Post History
#2: Post edited
by
Snoopy
·
2022-07-25T14:30:34Z (over 2 years ago)
- **Question:**
$\def\gcd{\operatorname{gcd}}$Let $R$ be a [gcd domain](https://en.wikipedia.org/wiki/GCD_domain). Does it always holds that $\gcd(x^m,y^m)=\gcd(x,y)^m$?- ---
- **Context.**
- If the ring is a [Bezout domain](https://en.wikipedia.org/wiki/B%C3%A9zout_domain), then we can apply [this method](https://math.stackexchange.com/questions/10400/comaximal-ideals-in-a-commutative-ring). However, a gcd domain may not be a Bezout domain. I don't know how I can go on.
- **Question:**
- $\def\gcd{\operatorname{gcd}}$Let $R$ be a [gcd domain](https://en.wikipedia.org/wiki/GCD_domain). Does it always hold that $\gcd(x^m,y^m)=\gcd(x,y)^m$?
- ---
- **Context.**
- If the ring is a [Bezout domain](https://en.wikipedia.org/wiki/B%C3%A9zout_domain), then we can apply [this method](https://math.stackexchange.com/questions/10400/comaximal-ideals-in-a-commutative-ring). However, a gcd domain may not be a Bezout domain. I don't know how I can go on.
#1: Initial revision
by
Snoopy
·
2022-07-22T18:12:11Z (over 2 years ago)
The gcd of powers in a gcd domain
**Question:**
$\def\gcd{\operatorname{gcd}}$Let $R$ be a [gcd domain](https://en.wikipedia.org/wiki/GCD_domain). Does it always holds that $\gcd(x^m,y^m)=\gcd(x,y)^m$?
---
**Context.**
If the ring is a [Bezout domain](https://en.wikipedia.org/wiki/B%C3%A9zout_domain), then we can apply [this method](https://math.stackexchange.com/questions/10400/comaximal-ideals-in-a-commutative-ring). However, a gcd domain may not be a Bezout domain. I don't know how I can go on.