GCDs in integral domains are unique up to unit factors (associates)

Question

The book I am reading mentioned that the greatest common divisor of $a_1, \dots, a_n$ in an integral domain $R$, whenever it exists, is unique up to arbitrary invertible factors. What does "up to arbitrary invertible factors" mean?

Definition: Let $a_1, \dots, a_n$ be nonzero elements of the ring $R$. An element $d \in R$ is a greatest common divisor of $a_1, \dots, a_n$ if it possesses the properties:

$d \vert a_i$, for $i=1, \dots, n$
$c \vert a_i$, for $i=1, \dots, n$ implies $c \vert d$.

The definition of $\gcd$ varies from a text to another, what is yours? — ajotatxe, May 29 '19 at 15:10
Note that the definition can be restated as $,c\mid d!!!\overset{\rm\ \ def}\iff! c\mid a_1,\ldots ,a_n,,$ which make this (and other) proofs very simple - see my answer. — Bill Dubuque, May 29 '19 at 15:52

Andreas Caranti · Answer 1 · 2019-05-29T15:19:50.347

$\newcommand{\Z}{\mathbb{Z}}$$\renewcommand{\epsilon}{\varepsilon}$It means

if $d$ is a GCD, then for each unit $\epsilon$ we have that $\epsilon d$ is a GCD, and all GCDs have this form.

For instance in $\Z$ the pair $2, 3$ has the two GCDs $1, -1$.

I am taking the following definition. Let $R$ be an integral domain, and $a_{1}, \dots, a_{n} \in R$. An element $d \in R$ is said to be a GCD of $a_{1}, \dots, a_{n}$ if

$d \mid a_{i}$ for all $i$;
for each $e \in R$, if $e \mid a_{i}$ for all $i$, then $e \mid d$.

Of course a GCD need not exist for all choices of $a_{1}, \dots , a_{n}$.

score 1 · Answer 2 · answered May 29 '19 at 15:16

Well, if $d$ is a common divisor of $a,b$ and $e$ is a unit, then $ed$ is a common divisor of $a,b$, since if $a=a'd$ and $b=b'd$ for some $a',b'$, then $a=(a'e^{-1})(ed)$ and $b = (b'e^{-1})(ed)$. This indeed means that the gcd is unique up to invertible elements.

Bill Dubuque · Accepted Answer · 2020-12-06T03:54:33.760

1

By the universal $\rm\color{#c00}{def}$inition of the gcd, If $\,d,d'$ are both gcds of all of the $\,a_i\,$ then $\,c\mid d\!\!\!\overset{\rm\ \ \color{#c00}{def}}\iff\! c\mid a_1,\ldots ,a_n\!\!\!\overset{\rm\ \ \color{#c00}{def}}\iff\! c\mid d',\,$ so for $\,c=d\,$ we get $\,d\mid d'$ and for $\,c = d'\,$ we get $\,d'\mid d,\,$ so $\,d,d'$ divide each other so are associate (thus $\,d' = u d\,$ for some unit $u,\,$ being in a domain).

edited Dec 06 '20 at 03:54

answered May 29 '19 at 15:49

Bill Dubuque

282,220

this is exactly what I was looking for. It does not seem trivial(the link), though the book treated it like it was. thanks! – TheLast Cipher May 29 '19 at 16:04
The second answer, by Timothy Wagner, from the link mentioned that $s$ and $r$ are implied to be units. How so? – TheLast Cipher May 29 '19 at 16:14
@TheLastCipher In a domain: $,\large d\mid d'!\mid d,\Rightarrow, d\overset{\Large ,r}\longrightarrow d'\overset{\Large ,s}\longrightarrow d, $ so $,\large ,rs, d = d,$ so $,\large ,rs = 1,$ by cancelling $\large ,d\neq 0\ \ $ – Bill Dubuque May 29 '19 at 16:15
I have arrive at that same computation independently. What I am asking is how does $rs = 1$ imply $r=1=s$? – TheLast Cipher May 29 '19 at 16:17
$r$ and $s$ are units does mean $r=1=s$, right? – TheLast Cipher May 29 '19 at 16:18
@TheLastCipher No, a unit is an invertible element (vs. the unit $1)$. We are showing, that the gcds are unique up to being associates, i.e. up to a unit factor (the unit $,r,$ or $,s,$ above). – Bill Dubuque May 29 '19 at 16:24
gcds are unique up to being associates mean that a gcd is a 'multiple' of another gcd? – TheLast Cipher May 29 '19 at 16:24
@TheLastCipher By definition, in a domain, $a$ and $b$ are associate if they divide each other, i.e. $, a\mid b\mid a.,$ As we saw above, in a domain this is equivalent to $, a = ub,$ for some unit (invertible) $u,,$ so $,b = u^{-1} a,,$ i.e. they are both unit multiples of each other. – Bill Dubuque May 29 '19 at 16:28
I got it now. Thanks for claifying the definition. I unwittingly misunderstood the definition of unit. – TheLast Cipher May 29 '19 at 16:30
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@TheLastCipher For motivation see also posts on unit normalization of gcds. – Bill Dubuque May 29 '19 at 16:31
The link seems broken. – TheLast Cipher May 29 '19 at 16:32
@TheLastCipher Oops, should be this (had to change "me" to my userid = 242). – Bill Dubuque May 29 '19 at 17:11

GCDs in integral domains are unique up to unit factors (associates)

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