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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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First syzygy is unique up to direct summand

Let $R:=K[x_0,...,x_n]$ be the polynomial ring over a field $K$ and $M$ be finitely generated $R$ module. Let $(m_i)_{i=1,...,k}$ be a generating set for $M$. Then we can define the free module $F_0 :=...
Flynn Fehre's user avatar
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Free module and finitely generated submodule.

Let $A$ be a commutative ring, $M$ a free $A$-module and $N\subseteq M$ a finitely generated submodule. Prove that there is a Sub-Noetherian Ring $A_0 \subseteq A$, a free $A_0$-submodule $M_0 \...
George's user avatar
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When defining a free module, is the action of a ring $R$ on itself as a module implied to be multiplication?

In the free modules section of Atiyah & MacDonald, they define a free $R$-module $M$ as any $R$-module such that $M \cong \bigoplus_{i \in I} M_i$ where each $M_i \cong R$ as an $R$-module, and $I$...
a_bird's user avatar
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cokernel of acyclic chain complex of projective modules after tensoring

Let $R$ be a commutative Noetherian ring, $(\mathbf F,d)$ be an acyclic chain complex of projective $R$-modules and $M$ be a finitely generated $R$-module. Let $i$ be an integer and put $N:=\text{...
Alex's user avatar
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For $R=\mathbb{k}[x,y]$, $\operatorname{Hom}_R(M,R)$ is always free.

Let $R=\mathbb{k}[x,y]$ be the polynomial ring with two variables over a field $\mathbb{k}$. I claim that $\operatorname{Hom}_R(M,R)$ is always free for any finitely generated $R$-module $M$. I ...
Ramanasa's user avatar
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21 votes
1 answer
337 views

Existence of maximal free submodules

Let $A$ be a domain and $M$ be a finitely generated module. Is there a free submodule which is maximal among free submodules? Answer is yes for Noetherian rings, obviously. Also the rank of any such ...
Davide Pierrat's user avatar
1 vote
1 answer
94 views

Is $\mathbb{R}$ a free $\mathbb{Q}$-module

Is $\mathbb{R}$ a free $\mathbb{Q}$-module? $\mathbb{R}$ is the real numbers and $\mathbb{Q}$ is the rational numbers. I seem to get that the algebraic closure on $\mathbb{Q}$ is a free $\mathbb{Q}$-...
cute dunkey's user avatar
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Every finitely generated and torsion free $A$-module is free when $A$ is PID

If $A$ is a commutative ring with identity and a PID and $M$ is a finitely generated and torsion free $A$-module, then $M$ is free. I am trying to solve this excercise from my Commutative algebra ...
Superdivinidad's user avatar
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primitive idempotent

I’ve tried to prove: $R$ local, Artin ring $\phi:End(P) \rightarrow End(Q) $ ,$e\in End(P) $ is primivite idempotent and $P$ and $Q$ free module so $P= (\oplus R_i)$ and $Q=(\oplus R_j) $then how ...
wanheda's user avatar
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If $N$ contains a maximal submodule, does this imply that $ \text{Rad} N $ is a superfluous submodule of $N$?

If $M$ is a finitely generated module over a ring $R$ with unity, it can be readily demonstrated that the radical of $M$, denoted $ \text{Rad} M$, qualifies as a superfluous submodule of $M$. The ...
Liang Chen's user avatar
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Revisiting proof that all bases of a free module $M$ over a commutative unitary ring are equipotent

I am following closely the book of T.S. Blyth, Module Theory. There is a theorem which says that every free $R$-module $M$ where $R$ is a commutative unitary ring, has equipotant base. The process is ...
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Generating a free module

Let $R$ be a ring with $1$ and let $M$ be an $R$-module. Let $A \subseteq M$. Is it true that the following are equivilent. $A$ is linearly independent. $R \cong Ra$ and $$\sum_{a \in A}Ra = \...
Ethan Kharitonov's user avatar
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Noether normalization and free modules

Let $A$ be a finitely generated algebra over some field $k$, and assume that $A$ is an integral domain. By Noether normalization, we have a finite injective ring homomorphism $\varphi: k[x_1, \ldots, ...
Adelhart's user avatar
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Module Theory, Free Modules, Finitely Generated Modules

I want to prove that if R is a PID and M is a finitely generated R-module, then there exist a free R- module F such that $M \cong F \oplus M_{\text{tors}}$ So, in order to prove the above I set $M/M_{...
dota's user avatar
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Alternative examples of free modules without the IBN property

I'm trying to search for as many examples of non-IBN free modules as possible. But so far I can only find two examples. One is $End(V)$, where $V$ is a vector space of countable dimensions. Another is ...
user760's user avatar
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Hint for a Sequence of Free Modules

Could someone please give me a hint for this problem? The following sequence is exact: $0 \rightarrow A \xrightarrow{f} B \xrightarrow{g} C \xrightarrow{h} D \xrightarrow{i} E \xrightarrow{j} F \to 0$....
Tom's user avatar
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2 answers
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Theorem 11, Section 4.5 of Hungerford's Algebra

Let $R$ be a ring with identity. If $A$ is a unitary right $R$-module and $F$ is a free left $R$-module with basis $Y$, thell every element $u$ of $A\otimes_R F$ may be written uniquely in the form $u ...
user264745's user avatar
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Why is the submodule of K[x,y] generated by x and y not a free module?

If $\mathbb{K}$ is a field, and we consider the module $\mathbb{K}[x,y]$ over itself, I've seen others give the example that the ideal generated by $\langle x, y \rangle$ is not a free module to ...
Vibbz's user avatar
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1 answer
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Difference betwen universal properties

I'm studying the construction of the tensor product. To do this, I opted for the quotient space's approach, defining first the free vector space over the cartesian $V\times W$ of two vector spaces. ...
Paulo Estêvão's user avatar
1 vote
1 answer
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Understanding homomorphism on tensor product of module

I'm trying to convince myself one part of the proof for a theorem in chapter 10 in DUmmit and Foote. Let $R$ be a subring of $S$, let $N$ be a left $R$-module and let $\iota:N \to S\otimes_RN$ be the ...
Remu X's user avatar
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F is a free ZG-module implies F is also free as a ZK-module

Let $K \le G$ be a subgroup. I have to show that if $F$ is a free $\mathbb{Z}G$-module, then it is also free as a $\mathbb{Z}K$-module. The definition I have in mind is that $F=\left\{ \sum r_x x \mid ...
Todd Burnett's user avatar
1 vote
1 answer
86 views

Exercise 9, Section 4.2 of Hungerford’s Algebra

If $F_1$ and $F_2$ are free modules over a ring with the invariant dimension property, then $\text{rank} (F_1 \oplus F_2) = \text{rank} F_1 + \text{rank} F_2$. I have written proof of this exercise ...
user264745's user avatar
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Projective and free modules

In the graded context, if $R = K[x_1,...,x_n]$ where $K$ is a field, is a projective R-module a free R-module?
Cib's user avatar
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Lemma 10, Section 4.2 of Hungerford’s Algebra

Let $R$ be a ring with identity, $I$ ($\neq R$) an ideal of $R$, $F$ a free $R$-module with basis $X$ and $\pi : F\to F/IF$ the canonical epimorphism. Then $F/IF$ is a free $R/I$ module with basis $\...
user264745's user avatar
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127 views

Theorem 4, Section 4.2 of Hungerford’s Algebra

Every vector space $V$ over a division ring $D$ has a basis and is therefore a free $D$-module. More generally every linearly independent subset of $V$ is contained in a basis of $V$. Sketch of proof: ...
user264745's user avatar
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Bundle map from $Spec(R[x_{1}, \dots, x_{n}])$ to $Spec(R)$

My question: let $i: R \rightarrow R[x_{1}, \dots, x_{n}]$ be natural embedding, then $i^{*}: Spec(R[x_{1}, \dots, x_{n}]) \rightarrow Spec(R)$ is a bundle map. How should this conclusion be proven? I ...
jhzg's user avatar
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Polynomial rings and symmetric algebras on the dual module are isomorphic

Question: if $M$ is a finitely generated free R−module of rank n, then $Sym(M^{*}) \cong R[x_{n}, \dots, x_{n}]$, where $M^{*}$ is the dual module of $M$ and $Sym(M^{*}) = \frac{R \oplus M^{*} \oplus ...
jhzg's user avatar
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1 answer
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Free submodules of an integral $R.$

I was told by the author of the answer here Showing that the rank of $M$ is exactly $1.$ that: Free submodules of an integral domain $R$ are exactly the principal ideals of $R.$ I am wondering which ...
user avatar
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1 answer
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Showing that the rank of $M$ is exactly $1.$

Here is the question I am trying to solve: Let $R = \mathbb Z[x]$ and let $M = (2,x)$ be the ideal generated by $2$ and $x,$ considered as a submodule of $R.$ Show that $\{2,x\}$ is not a basis of $M.$...
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2 votes
0 answers
57 views

Generating set for submodule of free module over a PIR

Let $R$ be a commutative principal ideal ring (not necessarily a domain), $S \cong R^n$ a free $R$-module. We know that any submodule $M$ of $S$ has $m := \operatorname{length}_R(M) \leq n \cdot \...
JBuck's user avatar
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Finitely generated $R$-module has a free resolution of length at most $1$?

If $R$ is a PID and $M$ is a finitely generated $R$-module, How do I show that M has a free resolution of length at most $1$? So I have to show that M has a free resolution $....\to F_2 \to F_1 \to ...
some_math_guy's user avatar
1 vote
0 answers
85 views

Proving two modules are free based on their direct sum [duplicate]

I have given two modules $M$ and $N$ over a local ring $R$. I also know that $M \oplus N \cong R^n$ for some $n\in \mathbb{N}$. I then have to prove that both $M$ and $N$ are free modules. Since $M \...
MarlonButBetter's user avatar
1 vote
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132 views

Prove that the polynomial ring $R[x]$ is a flat $R$-module.

Here is the question I am trying to solve: Prove that the polynomial ring $R[x]$ in the indeterminate $x$ over the commutative ring $R$ is a flat $R$-module. My thoughts: We know by corollary 42 in D&...
Intuition's user avatar
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0 votes
2 answers
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Why a nonzero finite abelian group is not projective?

Here is the question I am trying to solve (I know it is answered here $A$ be a nonzero finite abelian group then $A$ is not a projective or injective $\Bbb Z$ module. but the answer is not very clear ...
Intuition's user avatar
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2 votes
0 answers
34 views

partial Quillen-Suslin on square matrix

The Quillen-Suslin theorem states that projective modules over $\mathbb{Q}[x_1, \dots , x_n]$ are free. (This also holds over more general fields or rings.) I have a square $m\times m$ matrix $M$ over ...
MatthysJ's user avatar
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4 votes
1 answer
115 views

Construction of the free operad

I am reading Fresse's "Koszul duality of operads and homology of partition posets", and trying to understand the construction of the free operad $F(M)$ on a symmetric sequence $M$ as ...
Margaret's user avatar
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2 votes
1 answer
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Generalized Notion of Krylov Subspaces

Let $\mathcal{X}$ be a vector space over a field $\mathbb{K}$ and let $x_0 \in \mathcal{X} \setminus \{0_{\mathcal{X}}\}$ (here, $0_{\mathcal{X}}$ denotes the zero vector in $\mathcal{X}$). We denote ...
Trouble Is Back's user avatar
2 votes
0 answers
384 views

Are projective modules free over a polynomial ring with infinite indeterminates over a field?

(1) In 1963, Bass proved that any nonfinitely generated projective module is free over a connected Noether ring. (2) Quillen–Suslin theorem states that any finitely generated projective module over a ...
Liang Chen's user avatar
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1 answer
41 views

Does this argument use the fact that $\{e_i\}$ is linearly independent at any point?

I'm looking through a proof and I'm trying to tell if it uses a certain fact. If it doesn't, then I think I've figured out a homework problem. Here is the lemma and the proof of it: Lemma: Every ...
blakedylanmusic's user avatar
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0 answers
54 views

How to find an injective ring homomorphism from $S$ to $M_n(R)$,where S is a commutative unitary ring as well as a free module over its subring $R$?

Suppose $S$ is a commutative unitary ring and $R$ is a subring of $S$,naturally S becomes a $R$-module. To prove: If S is a free $R$-module over R of rank $n$,then there is a ring isomorphism between $...
YSouSerious's user avatar
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0 answers
38 views

When $R$-Module of Rank $r$ has a submodule isomorphic to $R^r$?

Let $R$ be an integral domain,$M$ is a finitely generated $R$ module. Prove that rank$(M)=r$ iff $M$ has a free submodule $N \equiv R^r$ , such that $M/N$ torsion . If $R$ is a PID then $N$ may be ...
Infinity's user avatar
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0 answers
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Linear independence of a subset of elements in a finitely generated module to the number of generators of the module

Let $M$ be a finitely generated $R$-module. suppose $m_i, ..., m_k$ and $p_i, ..., p_n$ are elements in $M$ so that ${m_i, ..., m_k}$ is linearly independent over ring $R$, and ${p_i, ..., p_n}$ is a ...
Tessa's user avatar
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1 vote
0 answers
94 views

$P$ is projective implies $P℘$ is a free $R℘$-module

Statement: Let $R$ be a commutative Noetherian ring and let $P$ be a finitely generated $R$-module, then $P$ is projective if and only if $P℘$ is a free $R℘$-module for all $℘$ in $Spec(R)$. For the ...
Nilotpal Chakraborty's user avatar
1 vote
1 answer
62 views

When are the integers a free R-module?

The are some $R$ for which the integers are a free $R$-module. (For example, when $R = Z$.) But there are other cases when the integers are not. (E.g. take $R = Z[G]$ with the action induced by the ...
Cookie Monster's user avatar
1 vote
1 answer
160 views

Standard decomposition of vector-space endomorphism as direct sum of abelian groups

Let $V=\mathbb Q^7$, $\phi: V \rightarrow V$ the $\mathbb Q$-linear map given by the matrix: $A=\begin{align*} \begin{pmatrix} 1&0&0&0&1&0&2 \\0&2&1&0&0&-1&...
some_math_guy's user avatar
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0 answers
118 views

Show that if $Q$ is a non-trivial $\mathbb{Z}$-module, that is divisible, then $Q$ can not be a projective $\mathbb{Z}$-module. [duplicate]

As the question says, I want to show that If $Q$ is a non-trivial $\mathbb{Z}$-module, that is divisible, then $Q$ can not be a projective $\mathbb{Z}$-module. Now, here is my reasoning: Let $F$ be ...
Ben123's user avatar
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0 votes
1 answer
69 views

Flatness of $\mathbb{C}[x_1,\ldots,x_n]$ over $\mathbb{C}[f]$, $f \in \mathbb{C}[x_1,\ldots,x_n]$, $n \geq 1$

Let $f \in \mathbb{C}[x_1,\ldots,x_n]$, $n \geq 1$. Call $f$ 'good' if $\mathbb{C}[x_1,\ldots,x_n]$ is flat over $\mathbb{C}[f]$. Is it true that every $f \in \mathbb{C}[x_1,\ldots,x_n]$ is good? $f ...
user237522's user avatar
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0 votes
0 answers
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Finitely generated module equal to free module of rank $n$? Not just the quotient of free module.

(For commutative algebra) I have a Ring defined as the polynomial Ring over Field $K$: $R = K[x]$. I need to show that the finitely generated module $M$ is equal to a free module of rank $n$ for ...
mad_scientist's user avatar
2 votes
1 answer
91 views

Alternative proof for Structure theorem for finitely generated modules over a principal ideal domain

I'm thinking of an alternative proof for Structure theorem for finitely generated modules over a principal ideal domain (also called Fundamental theorem of finitely generated modules over a PID in ...
Ubik's user avatar
  • 488
6 votes
4 answers
277 views

Every $K[G]$-module is torsionless?

Let $G$ be a finite group and $K$ a field. Consider the group ring $R:=K[G]$. Let $M$ be a (left) $R$-module. Is it true that then there exists a set $S$ and an injective $R$-module homomorphism $M\...
Margaret's user avatar
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