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


This is from Steps in Commutative Algebra, page 107.

Thanks a lot

  • 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

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.


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