of the title of the post, but let me show you the context: Let $\mathbb K$ be an algebraic closed field and let $f\in \mathbb K[X,Y]$ be a polynomial such that is ...
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2 votes
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Finding a basis for a free module

Let q $\in$ $\Bbb Q^3$, such that the elements of q are positive numbers. Let M $\subset \Bbb Z^3$ be defined as $\hspace{3in}$M = $\{ \mathbf n \in \Bbb Z^3 | \mathbf n \cdot \mathbf q \in \Bbb Z \}$,...
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$M/xM$ free over $R/xR$ implies $M$ is free over $R$ when $R$ Noetherian

The following is from https://stacks.math.columbia.edu/tag/00NS. I'm having some difficulty understanding some steps of the proof. Let $R$ be a Noetherian local ring. Let $x \in \mathfrak m$. Let $M$ ...
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Under what conditions would $S^{-1}F$ be a free left $S^{-1}R$-module?

So I am thinking of a ring $R$ with multiplicative identity $1_{R}\ne 0_{R}$, $F$ is a free left $R$-module, and every element of $S$ is not a zero divisor, and $0_{R}\in S$. Under these conditions ...
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I proved that submodules of free modules are sums of ideals. What's wrong?

I'm probably overlooking something very trivial. Suppose that $A$ is some ring, $M \hookrightarrow \bigoplus_{j \in J} A$ a left $A$-submodule and $J$ is well-ordered. Let $F_j := \bigoplus_{i;i \leq ...
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1 answer
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Direct limit of n-presented modules?

A module $M$ is said to be $n$-presented If there exist an exact sequence $$F_{n}\to F_{n-1}\to \cdots \to F_{1}\to F_{0}\to M$$ with each $F_{i}$ is free finitely generated. For example $M$ is $0$-...
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Showing that a certain Tor group vanishes

I've been working through Atiyah & Macdonald's text, and came across the following problem from an old commutative algebra qualifying exam. Let $M,N$ be $R$-modules over a commutative ring $R$. ...
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Some conclusions with respect to character module $M^{\ast}=Hom_{\mathbb{Z}}(M, \mathbb{Q/Z})$

Given an $R$-module $M$, we define its character module as $M^{\ast}=Hom_{\mathbb{Z}}(M, \mathbb{Q/Z})$. I have already proved that $M$ is a flat module iff $M^{\ast}$ is an injective module. Now I am ...
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Finitely generated projective $\mathbb{Z}[[x]]$-modules are free?

$\newcommand{\Z}{\mathbb{Z}}$ Let $A = \Z[[x]]$ be the ring of power series. We know that this is not a PID. In particular a submodule of the free $A$-module $\bigoplus_{i=1}^nA$ does not have to be ...
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1 answer
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Basis for Module Over PID

Given $M$ is a free module over PID (let's say R). $$X = \{x_1, x_2, \cdots, x_m\}$$ span $M$. Can $X$ can be reduced, so that $X$ is a basis for $M$? Note: $X$ is a basis for module-$M$ if $X$ span $...
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Invariant basis number for matrix algebras

Let $\mathbb{k}$ be a field (we can suppose $k=\mathbb{R}$ or $\mathbb{C}$ if necessary). Let $M_n(\mathbb{k})$ denote the ring of matrices with entries in $\mathbb k$. Recall that a ring $R$ has the ...
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Proof that any element of a free abelian group can be extended to a basis

Let $A$ be a free abelian group of rank $n$, and let $\alpha_1 \in A\setminus \{0\}$ such that $\alpha_1 \not \in kA$ for all $k > 1$. Do there always exist $\alpha_2,\ldots, \alpha_n \in A$ such ...
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About exactness of $Hom$ functor?

We know that a module $P$ is projective if and only if the functor $Hom(P,-):R-Mod \to Ab$ is exact, i.e it preserves epimorphisms: If $\alpha: M \to N$ is an epimorphism of modules then $Hom(P,\alpha)...
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1 answer
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Can a matrix ring have a non-free ring extension?

Let $D$ be a skew field, and consider an arbitrary noncommutative unital ring $R$ extending the matrix ring $D^{n\times n}$. Must $R$ be free as a $D^{n\times n}$-module? If $R$ is finite-dimensional,...
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2 votes
1 answer
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Is the ring $R$ a free module over $S$?

$S$ is a simple Artinian ring with unity. $S \subset R$ is a ring extension. I know an additional condition that $R$ is finitely generated as left or right $S$-module. Is it true that $R$ is free as ...
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Phantom morphisms and Tor functor?

$\newcommand{\Tor}{\operatorname{Tor}}$In the category of R-modules, a morphism $f:M\to N$ is called a phantom morphism if for every finitely presented module $F$ and every morphism $g:F\to M$, $fg$ ...
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1 answer
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Ideal generated by the entries of a representative matrix of a map between free modules [closed]

Let $R$ be a commutative Noetherian ring. Let $f: R^{\oplus a}\to R^{\oplus b}$ be an $R$-linear map. Let $\{e_i\}_{i=1}^a$ and $\{e_i\}_{i=1}^b$ be the standard bases of $R^{\oplus a}$ and $R^{\oplus ...
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2 votes
1 answer
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Equivalent statements of a finitely generated module being locally free

Let $M$ be a finitely generated $R$-module. Prove that the following conditions on $M$ are equivalent: (a) $M$ is locally free over $R$ (i.e. $M_m$ is free over $R_{m}$ for all maximal ideals $m\...
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1 vote
1 answer
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Localization of Cohen-Macaulay module of finite projective dimension at non-maximal prime ideal

Let $(R,\mathfrak m)$ be a local Gorenstein domain of dimension $2$. Let $M$ be a finitely generated $1$-dimensional module with projective dimension $1$. Then by Auslander-Buchsbaum formula, $\mathrm{...
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On splitness of epimorphisms?

Let $\alpha: M \to N$ be an epimorphism of left R-modules. Let $Q$ be a left R-module of finite projective dimension d with a morphism $\beta : Q \to N$. We know that if d=0 i.e, $Q$ is projective ...
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Extending a theorem relating submodules of free modules over a PID to graded modules. [duplicate]

We have the following theorem: Suppose $M$ is a free module of finite rank over a PID, $R$, and $N$ is a submodule of $M$. Then there exists a basis for $B:=\{m_1,m_2,...,m_k\}$ of $M$ and $r_1,r_2,......
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1 answer
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Show if $M$ is free of rank $n$ as $R$-module, then $M/IM$ is free of rank $n$ as $R/I$ module

Let $R$ be a ring and $I \subset R $ a two-sided ideal and $M$ an $R$-module with $$IM = \{ \sum r_i x_i \mid r_i \in I, x_i \in M \}. $$ Show that if the $R$-module $M$ is free of rank $n$ then $M/IM$...
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1 vote
1 answer
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Reference request modules over a small category

I am currently writing up one of my results where I am using modules over small $\mathbb{k}$-linear categories. I so far sadly did not find a proper introduction of them to refer the reader to and was ...
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Quotient of a module is free implies the module is free

I'd like to ask your opinion whether this answer is correct. If not, under which conditions the following lemma is true? Lemma Let $p$ be a prime number and $B$ a noetherian torsion free $\mathbb{Z}_p$...
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2 votes
1 answer
92 views

Software to determine if one ring is free over another

Suppose the fates have given me a polynomial ring $C$ in four indeterminates over $\mathbb Z$ and subrings $A < B < C$. Suppose further that the expressions the fates have given me for ...
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-1 votes
1 answer
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If $N \leq M$, there exist free modules $F_N \leq N$, $F_M \leq M$ with $N = F_N \oplus N_{tor}$, $M = F_M \oplus M_{tor}$ with $F_N \leq F_M$

Let $R$ be a principal ideal domain and $M$ a finitely generated $R$-module. Furthermore, let $N$ be a submodule of $M$. Prove or disprove: there exist free submodules $F_N \leq N$, $F_M \leq M$ with $...
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0 answers
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If ring R can be expressed as direct sum of ideals then there must exist idempotent elements such that $R e_j = A_j$ [duplicate]

Let {$A_i$} be ideals of R The ring R can e written as a direct sum $R = {\bigoplus}_i A_i$ if and only if the exists a set $e_1, e_2, ...$ of orthogonal idempotent elements of R such that $A_j = R ...
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1 vote
1 answer
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Question related to "example of free module with bases of different cardinality"

In this Planet Math article, it shows an example where a free module has bases with different cardinality. I am having some problems with show why $\phi$ is an $R$-module homomorphism.
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6 votes
2 answers
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Question on $(2, 1+\sqrt{-5})$ as a submodule of $\mathbb{Z}[\sqrt{-5}]$.

Let $R$ be the ring $\mathbb{Z}[\sqrt{-5}]$ and $I$ the ideal generated by $2$ and $1+\sqrt{-5}$, $I=(2, 1+\sqrt{-5})$. Show that $I$ is not R-module isomorphic to $R$ but $I\bigoplus I$ is R-module ...
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1 vote
1 answer
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Prove $D$ is $R$-algebra

I'm studying the paper Galois Theory and Galois Cohomology of Commutative Rings by S. U. Chase, D. K. Harrison and Alex Rosenberg. A couple days ago I posted this question where I wanted to prove that ...
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1 vote
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Every left $KG$-module $M$ is a right $KG$-module

I don't know how to prove that a left $KG$-module is a right $KG$-module. What I have so far is that a left $G$-module is a right $G$-module always defining the operation $x\cdot'g=g^{-1}\cdot x$. But ...
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4 votes
3 answers
190 views

Doubt about the definition of free vector spaces.

Introduction: Suppose you wish to construct a set $F(X)$ of linear combination of elements of a given a set $X$: $$V = v_{1}x_{1}+\cdots +v_{n}x_{n} \tag{1}$$ where $a\in \mathbb{K}$ and $x^{i} \in X$....
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1 answer
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Explanation of Corollary 2.15. Hungerford's Algebra book

Corollary 2.15. If $V$ and $W$ are finite dimensional subspaces of a vector space over a division ring $D$, then $dim V + dim W = dim (V \cap W) + dim (V + W)$. Sketch of proof. Let $X$ be a basis of $...
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1 answer
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Being direct summand of free module implies having dual basis.

I need to provet equality of two definitions of projective module: being direct summand of free module (or equally: having embedding into free module) and having dual basis. Lets use Wikipedia ...
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1 answer
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How to prove $| V | = |A|\operatorname{rank}_A V$, where $V$ is a free $A$-module.

This question was asked in my assignment on modules and I couldn't solve this particular question. Let $V$ be a free $A$-module of infinite rank. Then $|V | =|A|\cdot\operatorname{rank}_A V= \sup \{ ...
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0 votes
1 answer
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The direct sum of $n$-copies of a $K$-vector space $V$ is a free $\operatorname{End}_{R}(V)$-module

I came across questions in the free module section of my abstract algebra text. In the text, the notation $\operatorname{End}_{R}(M)$ denotes the set of all $M$-module endomorphisms of $M$. On to the ...
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0 votes
2 answers
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Question about finite dimensional vector space over a field $K$ and free $K[x]$-module

I came across questions in the free module section of my abstract algebra text. In the text, the notation $End_{R}(V)$ denotes the set of all $R$-module endomorphisms of $M$. Algebra: Abstract and ...
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1 vote
1 answer
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A finite dimensional vector $V$ is not free as an $\operatorname{End}_{K}(V)$-module

I came across questions in the free module section of my abstract algebra text. As on p358 of the text Algebra: Abstract and Concrete, the notation $\operatorname{End}_{R}(M)$ denotes the set of all $...
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3 votes
1 answer
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Rings of integers of number field and free modules over a PID

If $K \subseteq L$ are number fields and $O_{L}, O_{K}$ are their ring of integers, how can I prove that there exists a basis of $O_{L}$ over $\mathbb{Z}$ that contains a basis of $O_{K}$ over $\...
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2 votes
0 answers
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Free Module - sufficient condition

Let $R$ be a commutative ring with unity. Let $I$ be an ideal of $R$. Let $M$ be a left $R$-module. I came across the result that for a free $R$-module $N$, any two bases are equipotent. While proving ...
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1 vote
1 answer
56 views

Why are stably free modules called stably free?

Being of the kind who easily get troubled by questions of semantics, I've been troubled for a while now about why "stably free" modules are called just "stably free". It seems to ...
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0 votes
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Family of generators of a module over a local ring [duplicate]

I have a local ring $(R,\mathfrak m)$ and a module $M$ which is free of finite rank ($M\simeq R^n$, for some $n\in\mathbb N$). Given a family of generators $v_1,\cdots,v_m$ I know that I can extract a ...
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  • 2,093
1 vote
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Submodules generated by relations

When definig the tensor product, the textbook says that: The tensor product $M \otimes_RN$ is constructed as the quotient of the free module $R^{\oplus(M\times N)}$ modulo the submodule generated by ...
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1 vote
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local bases at $q=0$

If $k$ is a field, I'm having troubles in understanding part of the definition of a local base for a $k(q)$-vector space. More precisely, suppose that $V$ is a $k(q)$-vector space, and let $A$ be the ...
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0 votes
1 answer
111 views

Torsion free modules, free modules

This particular question was asked in my Abstract Algebra Exam and I couldn't solve it in examination hall. Now, I thought I should try it again but I was unable to. Prove that every free module F ...
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0 votes
1 answer
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Basis for a quotient of a free module $F$ without tensor products

I am stuck with the following: let $(A,\mathfrak{m},k)$ be a commutative local ring with unity, and $F$ a free $A$-module of rank $n$. Then: $$\{e_i\}_{i=1}^n \text{ is a basis for } F \text{ as an $A$...
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0 votes
1 answer
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Exercise 6.6 in Eisenbud's Commutative Algebra on the Family of Projective Plane Curves of Degree d

This was already asked by someone else here: They answered their own question there but I'm not satisfied with their answer for part a. I'll reproduce the question from that link here: Consider $k\...
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