$\mathbb{Z}[\sqrt{2}]/(3-\sqrt{2}) \simeq \mathbb{Z}_n$

Question

I am trying to find integer $n$ such that $\mathbb{Z}[\sqrt{2}]/(3-\sqrt{2}) \simeq \mathbb{Z}_n$.

My first guess is using "the 1st ring isomorphism theorem"

Let $\phi : \mathbb{Z}[\sqrt{2}] \rightarrow \mathbb{Z}_n$. I want to construct homomorphism such that whose kernel is $3-\sqrt{2}$

I tried to think of Norm map, i.e., $N(x) = a^2 - 2b^2$ for $x =a+b\sqrt{2} \in \mathbb{Z}[\sqrt{2}]$ but this was not good.

Any ideas?

Answer 1

The answer is $7$. Maybe the easiest way to think about it is to map $$ \mathbb{Z}[x] \to \mathbb{Z}_7 $$ by sending $x$ to $3$. Then you can observe that $x^2 - 2$ is in the kernel of this map, so there is a well-defined map $$ \mathbb{Z}[\sqrt{2}] \to \mathbb{Z}_7 $$ which sends $\sqrt{2}$ to $3$. It's clear that $3-\sqrt{2}$ is in the kernel of that map, which is what you want.

Answer 2

We rewrite $\Bbb Z[\sqrt 2]$ as $\Bbb Z[T]/(T^2 - 2)$ via the isomorphism sending $T$ to $\sqrt 2$. Then we have: $$\Bbb Z[\sqrt 2]/(3 - \sqrt 2) \simeq \Bbb Z[T]/(T^2 - 2, 3 - T) \simeq \Bbb Z[T]/(7, 3 - T) \simeq \Bbb Z/7.$$