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Show that $\mathbb{Z}[\sqrt{3}]/(3+\sqrt{3})\cong \mathbb Z_6$

I’ve recently asked a similar question but am still having trouble understanding it. Here’s my attempt so far:

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

I’m not sure how to continue? What do I need to learn to see that isomorphism.

Elise9
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2 Answers2

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Informally:

Let $R = \Bbb Z[\sqrt3]/(3+\sqrt3)$. Then $3+\sqrt3=0$ in $R$, which implies that $$ 9 = 3^2 = (-\sqrt3)^2 = 3, $$ and then, substracting $3$ from both sides, $$ 6=0. $$ Also note that $6=0$ implies that $3+\sqrt3=0$.

Therefore, the elements of $R$ consists of the integers (because $a+\sqrt3b=a-3b$ for all $a,b \in \Bbb Z$) with the only condition that $6=0$. Hence: $R \cong \Bbb Z_6$.

azif00
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    I'd like to give this +1 but it needs more information. First, how do we know that this ring has [at least] $6$ distinct elements. And in the other direction the direction you have already addressed, that $6=0$: It also would be helpful to add in the intermediate steps $3+\sqrt{3} = 0 \implies 3 = -\sqrt{3}$ $\implies -3 = \sqrt{3}$ $\implies (-3)^2 = (\sqrt{3})^2$ $\implies 9 = 3 \implies 9-3=6 =3-3= 0$. – Mike Mar 28 '25 at 20:15
  • @Mike I think having the integers with the only condition that $6=0$ is enough to have a ring isomorphic to $\Bbb Z_6$. – azif00 Mar 28 '25 at 21:49
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$$ \phi: \mathbb{Z}[\sqrt{3}] \to \mathbb{Z}_6 , a+\sqrt{3}b \mapsto a+3b$$

So if we have a ring morphism. $\phi(1)=1$, so $\phi(3)=3$ and therefore $\phi(\sqrt{3})=3$ is the only possibilty, so we guess this as our morphism.

We now have to check if $\phi$ is a morphism and look if it is surjective and observe $\ker \phi$.

$\phi(a+c+b\sqrt{3}+d\sqrt{3}) = \phi(a+b\sqrt{3})+\phi(c+\sqrt{3}d)$ \ $ \phi((a+b\sqrt{3})(c+\sqrt{3}d))= \phi(ac+\sqrt{3}(bc+ad)+3bd))$ $ =ac+3bd+3bc+3ad$ $= (a+3b)(c+3d) $

That it is surjective is clear.
$\ker \phi=\{a+\sqrt{3}b, a,b\in \mathbb{Z}|a \mod 6=0 \lor b \mod 2=0 \lor (a \mod 6 =3\land b\mod 2=1)\}$

If we have $a=6k:6=-(3+\sqrt{3})^2+6(3+\sqrt{3})$
If we have $b=2k,2\sqrt{3}=2(3+\sqrt{3})-6=... $
If we have $a=3+6k$ and $b=1+2l$:$3+\sqrt{3}$ So $\ker \phi \subseteq (3+\sqrt 3)$, but also $\phi(3+\sqrt 3)=0$

jojo math
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