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.

  • $\begingroup$ Did you mean "..., then $A\subseteq P$ "? $\endgroup$ – Rasmus Oct 29 '11 at 12:25
  • $\begingroup$ @Rasmus: As it's told to me, $A$ is a proper one. $\endgroup$ – Basil R Oct 29 '11 at 16:05
  • $\begingroup$ The claim cannot hold as written: if we take $A=P$, then $A\subsetneq L$, but $A$ is clearly not properly contained in $P$. $\endgroup$ – Arturo Magidin Oct 29 '11 at 22:00
  • $\begingroup$ Also, do you mean this to be a definition of primitive, or do you mean it to be a sufficient condition for primitivity? $\endgroup$ – Arturo Magidin Oct 29 '11 at 22:24
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    $\begingroup$ @Basil: I would suggest correcting it, and then posting the reference/proof you found as an answer. $\endgroup$ – Arturo Magidin Oct 30 '11 at 18:24

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.


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