This is proposition 12.2.3 from Ireland and Rosen's "A Classical Introduction to Modern Number Theory". $D$ is the ring of algebraic integers of some algebraic number field, and $A$ is an ideal in $D$. How is it deduced in the 3rd from last line that $\gamma_i - \gamma_i'$ is divisible by a? I tried writing explicitly what it means for $\sum \gamma_i \omega_i, \sum \gamma_i' \omega_i$ to be equivalent modulo a but I don't know what to do next.
If $\sum \gamma_i \omega_i$ and $\sum \gamma_i' \omega_i$ are equivalent modulo $(a)$, $\sum (\gamma_i - \gamma_i')\omega_i = ad$, $d \in D$. Since $\omega_1, \dots, \omega_n$ is an integral basis for $D$, $ad = am_1\omega_1 + \dots + am_n\omega_n$, $m_1, \dots m_n \in \mathbb{Z}$. Thus, $\sum (\gamma_i - \gamma_i')\omega_i = ad$ implies that
$$(\gamma_1 - \gamma_1' - am_1)\omega_1 + \dots + (\gamma_n - \gamma_n' - am_n)\omega_n = 0.$$
By linear independence of the $\omega_i$, $(\gamma_i - \gamma_1' - am_1) = \dots = (\gamma_n - \gamma_n' - am_n) = 0$. So $\gamma_i - \gamma_i' = am$ for $1 \leq i \leq n$, i.e., $a \mid \gamma_i - \gamma_i'$.