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Q&A The gcd of powers in a gcd domain

0 answers  ·  posted 1y ago by Snoopy‭  ·  edited 1y ago by Snoopy‭

Question abstract-algebra
#2: Post edited by user avatar Snoopy‭ · 2022-07-25T14:30:34Z (over 1 year 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 user avatar Snoopy‭ · 2022-07-22T18:12:11Z (over 1 year 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.