Question

$D$ be a UFD having infinitely many maximal ideals , then does $D$ have infinitely many irreducible elements which are pairwise non-associate?

score 295 · Answer 1 · 2026-03-26 11:00:01

295

Views

$D$ be a UFD having infinitely many maximal ideals , then does $D$ have infinitely many irreducible elements which are pairwise non-associate?

Published on 26 Mar 2026 - 11:00

score 120 · Answer 2 · 2016-09-03 11:24:45

120

Views

Relations between multisets of irreducible factors when one element divides another in a UFD

Published on 03 Sep 2016 - 11:24

score 415 · Answer 3 · 2016-09-04 23:51:08

415

Views

Can we specifically prove that if $n\in\mathbb{Z^+}$ is composite and square-free then the ring of integers of $\mathbb{Q}(\sqrt{-n})$ is not a UFD?

Published on 04 Sep 2016 - 23:51

score 154 · Answer 4 · 2016-09-07 01:08:35

154

Views

Proof help involving prime factorization, exponents

Published on 07 Sep 2016 - 1:08

score 275 · Answer 5 · 2026-03-27 03:43:14

275

Views

Prove $f(z)=p(z)/q(z)=\alpha$ has n solutions

Published on 27 Mar 2026 - 3:43

score 255 · Answer 6 · 2016-09-12 00:10:46

255

Views

$(f) = (c)(g)$, where $R$ is a UFD, $c \in R$, $f,g \in R[x]$, and $g$ is primitive

Published on 12 Sep 2016 - 0:10

score 155 · Answer 7 · 2016-09-19 00:08:26

155

Views

A nuance in the definition of UFD

Published on 19 Sep 2016 - 0:08

score 88 · Answer 8 · 2016-09-25 12:49:09

88

Views

Irreducble factors of $a$ are some of the irreducible factors of $b$ if $a$ divides $b$ in a UFD

Published on 25 Sep 2016 - 12:49

score 76 · Answer 9 · 2016-10-03 02:30:36

76

Views

What is property of ring of factorization domain without uniqueness?

Published on 03 Oct 2016 - 2:30

score 865 · Answer 10 · 2026-03-26 08:14:54

865

Views

$k[[X_1,\ldots,X_n]]$ is UFD for $k$ field

Published on 26 Mar 2026 - 8:14

score 236 · Answer 11 · 2016-10-05 04:39:58

236

Views

The basic facts about $\mathbb{R}[x,y].$

Published on 05 Oct 2016 - 4:39

score 344 · Answer 12 · 2016-10-25 03:26:29

344

Views

Irreducible polynomial with coeficients in an UFD is primitive

Published on 25 Oct 2016 - 3:26

score 187 · Answer 13 · 2016-10-31 21:42:02

187

Views

Examples of a UFDs without factoring algorithms?

Published on 31 Oct 2016 - 21:42

score 539 · Answer 14 · 2016-11-01 01:35:15

539

Views

Finding all invertible elements of $S^{-1}D$, $D$ is a Noetherian integral domain

Published on 01 Nov 2016 - 1:35

score 306 · Answer 15 · 2016-11-01 21:02:39

306

Views

When irreducible elements of a UFD remain irreducible in a ring extension

Published on 01 Nov 2016 - 21:02

List Question

$D$ be a UFD having infinitely many maximal ideals , then does $D$ have infinitely many irreducible elements which are pairwise non-associate?

Relations between multisets of irreducible factors when one element divides another in a UFD

Can we specifically prove that if $n\in\mathbb{Z^+}$ is composite and square-free then the ring of integers of $\mathbb{Q}(\sqrt{-n})$ is not a UFD?

Proof help involving prime factorization, exponents

Prove $f(z)=p(z)/q(z)=\alpha$ has n solutions

$(f) = (c)(g)$, where $R$ is a UFD, $c \in R$, $f,g \in R[x]$, and $g$ is primitive

A nuance in the definition of UFD

Irreducble factors of $a$ are some of the irreducible factors of $b$ if $a$ divides $b$ in a UFD

What is property of ring of factorization domain without uniqueness?

$k[[X_1,\ldots,X_n]]$ is UFD for $k$ field

The basic facts about $\mathbb{R}[x,y].$

Irreducible polynomial with coeficients in an UFD is primitive

Examples of a UFDs without factoring algorithms?

Finding all invertible elements of $S^{-1}D$, $D$ is a Noetherian integral domain

When irreducible elements of a UFD remain irreducible in a ring extension

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