My notes on Euclidean domains state a notion of ring of remainders that I should connect with the quotient by certain ideal but I don't see quite how the connection is done.
Given an euclidean domain $D$ with unique quotient and remainder defined and $m \in D \setminus \{0\}$, we define the ring of remainders $D_m$ as the set $\{R_m(a):a \in D\}$ together with operations $a+b = R_m(a+b)$ and $ab = R_m(ab)$.
Where $R_m$ is the function that gives the unique euclidean remainder of the division by $m$. This is done having in mind $K[X],\mathbb{Z}_n$. Then, the notes continue stating that $R_m:D \to D_m$ is an homomorphism and therefore $D/mD \cong D_m$.
However, I'm missing properties for $R_m$ to be a ring homomorphism. For instance, $R_m(1)$ does not need to be $1$ as in $\mathbb{Z}$, $(-1)(-1) = 1$ and $R_m(1) = 0$ (where we assume that remainder in $\mathbb{Z}$ is unique, when defined as positive).
Have you encountered anything similar to this before? The only similar text I can find is this.
Note that $R_m(q)\equiv q\pmod{m}$, for any $q\in D$, so the map $\phi:D_m\to D/mD$ given by $R_m(q)\mapsto [R_m(q)]=[q]$, the equivalence class of $R_m(q)$ in $D/mD$ is well defined. Then by definition of addition, $\phi(a+b)=[R_m(a+b)]=[a+b]=[a]+[b]=\phi(a)+\phi(b)$. Similarly $\phi(ab)=\phi(a)\phi(b)$. Then since $\phi$ is a bijection and preserves addition and multiplication, this identifies $D_m$ with $D/mD$, so $D_m$ must have a unit $a$ given by $\phi(a)=[1]$, note that $\phi(R_m(1))=[1]$, so $R_m(1)$ is the unit regardless of whether or not its equal to 1.