Proof: In a commutative monoid $M$ with cancellation laws, if $a,b\in M$ are relatively prime then $\gcd(a,b)=\mathcal{U}(M)$.

Question

Let $M$ be a commutative monoid for which the cancellation laws hold. Given $a,b\in M$, show that if $a$ and $b$ are relatively prime then $\gcd(a,b)=\mathcal{U}(M)$, where $\mathcal{U}(M)$ is the group of units of the monoid $M$.

I find the analogous proof for integers and integral domains everywhere, but I need to prove this for monoids. Thanks in advance.

For $a,b\in M$, $a$ is relatively prime to $b$ if for every $x\in M$ such that $a\mid bx$ then $a\mid x$. In the statement that I gave, $a$ is relatively prime to $b$ and $b$ to $a$. — Nuno R. Peres, Oct 28 '19 at 20:03

Bill Dubuque · Accepted Answer · 2019-10-28T22:32:56.357

-1

Hint $\,\ c\mid a,b\,\Rightarrow\, a\mid b(a/c)\,\Rightarrow\, a\mid a/c\,\Rightarrow\, c\mid 1$

Remark $ $ There are various notions of $\,a,b\,$ are "relatively prime / coprime" in rings, e.g. below excerpted from the linked post. You might find it instructive to investigate their relationships also.

[0] $\ \ \ x\mid a,b\,\Rightarrow\, x\mid 1$

[a] $\ \ \ a\mid bx \,\Rightarrow\, a\mid x $

[b] $\ \ \ a,b\mid x \,\Rightarrow\, ab\mid x $

[c] $\ \ \ (a)\cap (b) = (ab)$

edited Oct 28 '19 at 22:32

answered Oct 28 '19 at 20:11

Bill Dubuque

272,048

What does the notation $(a)\cap (b)=(ab)$ mean? sorry I'm sorta new to abstract algebra. Also, what did you mean when you wrote "[0], [a],[b], and [c]"? – Simon Fraser Oct 28 '19 at 23:11
@SimonFraser $\ (a),$ denotes the ideal $aR$ in a commutative ring $R$, i.e. all multiples of $a$, The brackets are just labels (consistent with the linked post). S – Bill Dubuque Oct 28 '19 at 23:15
@SimonFraser The definition of gcd doesn't vary but "relatively prime" may. – Bill Dubuque Oct 28 '19 at 23:22
@Simon No, it is correct as written, see here. – Bill Dubuque Oct 29 '19 at 17:28
@BillDubuqueI completely misunderstood. I was thinking of something completely illogical. You are correct. – Simon Fraser Oct 29 '19 at 18:06

Proof: In a commutative monoid $M$ with cancellation laws, if $a,b\in M$ are relatively prime then $\gcd(a,b)=\mathcal{U}(M)$.

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