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. $\blacksquare$

• 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. – darij grinberg Nov 27 '14 at 2:05

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.