Prove that $\mathbb{Z}[\sqrt 2]/(3+\sqrt 2)$ isomorphic to $\mathbb{Z}_7$

Question

I am really stuck here, could you please give an insight or show the proof step by step.

Answer 1

Consider the surjective map $\Bbb Z[\sqrt{2}] \to \Bbb Z_7$ sending $a+b\sqrt{2}\in \Bbb Z[\sqrt{2}]$ to $a - 3b\pmod{7}$. Show that this map is a ring homomorphism. If $x = a + b\sqrt{2}$ is in the kernel of this map, then $a\equiv 3b\pmod{7}$, and thus $a = 3b + 7k$ for some integer $k$. Since $(3 + \sqrt{2})(3 - \sqrt{2}) = 7$, then $$x = [b+k(3-\sqrt{2})](3 + \sqrt{2})\in (3 + \sqrt{2})$$ It follows that the kernel is $(3 + \sqrt{2})$. The result then follows from the first isomorphism theorem.

Answer 2

Consider the unique ring homomorphism $\chi\colon\mathbb{Z}\to\mathbb{Z}[\sqrt{2}]/(3+\sqrt{2})$, $\chi(z)=z+I$, where $I=(3+\sqrt{2})$.

_{If what you need to prove is true, then this homomorphism must be surjective and its kernel must be $7\mathbb{Z}$.}

We want to prove $\chi$ is surjective. Indeed, if $a+b\sqrt{2}\in\mathbb{Z}[\sqrt{2}]$, then $$ a+b\sqrt{2}+I=z+I $$ for some $z\in\mathbb{Z}$.

We just need to show that the equation $a-z+b\sqrt{2}=(x+y\sqrt{2})(3+\sqrt{2})$ is solvable with $x,y,z\in\mathbb{Z}$. This amounts to $$ \begin{cases} 3x+2y+z=a \\ x+3y=b \end{cases} $$ and the result is trivial, because we can take $y=0$, $x=b$ and $z=a-3b$.

Now $7\in\ker\chi$, because $\chi(7)=7+I=(3-\sqrt{2})(3+\sqrt{2})+I=I$.

Hence $\ker\chi\supseteq 7\mathbb{Z}$. Since $1\notin I$, the kernel of $\chi$ must be $7\mathbb{Z}$.