Question

Free module over a ring with identity with a basis of size $m, \ \forall m \ge n$

score 458 · Answer 1 · 2014-12-17 06:36:00

Free module over a ring with identity with a basis of size $m, \ \forall m \ge n$

Published on 17 Dec 2014 - 6:36

score 109 · Answer 2 · 2014-12-19 08:09:44

$M$ noetherian, $f$ endomorphism of $M$, $\operatorname{coker}f$ has finite length, then $\operatorname{coker}f^n$ and $\ker f^n$ have finite length.

score 167 · Answer 3 · 2014-12-19 10:21:47

167

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score 3.2k · Answer 4 · 2026-03-29 18:31:37

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score 899 · Answer 6 · 2014-12-20 16:53:12

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score 50 · Answer 15 · 2014-12-23 10:24:32

50

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List Question

Free module over a ring with identity with a basis of size $m, \ \forall m \ge n$

$M$ noetherian, $f$ endomorphism of $M$, $\operatorname{coker}f$ has finite length, then $\operatorname{coker}f^n$ and $\ker f^n$ have finite length.

When is a homomorphism an epimorphism?!

How to show $\mathbf{Q} $ is not free

Are the two ways of creating an $S^{-1}A$ algebra equivalent?

Example of a ring without Invariant Basis Property

Example of non noetherian ring and noetherian $\Bbb Z$-module

Homomorphism from a finitely generated module to a direct sum of modules

Isomorphism of quotient of direct sum modules

$\Bbb{Q}$ is not a finitely generated $\Bbb{Z}$-module

$\Bbb{Z}[X]/\langle f\rangle$ is a finitely generated $\Bbb{Z}$-module

Proof of the properties of tensor product

Tensor-Hom Adjunction In Monoidal Categories?

Understanding the Definition of the Tensor Product of Chain Complexes

How can $\Bbb{Z}/10$ be viewed as a $\Bbb{Z}$-module?

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