# $\mathbb{Z}[\sqrt{2}]/(3-\sqrt{2})$ is ring isomorphic to $\mathbb{Z}_n$.

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?

• $a+3b=7k;a+b\sqrt{2}=-3b+7k+b\sqrt{2}=-b\left(3-\sqrt{2}\right)+7k\in I$ – Lozenges Apr 10 '15 at 20:27

$\mathbb{Z}[\sqrt{2}]/(3-\sqrt{2})\simeq\mathbb{Z}[X]/(X^2-2,3-X)\simeq\mathbb{Z}/(3^2-2)=\mathbb{Z}_7$
Given a ring $R$ and two ideals $I\leqslant J$ of $R$, then $(R/I)/(J/I)\simeq R/J$.
• @Rubertos The isomorphism identifies $\sqrt{2}$ and $X$. To apply the third isomorphism theorem, check $\mathbb Z[\sqrt 2] \cong \mathbb Z[X]/(X^2-2)$ first. – Christoph Apr 10 '15 at 20:33
• @Christoph I only know that conjugation for fields.. For what kind of ring $R$, does that conjugation hold? And is there a text introducing this technique? Oh. And I don't get the third isomorphic relation too. Why $(X^2-2,X-3)/(X)\cong \mathh{Z}_7$? – Rubertos Apr 10 '15 at 20:37
• @Rubertos To see why $\mathbb{Z}[\sqrt{2}]/(3-\sqrt{2})\simeq\mathbb{Z}[X]/(X^2-2,3-X)$, let $R=\mathbb{Z}[X]$, $J=(X^2-2,3-X)$, $I=(X^2)$; to see why $\mathbb{Z}[X]/(X^2-2,3-X)\simeq\mathbb{Z}/(3^2-2)$, let $R=\mathbb{Z}[X]$, $J=(X^2-2,3-X)$, $I=(3-X)$. – Censi LI Apr 11 '15 at 5:54