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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.

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    $\begingroup$ "does it mean that $R/I$ is finite?" Yes, it does. $\endgroup$ – Arthur Feb 19 at 12:08
  • $\begingroup$ Thanks, any hint? $\endgroup$ – Philip L Feb 19 at 12:09
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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 $0\leq x, y<a$.

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.

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  • $\begingroup$ elements of $R/I$ are sets, you mean I should work with representatives? sorry for the dumb question :) $\endgroup$ – Philip L Feb 19 at 12:26
  • $\begingroup$ @PhilipL Yes, that's what I mean. I'm so used to not really caring about the distinction in writing, I'm sorry. $\endgroup$ – Arthur Feb 19 at 12:34
  • $\begingroup$ Thanks, I managed to show that using euclidean division. :) $\endgroup$ – Philip L Feb 19 at 13:14

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