Let $A$ be an ideal of $\mathcal O$ (Ring of integers of some algeibraic number field) and assume that $gcd([\gamma],A)= [1]$. How to show that if $\gamma \alpha=0$ (mod $A$) then $\alpha = 0$ (mod $A$). I know that the assumption $gcd([\gamma],A)= [1]$ means that $[\gamma]+A= \mathcal O$ and that $\gamma \alpha = 0$ (mod $A$) means that $\gamma \alpha \in A$. So the question now becomas: Show that if $[\gamma]+A= \mathcal O$ and $\gamma \alpha \in A$ then $\alpha \in A$.
I'm pretty new to ideals and everything I try doesn't make sense really.
Another little question: When we define that $A|B$ means that $AC=B$ for some ideals $A,B,C$ and $a = b$ (mod $A$) s.t. $a-b \in A$. Can we say that $a=b$ (mod $A$) also means that $A|a-b$ and then $AC=a-b$ for some ideal C?
Suppose that $\gamma\alpha=0\mod A$. Then, $\gamma\alpha\in A$, or, said differently, $(\gamma)(\alpha)\subseteq A$. But, this is equivalent to $A\mid (\gamma)(\alpha)$. But, since $((\gamma),A)=1$ this implies that $A\mid(\alpha)$ and so $(\alpha)\subseteq A$. But, this is equivalent to $\alpha\equiv 0\mod A$.