Is there another meaning of this notation?

Question

In a book I found the following notation:

Let $c,d\in \mathbb{Z}$ such that $c\mathbb{Z}+d\mathbb{Z}=\mathbb{Z}$.

For me, this means that $\gcd(c,d)=1$. If $\gcd(c,d)=1$, then there is $z,u\in \mathbb{Z}$ such that $cz+du=1$. Then any integer $b\in\mathbb{Z}$ can be written as $$b=b(cz+du)=czb+dub \in c\mathbb{Z}+d\mathbb{Z}.$$

My question: is the author means something more or just $\gcd(c,d)=1$.

Thanks

Answer 1

You are correct. $\ c\,\Bbb Z + d\,\Bbb Z = b\,\Bbb Z\iff \gcd(c,d) = \pm b.\,$ This follows essentially from Bezout's identity for the gcd - see this answer. The former is a natural ideal-theoretic way to write it in Bezout/PIDs like $\,\Bbb Z,\,$ which may explain the author's choice of formulation.

Answer 2

More or less: you've shown that if $\gcd(c,d) = 1$, then $c\mathbb Z + d \mathbb Z = \mathbb Z$, but you should also check that if $c\mathbb Z + d \mathbb Z = \mathbb Z$, then $\gcd(c,d) = 1$. This follows by the same logic, since then in particular $1 = cz+du$ for suitable $z$ and $u$, which on the other hand means that $1 = \gcd(c,d)$.

Answer 3

In the ring $(\Bbb Z,+,\times)$ the ideals have the form $d\Bbb Z$ for $d\ge0$ and $c\Bbb Z+d\Bbb Z$ is an ideal generated by the set $\{c,d\}$ and we prove that $$c\Bbb Z+d\Bbb Z=\gcd(c,d)\Bbb Z$$