Inclusion of quotient ideals. if $I \subseteq J$, then $I/qI \subseteq J/qJ$?

Question

Let $\mathbb{K}$ be an algebraic number field.

Let $I$ and $J$ be two fractional ideals of $\mathbb{K}$ and $q \in \mathbb{N}$ a positive integer.

Is it true that if $I \subseteq J$, then $I/qI \subseteq J/qJ$?

Answer 1

Let $\mathbb{K}=\mathbb{Q}$, and $q=5$.

Take $I=4\mathbb{Z}$, and $J=2\mathbb{Z}$; hence $I \subseteq J$. But

$I/5I= 4\mathbb{Z}/20\mathbb{Z}=\{ 0,4,8,12,16\}$, and

$J/5J= 2\mathbb{Z}/10\mathbb{Z}=\{ 0,2,4,6,8\}$. Clearly we do not have an inclusion in any sens.

Question: can we find always a bijection?

Answer 2

Note that from $I\subseteq J$ follows $qI\subseteq qJ$ hence we have a group homomorphism $I/qI\to J/qJ$. This is injective if and only if $qI=I\cap qJ$ and, in that case, you can identify $I/qI$ with a subgroup of $J/qJ$.