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.


  • 1
    $\begingroup$ "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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.