# nLab GCD ring

Contents

### Context

#### Algebra

higher algebra

universal algebra

# Contents

## Definition

Given a commutative monoid $(M, \cdot, 1)$, we say that element $a \in M$ divides $b \in M$ ($a \vert b$) if there exists an element $c \in M$ such that $a \cdot c = b$ and $c \cdot a = b$. If the commutative monoid has an absorbing element $0$, then for all $a \in M$, $a \vert 0$.

###### Definition

A commutative ring $R$ is a GCD ring if for every element $a \in R$ and $b \in R$, there is an element $c \in R$ such that $c \vert a$ and $c \vert b$, and for every other element $d \in R$ such that $d \vert a$ and $d \vert b$, $d \vert c$.

###### Definition

A commutative ring $R$ is a GCD ring if there is a function $\gcd:R \times R \to R$ such that for every element $a \in R$ and $b \in R$, $\gcd(a, b) \vert a$ and $\gcd(a, b) \vert b$, and for every other function $f:R \times R \to R$ such that $f(a, b) \vert a$ and $f(a, b) \vert b$, $f(a, b) \vert \gcd(a, b)$.

###### Definition

A commutative ring $R$ is a GCD ring if there are functions $\gcd:R \times R \to R$, $q_0:R \times R \to R$, and $q_1:R \times R \to R$ such that for every element $a \in R$ and $b \in R$, $q_0(a, b) \cdot \gcd(a, b) = a$ and $q_1(a, b) \cdot \gcd(a, b) = b$, and for every other triple of functions $f:R \times R \to R$, $r_0:R \times R \to R$, $r_1:R \times R \to R$ such that $r_0(a, b) \cdot f(a, b) = a$ and $r_1(a, b) \cdot f(a, b) = b$, there is a function $s:R \times R \to R$ such that $s(a, b) \cdot f(a, b) = \gcd(a, b)$.

###### Definition

A commutative ring $R$ is a GCD ring if there are functions $\gcd:R \times R \to R$, $q_0:R \times R \to R$, and $q_1:R \times R \to R$ such that for every element $a \in R$ and $b \in R$, $q_0(a, b) \cdot \gcd(a, b) = a$ and $q_1(a, b) \cdot \gcd(a, b) = b$, and $\gcd:R \times R \to R$ is a semilattice with unit element $0$.

The last definition implies that GCD rings are algebraic.