1
$\begingroup$

By definition, a primitive ideal $P$ exists if there is a simple $R$-module $S$ such that $Ann(S)$=$P$. I saw another statement as follows:

"$P$ is a primitive ideal of a ring if there is a left maximal ideal $L$ such that $P \subsetneq L \ $ and for any ideal $A$ of $R$, $A \subsetneq L\ $, then $A\subseteq P$ "

If this claim is right please note me some good references. Thanks.

  • 0
    Did you mean "..., then $A\subseteq P$ "?2011-10-29
  • 0
    @Rasmus: As it's told to me, $A$ is a proper one.2011-10-29
  • 0
    The claim cannot hold as written: if we take $A=P$, then $A\subsetneq L$, but $A$ is clearly not properly contained in $P$.2011-10-29
  • 0
    Also, do you mean this to be a *definition* of primitive, or do you mean it to be a *sufficient condition* for primitivity?2011-10-29
  • 0
    @ArturoMagidin: As you pointed, it cannot be the proper one. Yesterday, I found, very hard here,another defenition "An ideal $P$ in a ring is called (left)primitive if it is the largest ideal contained in some maximal (left)ideal $M$". And there is a therorem (by Jacobson) which shows these two definetions are the same (Lectures on Rings and Modules by J.Lambek). So; I think this question was answered before formally. Thanks Arturo, Rasmus, for any help and the attention. :)2011-10-30
  • 0
    @ArturoMagidin: Sorry but do you think I should delete this question?2011-10-30
  • 1
    @Basil: I would suggest correcting it, and then posting the reference/proof you found as an answer.2011-10-30

1 Answers 1

1

this is a community wiki answer intended to get this out of the unanswered queue.


As discovered in the comments, the user later discovered the statement as written is flawed, and that the most likely intended statement was that a primitive ideal is an ideal maximal with respect to containment in a maximal left ideal.