5
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ $a+3b=7k;a+b\sqrt{2}=-3b+7k+b\sqrt{2}=-b\left(3-\sqrt{2}\right)+7k\in I$ $\endgroup$ – Lozenges Apr 10 '15 at 20:27
6
$\begingroup$

$\mathbb{Z}[\sqrt{2}]/(3-\sqrt{2})\simeq\mathbb{Z}[X]/(X^2-2,3-X)\simeq\mathbb{Z}/(3^2-2)=\mathbb{Z}_7$

The above isomorphism only use the Third isomorphism theorem, that is:

Given a ring $R$ and two ideals $I\leqslant J$ of $R$, then $(R/I)/(J/I)\simeq R/J$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Why does the first isomorphic relation hold? $\endgroup$ – Rubertos Apr 10 '15 at 20:25
  • $\begingroup$ @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. $\endgroup$ – Christoph Apr 10 '15 at 20:33
  • $\begingroup$ @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 $(X^2-2,X-3)/(X)\cong \mathh{Z}_7$? $\endgroup$ – Rubertos Apr 10 '15 at 20:37
  • $\begingroup$ @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 $R=\mathbb{Z}[X]$, $J=(X^2-2,3-X)$, $I=(3-X)$. $\endgroup$ – Censi LI Apr 11 '15 at 5:54
  • $\begingroup$ @Rubertos And I think every standard algebra textbook will introduce three isomorphism theorem (of groups and rings, etc.) $\endgroup$ – Censi LI Apr 11 '15 at 5:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.