Prove that in ring $\mathbb{Z}, \forall a,b \in \mathbb{Z} $, $(a,b) = (d)$, where $ d = \mathrm{gcd}(a,b)$

Question

Prove that in ring $\mathbb{Z}, \forall a,b \in \mathbb{Z} $, $$(a,b) = (d),$$ where $ d = \mathrm{gcd} (a,b)$. Also, generalize this for all finite generated ideals in $\mathbb{Z}$.

I don't even know how to start. I am not sure if I got the definition of generated ideal. When I have $(a)$, what does that mean?

Answer 1

Hint:

• $x\in(a,b)\iff$ integers $n,m$ exist with $na+mb=x$
• $x\in(d)\iff$ integer $k$ exists with $kd=x$