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Questions: [See below]

$\rm\color{#c00}{(1)}$ What structure has $R/I$ when $I$ is a maximal left ideal? I know that if $I$ is a bilateral ideal then $R/I$ is a ring and, moreover, if $I$ is maximal then it is a field, in which case we get the simplicity of $R/I$.

This question naturally gives rise to the second question.

$\rm\color{#c00}{(2)}$ I know that if we define $J'(R)$ by changing the word left for the word right in $(\star)$, then $J(R)=J'(R)$ and the characterization $(\star\star)$ remains valid, with the word left changed for the word right. However, can we put the word bilateral instead of left or right in $(\star\star)$? In the French Wikipedia page Radical de Jacobson, $J(R)$ seems to be defined exactly this way and the proof of $(\star\star\star)$ presented there seems to make more sense since in that case $R/I$ is a field as usual.

Define the Jacobson radical of a ring (with unity) $R$ by $$ J(R):=\bigcap_{M\text{ a simple, left }R\text{-module}}\text{Ann}_R(M).\quad(\star) $$ Then one can show $$ J(R)=\bigcap_{I\text{ a maximal, left ideal of }R}I.\quad(\star\star) $$ Next, in order to show the $[\Longleftarrow]$ part of $$ r\in J(R)\iff\forall x\in R:1+xr\in R^{\times},\quad(\star\star\star) $$ one proceeds by contradiction as follows:

Proof of $[\Longleftarrow]$: Suppose that for all $x\in R$ we have $1+xr\in R^{\times}$ but that there is a maximal left ideal $I$ of $R$ for which $r\not\in I$. $\rm\color{#c00}{(1)}$ Then $R/I$ is simple, hence $R\overline{r}=I\text{ or }R/I$. Since $r\not\in I$, $R\overline{r}=R/I$ and so there is an $x\in R$ such that $1+xr\in I$. But $1+xr$ is invertible, hence $1\in I$ and $I=R$ which contradicts the hypothesis that $I$ is maximal. $\blacksquare$

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  • $\begingroup$ If this is your question: I do not believe that the intersection of all maximal two-sides ideals of $R$ is the Jacobson radical of $R$ in general. But if I remember it correctly, it is true when $R$ is Artinian. I do not know how you would define the annihilator of a two-sided $R$-module. $\endgroup$ – darij grinberg Nov 27 '14 at 2:05
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What structure has R/I when I is a maximal left ideal?

The best we can say is that it is a left $R$ module and not necessarily anything more.

moreover if $I$ is maximal then it is a field,

No, in general you will only get a simple ring.

However, can we put the word bilateral instead of left or right in (⋆⋆)?

It will not yield the Jacobson radical. Let $R$ be the ring of linear transformations of a countable dimensional vector space. Then $R$ has exactly one nontrivial bilateral ideal: the set of transformations with finite dimensional range. However the whole ring is Von Neumann regular with Jacobson radical zero.

In the French Wikipedia page Radical de Jacobson, J(R) seems to be defined exactly this way

I notice that the article says "le radical de Jacobson d'un anneau commutatif est..." Of course if the ring is commutative then we don't need left or right ideals anymore.

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