Question

Verify my solution - simple right submodule of Weyl algebra is inifnite dimensional as complex vector space.

score 122 · Answer 1 · 2019-11-08 20:33:34

122

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Verify my solution - simple right submodule of Weyl algebra is inifnite dimensional as complex vector space.

Published on 08 Nov 2019 - 20:33

score 130 · Answer 2 · 2019-11-09 14:43:14

130

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Ideal of upper triangular matrices over non noetherian ring.

Published on 09 Nov 2019 - 14:43

score 50 · Answer 3 · 2019-11-10 13:47:15

50

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Primary decomposition in Noetherian rings

Published on 10 Nov 2019 - 13:47

score 94 · Answer 4 · 2019-11-13 11:35:00

94

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If $x \in \cap_{n=1}^\infty I^n$, then $x \in xI$.

Published on 13 Nov 2019 - 11:35

score 75 · Answer 5 · 2019-11-16 16:54:04

75

Views

Dividing a ring by the annihilator of a Noetherian module yields a Noetherian ring

Published on 16 Nov 2019 - 16:54

score 87 · Answer 6 · 2019-11-19 20:52:36

87

Views

I look for 2 commutative algebra counterexamples

Published on 19 Nov 2019 - 20:52

score 103 · Answer 7 · 2019-11-20 18:17:23

103

Views

Is noetherian ring finite over subring of invariants of a finite group?

Published on 20 Nov 2019 - 18:17

score 84 · Answer 8 · 2019-11-21 14:41:32

84

Views

Finite ring extensions doesn't preserve non-zero-divisors

Published on 21 Nov 2019 - 14:41

score 217 · Answer 9 · 2019-11-25 16:47:55

217

Views

Prove that $\mathrm{ht}(P/Ra)=\mathrm{ht}(P) -1$

Published on 25 Nov 2019 - 16:47

score 29 · Answer 10 · 2019-11-29 05:32:18

29

Views

Ideals the same after localization by associated primes

Published on 29 Nov 2019 - 5:32

score 135 · Answer 11 · 2019-12-06 22:23:49

135

Views

$\operatorname{Spec}(A)$ noetherian implies a.c.c. on prime ideals of $A$. The converse is true?

Published on 06 Dec 2019 - 22:23

score 130 · Answer 12 · 2019-12-13 23:36:02

130

Views

Subring of $\mathbb{Z}[x,y]$ generated by $\{yx^i | i\geq 1\}$ is not Noetherian

Published on 13 Dec 2019 - 23:36

score 168 · Answer 13 · 2019-12-17 08:39:13

168

Views

Faithful $S$-Noetherian module

Published on 17 Dec 2019 - 8:39

score 101 · Answer 14 · 2026-03-26 02:53:41

101

Views

Example of $S$-Noetherian rings

Published on 26 Mar 2026 - 2:53

score 159 · Answer 15 · 2019-12-27 19:20:56

159

Views

Let $A$ be a Noetherian domain of dimension 1 with infinitely many maximal ideals. Let $f \in A[y]$. Show that $(f)$ is not a maximal ideal of $A[y]$.

Published on 27 Dec 2019 - 19:20

List Question

Verify my solution - simple right submodule of Weyl algebra is inifnite dimensional as complex vector space.

Ideal of upper triangular matrices over non noetherian ring.

Primary decomposition in Noetherian rings

If $x \in \cap_{n=1}^\infty I^n$, then $x \in xI$.

Dividing a ring by the annihilator of a Noetherian module yields a Noetherian ring

I look for 2 commutative algebra counterexamples

Is noetherian ring finite over subring of invariants of a finite group?

Finite ring extensions doesn't preserve non-zero-divisors

Prove that $\mathrm{ht}(P/Ra)=\mathrm{ht}(P) -1$

Ideals the same after localization by associated primes

$\operatorname{Spec}(A)$ noetherian implies a.c.c. on prime ideals of $A$. The converse is true?

Subring of $\mathbb{Z}[x,y]$ generated by $\{yx^i | i\geq 1\}$ is not Noetherian

Faithful $S$-Noetherian module

Example of $S$-Noetherian rings

Let $A$ be a Noetherian domain of dimension 1 with infinitely many maximal ideals. Let $f \in A[y]$. Show that $(f)$ is not a maximal ideal of $A[y]$.

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