Question

Simple modules of a finite dimensional commutative $\mathbb{R}$-algebra

score 1k · Answer 1 · 2017-06-05 20:45:39

1k

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Simple modules of a finite dimensional commutative $\mathbb{R}$-algebra

Published on 05 Jun 2017 - 20:45

score 65 · Answer 2 · 2017-06-07 18:37:19

65

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is there anything special about this particular algebra?

Published on 07 Jun 2017 - 18:37

score 426 · Answer 3 · 2017-06-11 22:39:51

426

Views

Finitely generated $\mathbb{Z}$-algebra that is an integral domain has trivial Jacobson radical

Published on 11 Jun 2017 - 22:39

score 347 · Answer 4 · 2017-06-12 23:27:03

347

Views

Motivation to introduce tensor products

Published on 12 Jun 2017 - 23:27

score 67 · Answer 5 · 2017-06-18 17:17:28

67

Views

If $e_1, e_2$ are commuting idempotents whose difference lies in the Jacobson radical then $e_1 = e_2$

Published on 18 Jun 2017 - 17:17

score 417 · Answer 6 · 2017-07-07 06:23:16

417

Views

A Künneth Theorem for Hochschschild Homology

Published on 07 Jul 2017 - 6:23

score 106 · Answer 7 · 2017-07-12 08:08:33

106

Views

A question on locally finite fields

Published on 12 Jul 2017 - 8:08

score 251 · Answer 8 · 2017-07-12 19:27:11

251

Views

Real algebras of dimension two

Published on 12 Jul 2017 - 19:27

score 772 · Answer 9 · 2017-08-09 13:43:21

772

Views

What are the units of a tensor product of rings?

Published on 09 Aug 2017 - 13:43

score 65 · Answer 10 · 2017-08-14 04:42:39

65

Views

$A\otimes\bar{k}\simeq\bar{k}\times\cdots\times\bar{k}$, then the number of $k$-algebra maps $A\to\bar{k}$ is the dimension of $A$?

Published on 14 Aug 2017 - 4:42

score 159 · Answer 11 · 2017-08-14 15:06:57

159

Views

Dualizing simple module which is given by primitive idempotent

Published on 14 Aug 2017 - 15:06

score 64 · Answer 12 · 2017-08-14 18:43:22

64

Views

Let $\beta$ a $\Lambda$-homomorphism is nilpotent then $\operatorname{Ext}^k(\beta , M)$ is nilpotent, for all $M$ $\in$ $\operatorname{mod}(\Lambda)$

Published on 14 Aug 2017 - 18:43

score 737 · Answer 13 · 2017-09-02 20:01:14

737

Views

Algebra over vector space

Published on 02 Sep 2017 - 20:01

score 89 · Answer 14 · 2017-09-07 12:00:35

89

Views

Finite basis for the polynomial identities of two-step solvable Lie algebras

Published on 07 Sep 2017 - 12:00

score 577 · Answer 15 · 2017-10-06 22:16:49

577

Views

The generators in the definition of Grassmann Algebra

Published on 06 Oct 2017 - 22:16

List Question

Simple modules of a finite dimensional commutative $\mathbb{R}$-algebra

is there anything special about this particular algebra?

Finitely generated $\mathbb{Z}$-algebra that is an integral domain has trivial Jacobson radical

Motivation to introduce tensor products

If $e_1, e_2$ are commuting idempotents whose difference lies in the Jacobson radical then $e_1 = e_2$

A Künneth Theorem for Hochschschild Homology

A question on locally finite fields

Real algebras of dimension two

What are the units of a tensor product of rings?

$A\otimes\bar{k}\simeq\bar{k}\times\cdots\times\bar{k}$, then the number of $k$-algebra maps $A\to\bar{k}$ is the dimension of $A$?

Dualizing simple module which is given by primitive idempotent

Let $\beta$ a $\Lambda$-homomorphism is nilpotent then $\operatorname{Ext}^k(\beta , M)$ is nilpotent, for all $M$ $\in$ $\operatorname{mod}(\Lambda)$

Algebra over vector space

Finite basis for the polynomial identities of two-step solvable Lie algebras

The generators in the definition of Grassmann Algebra

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