Greatest Common Divisor in a ring definition

We were given this defintion in our Algebra $$2$$ course:

Definition $$4.1$$: Let $$R$$ be a commutative ring and $$a, b\in R$$. An element $$d \in R$$ is called a greatest common divisor, $$\operatorname{gcd}(a, b)$$, of $$a$$ and $$b$$ if:
$$d$$ divides both $$a$$ and $$b$$ $$($$i.e., $$\exists x$$, $$y \in R$$ such that $$a = dx$$, $$b = dy)$$,
• If $$e \in R$$ divides both $$a$$ and $$b$$, then $$e$$ divides $$d$$.

I don't really understand how the second point could work with say the integers, for example $$\operatorname{gcd}(10,12)$$, if you pick $$2$$ to be the first $$\operatorname{gcd}$$, then for $$e=1$$ you have that it does divide both $$10$$ and $$12$$, and it does divide $$2$$. But then $$2$$ doesn't divide $$1$$ so $$2$$ is no longer a gcd according to this surely?

• So then $\gcd(10, 12) \ne 1$. That seems fine to me! – Robert Lewis Nov 4 '18 at 19:02
• It's not $2$ which has to divide $1$, according to this definition, but the reverse. – Bernard Nov 4 '18 at 19:06
• Surely they'd both have to divide each other? – Andy010101 Nov 4 '18 at 19:23
• Equivalently, $\, c\mid a,b \iff c\mid \gcd(a,b).\$ This means the gcd is "greatest" w.r.t. divisibility partial order. – Bill Dubuque Nov 4 '18 at 19:44
• In your example $d\mid 10,12\iff d\mid 2\$ so $\,\gcd(10,12) = 2\$ – Bill Dubuque Nov 12 '18 at 3:28