Subgroup $(I, +)$ has finite index in $(R, +)$

I came across this question and I'll appreciate your help.

Let $$R = {\{a+b\sqrt{7} :a,b \in \mathbb{Z}}\}$$

Let $$I$$ be an ideal in $$R$$ and assume that there exist $$0\neq a\in \mathbb Z$$ s.t $$a\in I$$

Show that the subgroup $$(I, +)$$ has finite index in $$(R, +)$$.

first, I'm not sure what's the meaning of - "$$(I, +)$$ has finite index in $$(R, +)$$", does it mean that $$R/I$$ is finite?

What I managed to show is that if $$a\in I$$ then $$I + a= a+I = I$$ , but couldn't go further since I'm not sure what exactly I have to show.

The next step is to show that this statement holds for any $$I\neq0$$.

Guidance will be appreciated.

Thanks.

• "does it mean that $R/I$ is finite?" Yes, it does. – Arthur Feb 19 at 12:08
• Thanks, any hint? – Philip L Feb 19 at 12:09

Basic idea: $$a\in I$$ also means $$a\sqrt7\in I$$. Now show that any element of $$R/I$$ can be written as $$x + y\sqrt7 + I$$ with $$0\leq x, y.
For the second part, where you're given an arbitrary non-zero $$I$$ instead of an $$I$$ containing an integer, you have to take an arbitrary non-zero element of $$I$$ and use that to show that there is a non-zero integer in $$I$$. The rest follows from part one.
• elements of $R/I$ are sets, you mean I should work with representatives? sorry for the dumb question :) – Philip L Feb 19 at 12:26