what would be an $n$ such that $\mathbb{Z}[\sqrt{2}]/(3-\sqrt{2})$ is ring isomorphic to $\mathbb{Z}_n$?
This problem was on a qualification test. Here's how I solved it, but I'm not satisfied with my answer since it is too intuitive.
Let $I=(3-\sqrt{2})$. Since $3+I=\sqrt{2}+I$, $9+I=2+I$ hence $7+I=I$. Hence the characteristic of the quotient ring $\mathbb{Z}[\sqrt{2}]/(3-\sqrt{2})$ is not greater than $7$. For this reason, I expected $n$ would be $7$.
Define $\phi(1)=\bar{1}$ and $\phi(\sqrt{2})=\bar{3}$.
Since $2$ is square-free the function $\phi:\mathbb{Z}[\sqrt{2}]\rightarrow \mathbb{Z}_7$ is well defined.
Then, it can be directly checked that $\phi$ is a ring epimorphism.
Let $a+b\sqrt{2}\in \ker(\phi)$.
Then, $a+3b \equiv 0 \pmod 7$
Thus for some $k$, $a+3b=7k$.
Note that every element in $(3-\sqrt{2})$ is of the form $3c-2d+\sqrt{2}(3d-c)$.
Define $A=3-2k, B= 3k-1$.
Note that $a+3b=7k=A+3B$.
Thus, $a=A+3l$ and $b=B-l$ for some $l$.
Thus. $a=(3+3l) - 2k , b=3k - (1+l)$. Thus, $a+b\sqrt{2}\in I$.
This shows that $\ker(\phi)=I$. This proves the problem. Q.E.D.
Is there another way to prove this?
$\mathbb{Z}[\sqrt{2}]/(3-\sqrt{2})\simeq\mathbb{Z}[X]/(X^2-2,3-X)\simeq\mathbb{Z}/(3^2-2)=\mathbb{Z}_7$
The above isomorphism only use the Third isomorphism theorem, that is: