Both notions seem to measure "largeness" of an ideal in a ring. how are they related? does one imply the other and vice versa?

i am reading this book "Hereditary Noetherian Prime Rings and Idealizers" by Levy and Robson.

On Page 58, Lemma 13.6, it says:

(Here $R$ is a Hereditary Noetherian Prime (HNP) ring)

All simple right $R$-modules occur as submodules of $R_\mbox{quo}/R$, where $R_\mbox{quo}$ is a ring of quotients of $R$. (indeed $R_\mbox{quo}$ is the injective hull $E(R)$ of $R$)

Let me copy the first few lines of the proof.

Let $X = R/M$ be a simple $R$-module. Since, by convention, $R \neq R_\mbox{quo}$, the socle of $R$ is zero. Therefore $M$ is essential in $R$. (and the proof goes on ...)

Here $M$ is maximal, and we have Maximal implies Essential. but at the same time it seems to suggest that under general conditions, Maximal does not necessarily imply Essential.

so my questions are

(i) when would one imply the other?

(ii) In those few lines i have copied, is the HNP property of $R$ involved?

(iii) As a side question, how important is it that a simple $R$-module can be embedded into the quotient module $E(R)/R$, where $E(R)$ is the injective hull of $R$? what is it really saying?


Here is a proof that the authors probably had in mind: if $M$ is not essential in $R$ then there is a non-zero right ideal $J$ such that $M \cap J = 0$. But then $J$ embeds into the simple module $R/M$ and so $J$ must be simple itself, and then $J$ is contained in the socle of $R$. However this socle is zero because $R \neq R_\rm{quo}$.

Even for prime Noetherian rings it can happen that a maximal right ideal is not essential. This happens for example if $R = M_2(k)$ is the $2 \times 2$ matrix ring over a field $k$ and $M = \begin{pmatrix} k & k \newline 0 & 0 \end{pmatrix}$. Note that $R = R_\rm{quo}$ in this case.

  • $\begingroup$ :) i am happy to receive such a reply. but a quick question: in the proof of Lemma 13.6, we need M to be essential. but then you are now using Lemma 13.6 to show that M is essential, which is really begging the question! $\endgroup$ – Paslig Keir May 31 '13 at 14:40
  • $\begingroup$ it must have been obvious. but did we use HNP in your proof? $\endgroup$ – Paslig Keir May 31 '13 at 14:58
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    $\begingroup$ You need prime Noetherian to guarantee the existence of the quotient ring. Later on in the proof, you need the hereditary property to ensure that $M$ is projective, and the Noetherian property to ensure that $M$ is finitely generated. $\endgroup$ – user91132 May 31 '13 at 14:59
  • $\begingroup$ before asking any more questions, let me work out a few things myself. i am so happy and thank you, you have made my day! :) $\endgroup$ – Paslig Keir May 31 '13 at 15:12

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