Proof of the ring isomorphism: $\mathbb{Z}[\sqrt{7}] /(5+2 \sqrt{7}) \cong \mathbb{Z} / 3$

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 $$\phi (3) = 0$$.

• What is $(5+2\sqrt7)(5-2\sqrt7)$? – Lord Shark the Unknown Jun 1 at 17:23
• 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 $3 \in ker(\phi)$? – NotAbelianGroup Jun 1 at 17:30
• Thx AbelianGroup, Jerome will have one less question on wednesday ! – Marine Galantin Jun 10 at 20:55
• @MarineGalantin Ahah pas de soucis! – NotAbelianGroup Jun 11 at 6:41

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 $$5 +2 \sqrt{7}$$.

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 $$\mathbb{Z}[\sqrt{7}]$$.