(Edited)


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?



Maximal ideals and Essential ideals
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.


