# On my attempt in showing that $\mathbb Z[\sqrt{-2}] / (1 - \epsilon) \simeq\mathbb Z_3$.

Let $A = Z[\sqrt{-2}]$ be the ring with elements of the form $\{ a +b \epsilon\mid a,b \in\mathbb Z \}$ and where $\epsilon^2 = -2$.

Let $(1- \epsilon)$ denote the ideal generated by $1- \epsilon$.

I am tasked to show that $A / (1 - \epsilon) \simeq \mathbb Z_3$.

My attempt:

First I see that all elements in the ideal $(1- \epsilon)$ are given by $\{ (a+b\epsilon) (1- \epsilon) | a,b \in Z \}$ so a generic element has the form $a+2b + \epsilon(b -a )$.

Now I want to show that there are 3 equivalence classes in $A / (1 - \epsilon)$ where two elements $x,y$ are equivalent $\iff (y_1 +y_2\epsilon) -(x_1+x_2 \epsilon) \in (1 - \epsilon)$.

So I obtain the condition that $y_1 - x_1$ must be of the form $a+2b$ and $y_2 -x_2$ must be of the form $b-a$ but I can't seem to find a way to utilize this information.

• have you tried solving for $a$ and $b$ in terms of $x_1,x_2,y_1,y_2$ ? – mercio Jun 30 '17 at 13:29
• @mercio yes I tried but I could not get a solution in that way, is it possible? – Monolite Jun 30 '17 at 14:18

Let $u=1- \epsilon$. Then $a+b\epsilon=a+b-bu$. Moreover, $u\bar u=3$. Therefore, $a+b\epsilon \equiv c \bmod u$, where $c=a+b \bmod 3$. This proves that the quotient ring has at most $3$ elements. Since $0,1,-1$ are not equivalent mod $u$, the quotient ring has exactly $3$ elements and so is isomorphic to $\mathbb Z_3$.

Alternatively, you can just argue as follows: $$\mathbb Z[\sqrt{-2}] \cong \frac{\mathbb Z[x]}{(x^2+2)} \implies \frac{\mathbb Z[\sqrt{-2}]}{(1-\epsilon)} \cong \frac{\mathbb Z[x]}{(x^2+2,1-x)} = \frac{\mathbb Z[x]}{(3,1-x)} \cong \frac{\mathbb Z}{(3)}$$ using that $x^2+2=(x^2-1)+3=-(1-x)(1+x)+3$.

• Thank you greatly, why does the last isomorphism hold? – Monolite Jun 30 '17 at 14:22
• @Monolite, it's induced by $f(x) \mapsto f(1)$. – lhf Jun 30 '17 at 14:34
• There is a theorem (sadly, I don`t remember its proof) that $\mathbb{Z}[i]/(a+bi) \cong \mathbb{Z}_{a^2 + b^2} = \mathbb{Z}_{|a+bi|}$. – Alexander Markov Jun 30 '17 at 14:58
• I thought I understood initially but why do we need that $c=a+b \bmod 3$ what is the link with $u\bar u=3$? – Monolite Jun 30 '17 at 15:33
• @Monolite I posted a complete and rigorous form of the first argument. – Bill Dubuque Jun 30 '17 at 16:29

There is an image of $\,\Bbb Z\,$ in $R = \Bbb Z[\sqrt{-2}]/(1\!-\!\sqrt{-2})\,$ by mapping $\,n\,$ to $\ n \pmod{1\!-\!\sqrt{-2}}$ by composing the natural maps $\,\Bbb Z\to \Bbb Z[\sqrt{-2}]\to R.\,$ This map $\, h\color{#0a0}{ \ {\rm is\ surjective\ (onto)}}$ since $\,{\rm mod}\ \sqrt{-2}\!-\!1\!:\,\ \sqrt{-2}\equiv 1\,\Rightarrow\, a+b\sqrt{-2}\equiv a+b\in\Bbb Z,\,$ and has kernel $\,\color{#c00}{\ker h = 3\Bbb Z}\,$ by

\begin{align} n\in (1\!-\!\sqrt{-2})\,\Bbb Z[\sqrt{-2}] \iff &\ \dfrac{n}{1\!-\!\sqrt{-2}}\ \ \,\in\,\ \ \Bbb Z[\sqrt{-2}]\\[0.5em] \iff &\ \dfrac{n(1\!+\!\sqrt{-2})}{3}\in\Bbb Z[\sqrt{-2}] \ \ \ {\rm by\ \ rationalizing\ denominator}\\[0.5em] \iff &\ \color{#c00}{3\mid n}\end{align}

By the First Isomorphism Theorem: $\, \color{#0a0}{R = {\rm Im}\ h} \,\cong\, \Bbb Z/\color{#c00}{\ker h} \,=\, \Bbb Z/\color{#c00}{3\Bbb Z}$

• Thank you, was there a more intuitive reason I was missing before? – Monolite Jul 1 '17 at 0:47

From $y_1-x_1 = a+2b$ and $y_2-x_2 = b-a$,
you get by adding the two equations, $y_1+y_2-x_1-x_2 = 3b$.

If $y_1+y_2-x_1-x_2$ is a multiple of $3$ then your system has a unique solution in integers, $b = (y_1+y_2-x_1-x_2)/3$ and then $a=y_1-x_1-2b$, and so $x_1+\epsilon x_2$ and $y_1+\epsilon y_2$ are in the same class.

If $y_1+y_2-x_1-x_2$ is not a multiple of $3$ then the system has a unique solution in the rational numbers but they are not integers so they are not in the same class.

Therefore they are in the same class if and only if $x_1+x_2 \equiv y_1+y_2 \pmod 3$.

Since $x_1+x_2$ can be congruent to either $0,1$ or $2$ mod $3$, the three equivalence classes are $S_k = \{(x_1+\epsilon x_2) \mid x_1+x_2 \equiv k \pmod 3 \}$ with $k=0,1,2$