Question

$R$ integral domain where every prime ideal is principal then set of non-principal ideals has maximal element

score 748 · Answer 1 · 2017-03-24 17:20:25

748

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$R$ integral domain where every prime ideal is principal then set of non-principal ideals has maximal element

Published on 24 Mar 2017 - 17:20

score 143 · Answer 2 · 2026-03-26 17:46:12

143

Views

Prove that every non-zero ideal of ID is maximal.

Published on 26 Mar 2026 - 17:46

score 186 · Answer 3 · 2026-03-25 04:40:23

186

Views

Show that local rings on a curve are PID

Published on 25 Mar 2026 - 4:40

score 105 · Answer 4 · 2023-02-10 12:20:12

105

Views

Finitely-generated submodules of a finitely generated module over a PID have a complement?

Published on 10 Feb 2023 - 12:20

score 78 · Answer 5 · 2023-02-28 18:08:51

78

Views

The Ring of Locally Holomorphic Functions around a Fixed Point in $\mathbb{C}$ is a PID

Published on 28 Feb 2023 - 18:08

score 81 · Answer 6 · 2023-03-05 01:27:03

81

Views

prime ideals in non-integral-domain rings in which every proper ideal can be generated by a single element

Published on 05 Mar 2023 - 1:27

score 51 · Answer 7 · 2023-03-24 18:14:51

51

Views

Proof that these ideals in an integral closure are principal

Published on 24 Mar 2023 - 18:14

score 68 · Answer 8 · 2026-03-25 14:47:03

68

Views

Generator of ideal in $Q(\zeta_5)$

Published on 25 Mar 2026 - 14:47

score 26 · Answer 9 · 2023-04-30 20:54:21

26

Views

Showing an ideal $(x_1, ... ,x_n)$ of a PID $R$ contains 1 using a submodule of $R^n$

Published on 30 Apr 2023 - 20:54

score 82 · Answer 10 · 2026-03-26 12:36:57

82

Views

Let $\psi:\mathbb Z[x] \rightarrow \mathbb R$ be the homomorphism defined by $\psi(p(x))=p(\sqrt3)$.

Published on 26 Mar 2026 - 12:36

score 425 · Answer 11 · 2023-05-09 16:04:48

425

Views

If each subring of the fraction field of an integral domain $\mathcal{R}$ (and properly containing $\mathcal{R}$) is a PID, is $\mathcal{R}$ a PID?

Published on 09 May 2023 - 16:04

score 79 · Answer 12 · 2026-03-28 15:27:00

79

Views

Does sub-module of free module over PID is free require commutativity?

Published on 28 Mar 2026 - 15:27

score 38 · Answer 13 · 2026-03-28 15:18:43

38

Views

Finitely-Generated modules over a commutative ring with $1$.

Published on 28 Mar 2026 - 15:18

score 44 · Answer 14 · 2023-06-09 12:47:54

44

Views

Prove that if $\text{gcd}(v, w) \simeq av + bw$, then $\text{gcd}(a, b) \simeq 1$.

Published on 09 Jun 2023 - 12:47

score 52 · Answer 15 · 2026-03-25 08:51:44

52

Views

Prove that every ideal of $\Bbb{Z}$ has the form $m\Bbb{Z}=\{mk, k\in \Bbb{Z}\}$

Published on 25 Mar 2026 - 8:51

List Question

$R$ integral domain where every prime ideal is principal then set of non-principal ideals has maximal element

Prove that every non-zero ideal of ID is maximal.

Show that local rings on a curve are PID

Finitely-generated submodules of a finitely generated module over a PID have a complement?

The Ring of Locally Holomorphic Functions around a Fixed Point in $\mathbb{C}$ is a PID

prime ideals in non-integral-domain rings in which every proper ideal can be generated by a single element

Proof that these ideals in an integral closure are principal

Generator of ideal in $Q(\zeta_5)$

Showing an ideal $(x_1, ... ,x_n)$ of a PID $R$ contains 1 using a submodule of $R^n$

Let $\psi:\mathbb Z[x] \rightarrow \mathbb R$ be the homomorphism defined by $\psi(p(x))=p(\sqrt3)$.

If each subring of the fraction field of an integral domain $\mathcal{R}$ (and properly containing $\mathcal{R}$) is a PID, is $\mathcal{R}$ a PID?

Does sub-module of free module over PID is free require commutativity?

Finitely-Generated modules over a commutative ring with $1$.

Prove that if $\text{gcd}(v, w) \simeq av + bw$, then $\text{gcd}(a, b) \simeq 1$.

Prove that every ideal of $\Bbb{Z}$ has the form $m\Bbb{Z}=\{mk, k\in \Bbb{Z}\}$

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