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

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 $$0\leq r,r'. I showed that $$\\u=r+\sqrt7(kq'+r')+I\$$ Now I want to show that $$u=r+\sqrt7r'+I$$ but I don't know how.

(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.