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I am asked to prove that $$ \mathbb{Z}[\sqrt{7}] /(5+2 \sqrt{7}) \cong \mathbb{Z} / 3 $$

First, if we define the following homorphism : $\phi:\mathbb{Z} \rightarrow \mathbb{Z}[\sqrt{7}] /(5+2 \sqrt{7})$, we can try to show that this homomorphism has $\ker(\phi)=(3)$ and is surjective to conclude with the first isomorphism theorem.

It is kind of informal but I realised that in $(5+2 \sqrt{7})$, $$5+2 \sqrt{7} \equiv 0$$ $$5 \equiv -2 \sqrt{7}$$ $$25\equiv 28$$ $$3 \equiv 0$$

But then I struggle to show properly that $\phi (3) = 0$.

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    $\begingroup$ What is $(5+2\sqrt7)(5-2\sqrt7)$? $\endgroup$ – Lord Shark the Unknown Jun 1 at 17:23
  • $\begingroup$ I can see that $3 + (5+2 \sqrt{7})(5-2 \sqrt{7}) = 3-3 = 0$, so can I say that $0 = \phi (3 + (5+2 \sqrt{7})(5-2 \sqrt{7}) ) = \phi (3)$, thus $3 \in ker(\phi)$? $\endgroup$ – NotAbelianGroup Jun 1 at 17:30
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    $\begingroup$ Thx AbelianGroup, Jerome will have one less question on wednesday ! $\endgroup$ – Marine Galantin Jun 10 at 20:55
  • $\begingroup$ @MarineGalantin Ahah pas de soucis! $\endgroup$ – NotAbelianGroup Jun 11 at 6:41
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Consider the natural homomorphism $$\varphi \colon \mathbb{Z} \hookrightarrow \mathbb{Z}\left[ \sqrt{7} \right] \twoheadrightarrow \frac{\mathbb{Z}\left[ \sqrt{7} \right]}{\left\langle 5 +2 \sqrt{7} \right\rangle}$$ which sends any integer $n \in \mathbb{Z}$ to its class $n +\left\langle 5+ 2 \sqrt{7} \right\rangle \in \frac{\mathbb{Z}\left[ \sqrt{7} \right]}{\left\langle 5 +2 \sqrt{7} \right\rangle}$.

As you said, it is enough to prove that $\text{Ker}(\varphi) = 3 \mathbb{Z}$ and $\varphi$ is surjective.

We have $3 = -\left( 5 -2 \sqrt{7} \right) \left( 5 +2 \sqrt{7} \right) \in \left\langle 5 +2 \sqrt{7} \right\rangle$, and hence $3 \mathbb{Z} \subset \text{Ker}(\varphi)$. Therefore, either $\text{Ker}(\varphi) = \mathbb{Z}$ or $\text{Ker}(\varphi) = 3 \mathbb{Z}$.

Now, if we had $\text{Ker}(\varphi) = \mathbb{Z}$, there would exist $a, b \in \mathbb{Z}$ such that $1 = \left( a +b \sqrt{7} \right) \left( 5 +2 \sqrt{7} \right)$, which would yield $5 a +14 b = 1$ and $2 a +5 b = 0$ since $\sqrt{7}$ is irrational. This is impossible. Thus, $\text{Ker}(\varphi) = 3 \mathbb{Z}$.

Finally, note that $\sqrt{7} = 2 +\left( 8 -3 \sqrt{7} \right) \left( 5 +2 \sqrt{7} \right)$. Therefore, for all $a, b \in \mathbb{Z}$, we have $$a +b \sqrt{7} +\left\langle 5 +2 \sqrt{7} \right\rangle = \varphi(a +2 b) \, \text{,}$$ which proves that $\varphi$ is surjective.

P.S.: In order to avoid possible confusions, I denote by $\left\langle 5 +2 \sqrt{7} \right\rangle$ the ideal of $\mathbb{Z}\left[ \sqrt{7} \right]$ generated by $5 +2 \sqrt{7}$.

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You're on the right path. However, I'd observe that $$ (5+2\sqrt{7})(5-2\sqrt{7})=25-28=-3 $$ which implies that $3\in\langle5+2\sqrt{7}\rangle$. Thus the unique ring homomorphism $\varphi\colon\mathbb{Z}\to\mathbb{Z}[\sqrt{7}]/\langle5+2\sqrt{7}\rangle$ defined by $\varphi(n)=n+\langle5+2\sqrt{7}\rangle$ has $\varphi(3)=0+\langle5+2\sqrt{7}\rangle$, meaning $3\in\ker\varphi$.

The problem is now to show that $\varphi$ is surjective. If you take $a+b\sqrt{7}\in\mathbb{Z}[\sqrt{7}]$ with even $b=2c$, then $$ a+b\sqrt{7}=a+2c\sqrt{7}=(a-5c)+c(5+2\sqrt{7}) $$ and therefore $a+b\sqrt{7}+\langle5+2\sqrt{7}\rangle=\varphi(a-5c)$.

Can you work out the problem for odd $b$? Hint: compute $(5+2\sqrt{7})(-2+\sqrt{7})$.

Note. I used $\langle5+2\sqrt{7}\rangle$ to denote the ideal, in order to avoid confusions with parenthesized expressions in $\mathbb{Z}[\sqrt{7}]$.

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