If $\mathbb{Z}\ni \underbrace{z}_{\ne 0} \in I \trianglelefteq R:=\{a+\sqrt7b:a,b\in\mathbb{Z}\}$ then $[R:I] < \infty$

Question

Let a ring $R:=\{a+\sqrt7b:a,b\in\mathbb{Z}\}$. Let an ideal $I\trianglelefteq R$ such that $z\in I$ for some $0\ne z\in\mathbb{Z}$. Prove that $[R:I]<\infty$.

I showed that $z\in I$ for some $z>0$. I defined $$\\ M:=\{a+\sqrt7b+I\in R/I:0\leq a,b< z\} \ $$ We want to show $M=R/I$. Let $u:=a+\sqrt7b+I\in R/I$. We know that $$\\ a =kq+r,b=kp'+r'\ $$ for some $0\leq r,r'<k$. I showed that $$ \\u=r+\sqrt7(kq'+r')+I\ $$ Now I want to show that $u=r+\sqrt7r'+I$ but I don't know how.

Answer 1

You're almost there.

(I assume $k=z$ in your calculations.)
Then just use the definition of ideal: $z\in I\implies\sqrt7q'z\in I$, and we also have $x+I=I$ if $x\in I$, thus $$u\ =\ r+\sqrt7r'+\sqrt7q'z+I\ =\ r+\sqrt7r'+I$$ as required.