1
$\begingroup$

I didn't understand why in this definition $I$ has to be an ideal to make sense.

REMARK

This is from Steps in Commutative Algebra, page 107.

Thanks a lot

$\endgroup$
2
  • 1
    $\begingroup$ It would still make sense and $(G : I)$ would still be a submodule, but you could make this complaint about a lot of definitions in (commutative) algebra. $\endgroup$
    – TTS
    Jun 26 '13 at 22:27
  • 1
    $\begingroup$ If you spend the time to check the details about why it's a submodule, you'll see clearly you don't need anything about $I$'s idealness. However, I agree that phrasing it in the way this book is makes one a little paranoid :) $\endgroup$
    – rschwieb
    Jun 27 '13 at 19:46
4
$\begingroup$

The definition doesn't claim that $I$ has to be an ideal, and in fact it doesn't, but if $S$ is any subset of $R$ then $(G :_M S) = (G :_M I)$ where $I$ is the ideal generated by $S$, so $I$ might as well be an ideal.

$\endgroup$

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.