# 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.

553 questions
Filter by
Sorted by
Tagged with
26 views

• 549
24 views

### 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 $I$...
1 vote
35 views

• 387
1 vote
36 views

• 4,249
97 views

### 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 ...
• 89
59 views

### 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. ...
1 vote
66 views

### 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 ...
• 1,071
40 views

• 4,249
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: ...
• 4,249
1 vote
15 views

### 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 ...
• 301
33 views

• 743
1 vote
62 views

1 vote
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&...
• 3,139
136 views

### 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 ...
• 3,139
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 ...
• 111
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 ...
• 1,769
86 views

### 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 ...
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 ...
• 923
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 ...
54 views

• 3,162
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 ...
• 1,308