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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.

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    Did you mean "..., then $A\subseteq P$ "?2011-10-29
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    @Rasmus: As it's told to me, $A$ is a proper one.2011-10-29
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    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
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    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
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    @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
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    @ArturoMagidin: Sorry but do you think I should delete this question?2011-10-30
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    @Basil: I would suggest correcting it, and then posting the reference/proof you found as an answer.2011-10-30

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