Question

Error in a "proof" that all rings have IBN (invariant basis number)

score 584 · Answer 1 · 2016-08-31 23:50:04

584

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Error in a "proof" that all rings have IBN (invariant basis number)

Published on 31 Aug 2016 - 23:50

score 137 · Answer 2 · 2016-10-11 07:13:43

137

Views

$M$ be a finitely gnerated module with a submodule $N$ such that $M/N$ is free , then is $N$ finitely generated?

Published on 11 Oct 2016 - 7:13

score 4.2k · Answer 3 · 2016-10-16 03:56:43

4.2k

Views

Prove that a ring R having the property that every ﬁnitely generated R-module is free is either a ﬁeld or the zero ring.

Published on 16 Oct 2016 - 3:56

score 596 · Answer 4 · 2016-10-19 21:37:23

596

Views

Finding basis of a submodule in $\mathbb{Z}^2$?

Published on 19 Oct 2016 - 21:37

score 67 · Answer 5 · 2016-10-29 12:59:49

67

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$A\in \mathrm{Mat}_n(\mathbb{Z})$ is the matrix representing $f$. Show that if $\mathrm{Coker}(f)$ is finite, then $|\det A|=|\mathrm{Coker}(f)|$.

Published on 29 Oct 2016 - 12:59

score 167 · Answer 6 · 2016-11-02 19:50:13

167

Views

Free objects in a categories of groups, module and other algebraic structures, category-theoretic considerations

Published on 02 Nov 2016 - 19:50

score 58 · Answer 7 · 2016-11-04 18:43:09

58

Views

$\text{End}_{\mathbb{Z}}\left(\prod_{i\in\mathbb{N}}\mathbb{Z}\right)$ is a free module over itself

Published on 04 Nov 2016 - 18:43

score 1.3k · Answer 8 · 2016-11-08 20:43:28

1.3k

Views

Let $M$ be a free $R$-module, and let it have an infinite basis. Then all bases of $M$ have the same cardinality

Published on 08 Nov 2016 - 20:43

score 185 · Answer 9 · 2026-03-25 11:16:16

185

Views

An exchange property for bases of a free module over a division ring $R$

Published on 25 Mar 2026 - 11:16

score 1.2k · Answer 10 · 2016-11-15 17:37:25

1.2k

Views

If $F_1$ and $F_2$ are free modules on the same set $A$, then $F_1 \cong F_2$

Published on 15 Nov 2016 - 17:37

score 259 · Answer 11 · 2016-11-20 14:33:04

259

Views

$I$ be an ideal of $R$ containing non-zero zero-divisors , then $R/I$ cannot be a projective $R$-module?

Published on 20 Nov 2016 - 14:33

score 99 · Answer 12 · 2016-11-21 05:25:31

99

Views

$R$ is a domain such that for $x,y \in R$, $(x)\subseteq (y)$ or $(y)\subseteq (x)$ . Is every f.g. torsion free module over $R$ free?

Published on 21 Nov 2016 - 5:25

score 77 · Answer 13 · 2016-12-05 17:30:50

77

Views

Zero elements of tensor product of two free modules

Published on 05 Dec 2016 - 17:30

score 320 · Answer 14 · 2017-01-15 21:43:20

320

Views

Quotient module and change of basis

Published on 15 Jan 2017 - 21:43

score 1.7k · Answer 15 · 2017-02-01 09:58:15

1.7k

Views

Finitely generated free $A$-module is isomorphic to $A^n$

Published on 01 Feb 2017 - 9:58

List Question

Error in a "proof" that all rings have IBN (invariant basis number)

$M$ be a finitely gnerated module with a submodule $N$ such that $M/N$ is free , then is $N$ finitely generated?

Prove that a ring R having the property that every ﬁnitely generated R-module is free is either a ﬁeld or the zero ring.

Finding basis of a submodule in $\mathbb{Z}^2$?

$A\in \mathrm{Mat}_n(\mathbb{Z})$ is the matrix representing $f$. Show that if $\mathrm{Coker}(f)$ is finite, then $|\det A|=|\mathrm{Coker}(f)|$.

Free objects in a categories of groups, module and other algebraic structures, category-theoretic considerations

$\text{End}_{\mathbb{Z}}\left(\prod_{i\in\mathbb{N}}\mathbb{Z}\right)$ is a free module over itself

Let $M$ be a free $R$-module, and let it have an infinite basis. Then all bases of $M$ have the same cardinality

An exchange property for bases of a free module over a division ring $R$

If $F_1$ and $F_2$ are free modules on the same set $A$, then $F_1 \cong F_2$

$I$ be an ideal of $R$ containing non-zero zero-divisors , then $R/I$ cannot be a projective $R$-module?

$R$ is a domain such that for $x,y \in R$, $(x)\subseteq (y)$ or $(y)\subseteq (x)$ . Is every f.g. torsion free module over $R$ free?

Zero elements of tensor product of two free modules

Quotient module and change of basis

Finitely generated free $A$-module is isomorphic to $A^n$

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