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

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.

• What is your definition of "relatively prime"? Oct 28 '19 at 19:55
• 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 $a$. Oct 28 '19 at 20:03

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

• 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]"? 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 Oct 28 '19 at 23:15
• @SimonFraser The definition of gcd doesn't vary but "relatively prime" may. Oct 28 '19 at 23:22
• @Simon No, it is correct as written, see here. Oct 29 '19 at 17:28
• @BillDubuqueI completely misunderstood. I was thinking of something completely illogical. You are correct. Oct 29 '19 at 18:06