Equivalent conditions on $d$ being a common divisor of $a$ and $b$ in an integral domain: $(a,b) \subseteq (d) \Leftrightarrow d|a$ and $d|b$

Question

Let $R$ be an integral domain. I want to show that $(a,b) \subseteq (d) \iff d|a$ and $d|b$.

The implication $\Leftarrow$ is clear: if $a = dc, b = dr$, then $r_1a + r_2b = r_1cd + r_2rd = (r_1c+r_2r)d \in (d)$.

But I'm not sure what do with $\Rightarrow$ . Well, if $(a,b) \subseteq (d)$, then $\forall r_1, r_2 \in R \ \ \exists r_3 \in R: \ r_1a + r_2b = r_3d$. In particular, $a + b = rd$ for some $r \in R$.

Answer 1

By definition, $(a,b)$ is the smallest ideal containing $a$ and $b$. Thus if $(a,b)\subseteq (d)$, then in particular $a\in(d)$ and $b\in (d)$.

Answer 2

You are right. If you take $r_1=1,r_2=0$, you know that there is some $r_3 \in R$ such that $a=r_3d$, which means that $d$ divides $a$ in $R$.

You can probably find what $r_1,r_2$ you should take to conclude that $d \mid b$.