Question

Difference of a ring and its module over itself

score 1k · Answer 1 · 2014-06-03 23:21:19

1k

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Difference of a ring and its module over itself

Published on 03 Jun 2014 - 23:21

score 37 · Answer 2 · 2026-03-27 23:57:28

37

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Quick question: SES where base ring is a PID

Published on 27 Mar 2026 - 23:57

score 325 · Answer 3 · 2026-03-25 07:43:27

325

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Sufficient conditions for quotient ring to be Cohen-Macaulay

Published on 25 Mar 2026 - 7:43

score 2.4k · Answer 4 · 2014-06-04 19:10:24

2.4k

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Dimension of a module (as vector space)

Published on 04 Jun 2014 - 19:10

score 248 · Answer 5 · 2026-03-25 09:28:35

248

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Direct product of Cohen-Macaulay rings/Eisenbud, Exercise 18.6

Published on 25 Mar 2026 - 9:28

score 366 · Answer 6 · 2026-03-25 07:42:38

366

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Suggest a good book or reference on graded modules over polynomial rings

Published on 25 Mar 2026 - 7:42

score 116 · Answer 7 · 2014-06-05 18:09:09

116

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$M \otimes_R K$ is uniquely divisible as $R$-module

Published on 05 Jun 2014 - 18:09

score 197 · Answer 8 · 2026-03-25 22:10:18

197

Views

$Q(\mathbb{Z}[t]) / \mathbb{Z}[t]$ is not injective

Published on 25 Mar 2026 - 22:10

score 249 · Answer 9 · 2014-06-06 12:55:00

249

Views

$\biggl( \prod_{i=1}^{+ \infty} \mathbb{Z}/2^i \mathbb{Z} \biggr)\otimes_{\mathbb{Z}}\mathbb{Q} \neq 0$ without using $p$-adic integers

Published on 06 Jun 2014 - 12:55

score 1k · Answer 10 · 2014-06-07 07:11:36

1k

Views

$2 \otimes_{R} 2 + x \otimes_{R} x$ is not a simple tensor in $I \otimes_R I$

Published on 07 Jun 2014 - 7:11

score 31 · Answer 11 · 2014-06-08 11:49:46

31

Views

Question about modules

Published on 08 Jun 2014 - 11:49

score 111 · Answer 12 · 2014-06-08 22:29:20

111

Views

map $\iota: A \rightarrow \mathbb{Q}\otimes_{\mathbb{Z}} A$, where $A$ is a finite abelian group of order n

Published on 08 Jun 2014 - 22:29

score 103 · Answer 13 · 2014-06-08 22:37:52

103

Views

Divisible modules over $\mathbb Z[x]$

Published on 08 Jun 2014 - 22:37

score 1k · Answer 14 · 2026-03-25 22:09:47

1k

Views

Integral domain with a finitely generated non-zero injective module is a field

Published on 25 Mar 2026 - 22:09

score 16 · Answer 15 · 2026-03-11 01:40:28

16

Views

Changing the product of $V$ a $K$-space.

Published on 11 Mar 2026 - 1:40

List Question

Difference of a ring and its module over itself

Quick question: SES where base ring is a PID

Sufficient conditions for quotient ring to be Cohen-Macaulay

Dimension of a module (as vector space)

Direct product of Cohen-Macaulay rings/Eisenbud, Exercise 18.6

Suggest a good book or reference on graded modules over polynomial rings

$M \otimes_R K$ is uniquely divisible as $R$-module

$Q(\mathbb{Z}[t]) / \mathbb{Z}[t]$ is not injective

$\biggl( \prod_{i=1}^{+ \infty} \mathbb{Z}/2^i \mathbb{Z} \biggr)\otimes_{\mathbb{Z}}\mathbb{Q} \neq 0$ without using $p$-adic integers

$2 \otimes_{R} 2 + x \otimes_{R} x$ is not a simple tensor in $I \otimes_R I$

Question about modules

map $\iota: A \rightarrow \mathbb{Q}\otimes_{\mathbb{Z}} A$, where $A$ is a finite abelian group of order n

Divisible modules over $\mathbb Z[x]$

Integral domain with a finitely generated non-zero injective module is a field

Changing the product of $V$ a $K$-space.

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