Proving that for certain ring of algebraic integers $R$, $R/bR$ is finite This is a part of proof I try to understand.
The situation is the following:
Suppose that $a,b,x,y$ are algebraic integers  such that $b \neq 0$ and $ax+by=1$. Set $K:=\mathbb{Q}(a,b,x,y)$ and $R:=O_K,$ that is, a subring of all algebraic integers contained in $K$.
Next statement of the proof is (whithout any further comments):
"Then $R/bR$ is a finite ring."
Hence my question:
How can I prove that $R/bR$ is finite? Or is it somewhat obvious?
(I should probably add that my knowledge of alg. number theory is very limited.)
My attempt so far:
From the given relation, it is not difficult to see that every element of $R/bR$ can be expressed as a $\mathbb{Q}-$linear combination of elements of the type $a^jy^i+bR$, $x^jy^i+bR$ for some $i,j \leq N$, where $N$ is a sufficiently large integer. I would also guess that there are not many possibilities for the values of the rational coefficients in those linear combinations. But that seems to be far from the desired conclusion.
 A: This is just a general thing about rings of integers, it has nothing to do with your problem...
Let $K$ be any number field, and $R$ its ring of integers. Then you can check that $R$ is free as a $\mathbb Z$-module, in particular $R \simeq \mathbb Z^n$ where $n$ is the degree of $K$ over $\mathbb Q$. 
Therefore if you have any integer $N \in \mathbb Z$, then $R / N R $ is clearly finite (it has cardinality $N^n$!).
Now let $b \in R$, so that $b$ satisfies a monic polynomial equation with coefficients in $\mathbb Z$:
$$ b^n + a_{n-1}b^{n-1} + \dots + a_0 = 0$$
of minimal degree. In particular, $a_0 \neq 0$, and 
$$ a_0 = - a_1 b - a_2 b^2 - \dots - b^n = b ( -a_1 - a_2 b - \dots - b^{n-1}) \in b R,$$
i.e., $a_0 R \subseteq bR$.
So $|R / bR | \leq |R / a_0R|$, and this last ring is finite as we showed above. 
A: Let us prove the following theorem:  

Theorem
  In a finite algebraic extension $K/\mathbb Q$ with the ring of integers $R$, if $p\in R$, then $\mid R/pR\mid=\text{N}_{K/\mathbb Q}(p)$.  

We shall prove this by use of the following two lemmas:  

Lemma I (Chinese Remainder Theorem)
  If an ideal $\mathfrak A=\mathfrak B\mathfrak  C$ in a ring $R$, and if $\mathfrak B+\mathfrak C=R$, then $R/\mathfrak A\cong R/\mathfrak B\oplus R/\mathfrak C$.  

Proof
Define $\phi: R/\mathfrak A\rightarrow R/\mathfrak B\oplus R/\mathfrak C$ by sending $x+\mathfrak A$ to $(x+\mathfrak B,x+\mathfrak C)$. Then by assumption, there are $b\in \mathfrak B, c\in\mathfrak C$ such that $b+c=1$. And we define $\psi:R/\mathfrak B\oplus R/\mathfrak C\rightarrow R/\mathfrak A$ by sending $(x+\mathfrak B,y+\mathfrak C)$ to $(xc+yb)+\mathfrak A$. Clearly $\psi$ is the inverse of $\phi$, hence the conclusion.  

Lemma II
  It is our theorem, in the case of prime $\mathfrak p\in R$.  

Proof
Since $\mathfrak p$ is prime, the ideal $\mathfrak pR$ is maximal, as $R$ is a Dedekind domain. And for $\mathfrak pR\cap \mathbb Z$ is a prime ideal of $\mathbb Z$, it is generated by a prime integer $p\in\mathbb Z$. Now $\kappa=R/\mathfrak pR$ is a finite extension of the finite field $\mathbb Z/p\mathbb Z$, so $\kappa$ is also a finite field, with its degree of extension $=f$. Then we should prove that $\text{N}(\mathfrak p)=p^f=\mid R/\mathfrak pR\mid$. (Up to this point, the desired finiteness has been proved, without using the lemmas at all.)
Now let $L/K$ be a normal closure of $K$, and let $[L:K]=m, [K:\mathbb Q]=n$. Let $(p)=\prod_{i=1}^k\mathfrak P_i^{e_i}$ be the prime ideal factorisation of $(p)=pR'$, where $R'$ is the ring of integers of $L$, and where $\mathfrak P_1=(\mathfrak p)$. Since $L/\mathbb Q$ is galois, all the ramification indices $e_i$ are equal, denoted by $e$. Further, all the $p^{f_i}=\mid R'/\mathfrak P_i\mid$ are equal to some $p^{f'}$. So we have $ef'k=nm$, by means of lemma I, and of the isomorphism $R/\mathfrak P\cong \mathfrak P/\mathfrak P^2$.
Now $\text{N}_{L/\mathbb Q}\mathfrak P_1=\prod_{i=1}^k\mathfrak P_i^{nm/k}=\prod_{i=1}^k\mathfrak P_i^{ef'}=(p)^{f'}$. Writing $f''=[R'/\mathfrak P_1:R/\mathfrak p]$, we find: $f'=ff''$. Thus, from $\text{N}_{K/\mathbb Q}\circ\text{N}_{L/K}=\text{N}_{L/\mathbb Q}$, we obtain the claim.  

Proof of the theorem
  First we observe that the norm of $p$ is equal to the norm of the ideal $pR$ generated by $p$. Then this ideal can be written as a product of prime ideals. By the above two lemmas, we obtain the theorem as an immediate consequence.  Q.E.D 

If something does not pertain here, tell me. Thanks in advance.  
