List questions from principal-ideal-domains

Published on 08 Apr 2015 - 18:20

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gcd in principal ideal domain

Published on 09 Apr 2015 - 0:14

168

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Is this module injective?

Published on 12 Apr 2015 - 10:40

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$R/Ra$ is an injective module over itself

Published on 26 Mar 2026 - 22:17

132

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Let $R$ be a PID and $I$ a prime ideal of $R$ s.t. $0 \subset I \subset 1_R$

Published on 19 Apr 2015 - 18:17

310

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How to classify the rings of fractions of a principal ideal domain?

Published on 19 Apr 2015 - 20:48

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A UFD is a PID if it has dimension $\le 1$ (nonzero prime ideals are maximal)

Published on 26 Mar 2026 - 3:11

739

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Example of a domain where all irreducibles are primes and that is not a GCD domain

Published on 29 Mar 2026 - 7:40

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Let $q$ be a prime congruent to 3 mod 4, prove the quotient ring $\mathbb{Z}[i]/(q)$ is a field with $q^2$ elements

Published on 27 Mar 2026 - 18:27

1k

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Is torsion-free equivalent to free for non-finitely generated modules over a PID?

Published on 23 Apr 2015 - 3:52

1.6k

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Why is a submodule of a free module over a PID is free?

Published on 25 Apr 2015 - 23:04

412

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Show that $S$ is a field

Published on 27 Mar 2026 - 4:15

191

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Prove that $a$ is a prime element of $R$

Published on 27 Apr 2015 - 21:53

105

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If prime p doesn't divide the class number, then if I is an ideal of $O_K$, and $I ^{p}$ is principal, then I is principal

Published on 28 Apr 2015 - 8:55

1.5k

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Intersection of ideals generated by two relatively prime elements

Published on 29 Apr 2015 - 12:14

List Question

Show that $R[x,y]$ is not a Principal Ideal Domain

gcd in principal ideal domain

Is this module injective?

$R/Ra$ is an injective module over itself

Let $R$ be a PID and $I$ a prime ideal of $R$ s.t. $0 \subset I \subset 1_R$

How to classify the rings of fractions of a principal ideal domain?

A UFD is a PID if it has dimension $\le 1$ (nonzero prime ideals are maximal)

Example of a domain where all irreducibles are primes and that is not a GCD domain

Let $q$ be a prime congruent to 3 mod 4, prove the quotient ring $\mathbb{Z}[i]/(q)$ is a field with $q^2$ elements

Is torsion-free equivalent to free for non-finitely generated modules over a PID?

Why is a submodule of a free module over a PID is free?

Show that $S$ is a field

Prove that $a$ is a prime element of $R$

If prime p doesn't divide the class number, then if I is an ideal of $O_K$, and $I ^{p}$ is principal, then I is principal

Intersection of ideals generated by two relatively prime elements

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