## You are here

Homean integral domain is lcm iff it is gcd

## Primary tabs

# an integral domain is lcm iff it is gcd

###### Proposition 1.

Let $D$ be an integral domain. Then $D$ is a lcm domain iff it is a gcd domain.

This is an immediate consequence of the following

###### Proposition 2.

Let $D$ be an integral domain and $a,b\in D$. Then the following are equivalent:

1. $a,b$ have an lcm,

2. for any $r\in D$, $ra,rb$ have a gcd.

###### Proof.

For arbitrary $x,y\in D$, denote $\operatorname{LCM}(x,y)$ and $\operatorname{GCD}(x,y)$ the sets of all lcm’s and all gcd’s of $x$ and $y$, respectively.

$(1\Rightarrow 2)$. Let $c\in\operatorname{LCM}(a,b)$. Then $c=ax=by$, for some $x,y\in D$. For any $r\in D$, since $rab$ is a multiple of $a$ and $b$, there is a $d\in D$ such that $rab=cd$. We claim that $d\in\operatorname{GCD}(ra,rb)$. There are two steps: showing that $d$ is a common divisor of $ra$ and $rb$, and that any common divisor of $ra$ and $rb$ is a divisor of $d$.

1. 2. Next, let $t$ be any common divisor of $ra$ and $rb$, say $ra=ut$ and $rb=vt$ for some $u,v\in D$. Then $uvt=rav=rbu$, so that $z:=av=bu$ is a multiple of both $a$ and $b$, and hence is a multiple of $c$, say $z=cw$ for some $w\in D$. Then the equation $axw=cw=z=av$ reduces to $xw=v$. Multiplying both sides by $t$ gives $xwt=vt$. Since $vt=rb=xd$, we have $xd=xwt$, or $d=wt$, so that $d$ is a multiple of $t$.

As a result, $d\in GCD(ra,rb)$.

$(2\Rightarrow 1)$. Suppose $k\in\operatorname{GCD}(a,b)$. Write $ki=a$, $kj=b$ for some $i,j\in D$. Set $\ell=kij$, so that $ab=k\ell$. We want to show that $\ell\in\operatorname{LCM}(a,b)$. First, notice that $\ell=aj=bi$, so that $a\mid\ell$ and $b\mid\ell$. Now, suppose $a\mid t$ and $b\mid t$, we want to show that $\ell\mid t$ as well. Write $t=ax=by$. Then $ta=aby$ and $tb=abx$, so that $ab\mid ta$ and $ab\mid tb$. Since $\operatorname{GCD}(ta,tb)\neq\varnothing$, we have $tk\in\operatorname{GCD}(ta,tb)$ (see proof of this here), implying $ab\mid tk$. In other words $tk=abz$ for some $z\in D$. As a result, $tk=abz=k\ell z$, or $t=\ell z$. In other words, $\ell\mid t$, as desired. ∎

Since the first statement is equivalent to $D$ being an lcm domain, and the second statement is equivalent to $D$ being a gcd domain, Proposition 1 follows.

Another way of stating Proposition 1 is the following: let $L$ be the set of equivalence classes on the integral domain $D$, where $a\sim b$ iff $a$ and $b$ are associates. Partial order $L$ so that $[a]\leq[b]$ iff $ac=b$ for some $c\in D$. Then $L$ is a semilattice (upper or lower) implies that $L$ is a lattice.

## Mathematics Subject Classification

13G05*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections