Where in the proof did Herstein use the fact that $A$ is a two-sided ideal of $R$? I'm reading Noncommutative Rings by I. N. Herstein. The theorem I'm having trouble with is 1.2.5, on page 16 of the book.
Some definition
1. Regular ideal

An ideal $\rho \subset R$ is called a regular ideal right ideal of $R$ iff There exists a $r \in R$, such that $x - rx \in \rho, \forall x \in R$.

2. Right quasi-regular element

$a$ is called the right-quasi element of $R$, is we can find $r \in R$, such that $a + r + ar = 0$. Such $r$ is called a right-quasi inverse of $a$.

3. Right quasi-regular ideal

An ideal $\rho \subset R$ is called a right-quasi regular ideal iff every element of it is right-quasi regular.

3. Simple module

A right $R-$module $M$ is called simple iff the two requirements below hold:
  
  
*
  
*$MR \neq 0$.
  
*$M$ has no non-trivial submodule.

4. Jacobson radical

The Jacobson radical is the set of all elements in $R$ that annihilate all simple $R-$modules.

Some properties


*

*$J(R) = \bigcap\limits_{M \text{ simple $R-$module}}\text{Ann}(M) = \bigcap\limits_{\rho \text{ regular, maximal right ideal of } R} \rho$

*$M$ is a right, simple $R-$module iff there's some maximal, regular right ideal $\rho \subset R$, such that $M \cong R/\rho$.

*Every right-quasi regular ideal is contained in $J(R)$, and $J(R)$ is the maximal ideal amongst the set of right-quasi ideals of $R$.

And this is a theorem I'm having trouble with.
Theorem 1.2.5 (page 16)

If $A$ is a two-sided ideal of $R$, then $J(A) = A \cap J(R)$.

Proof


*

*We'll now prove that $A \cap J(R) \subset J(A)$:
Let $a \in A \cap J(R)$, since $a \in J(R)$, as an element of $J(R)$, $a$ is right-quasi, hence there exists an $a'$, such that $a + a' + aa' = 0$, so $a' = -a -aa' \in A$, since $A$ is an ideal of $R$.
Being a right-quasi ideal of $A$, $A \cap J(R) \subset J(A)$.

*We'll now prove that $A \cap J(R) \supset J(A)$:
For every maximal, regular right ideal $\rho$ of $R$, let $\rho_A = A \cap \rho$. Now, there can only be 2 cases:


*

*$A \not \subset \rho$, since $\rho$ is maximal, $A + \rho = R$, and combining the two, we'll have:
$$R/\rho \cong (A + \rho) / \rho \cong A /(\rho \cap A) = A/\rho_A$$
Now, since $R / \rho$ is $R-$simple, we get that $\rho_A$ is a maximal right ideal of $A$ (?).
Since $\rho$ is regular, there exists some $b \in R$, such that $x - bx \in \rho, \forall x \in R$. Now $A + \rho = R$, hence $b = a + r$, for $a \in A$, and $r \in \rho$. So, we'll have $x - bx = x - (a + r)x = a - ax -rx \in \rho, \forall x \in R$. Since $r \in \rho$, we must have $rx \in \rho$. Hence $x - ax \in \rho, \forall x \in R$, hence $x - ax \in \rho_A, \forall x \in A$, which makes $\rho_A$ $A-$regular. Since $J(A)$ is the intersection of all maximal, regular ideal of $A$, we'll have $J(A) \subset \rho_A$.


*

*If $A \subset \rho$, then $\rho_A = A \cap \rho = A$, so obviously $J(A) \subset A =  \rho_A$.



Combining the two cases above, we'll have $J(A) \subset \bigcap \rho_A = \left(\bigcap \rho \right) \cap A = J(R) \cap A$. Hence, yielding the desired result.

After this theorem, Herstein point out that if $A$ is not two-sided, then the theorem's result will fail. But as far as I can see, there's no place in the theorem that Herstein actually used $A$ as a two-sided ideal of $R$.
And there's one thing that I cannot get, it's the (?) part. I know $R / \rho$ is $R-$simple, hence $A / \rho_A$ is also $R-$simple, which means that there is no right ideal of $R$ (not $A$) lies in between $\rho_A$, and $A$. How come he concluded that $\rho_A$ is a maximal ideal of $A$? Is it where I must use the fact that $A$ is two-sided to prove?
Thank you guys a lot,
And have a good day.
 A: I think your question many days.
But I still can't solve it.
Here are some idea.
I use the example of the author given.
$R=M_2(\Bbb{Z}_2)$, 
$A=\{\bigl( \begin{smallmatrix} \alpha & \beta \\
0 & 0 \end{smallmatrix} \bigr)
\mid \alpha,\beta\in \Bbb{Z}_2\}$,
$\rho=\{\bigl( \begin{smallmatrix} 0 & 0 \\ 
\sigma & \tau \end{smallmatrix} \bigr)
\mid \sigma,\tau\in \Bbb{Z}_2\}$,
then 
$J(A)\neq (0)=A\cap J(R)$.
$R$ has no nontrivial ideals 
see Fraleigh_A first course in abstract algebra, page 254, section 27, exercise 38.
I verify directly that $R$ has only two nonzero proper maximal right ideals, 
they are $A$ and $\rho$.
Both are regular.
I use this example to check each statement in the proof.
The statement "we get that $\rho_A$ is a maximal right ideal of $A$"
fail.
Because there is a right ideal $S=\{\bigl( \begin{smallmatrix} 0 & \gamma \\
0 & 0 \end{smallmatrix} \bigr)
\mid \gamma \in \Bbb{Z}_2\}$ 
such that $0\subsetneq S\subsetneq A$.
$(0)=\rho_A=\rho\cap A$ is not a maximal right ideal of $A$.
Hence I guess the condition $A$ is a "left" ideal is used to ensure that 
"$R/\rho$ is irreducible" implies that "$\rho_A$ is a maximal right ideal of $A$".
My classmate doubt this assert "Since $R/\rho$ is irreducible we get that $\rho_A$ 
is a maximal right ideal of $A$."

THEOREM 1.2.5.
If $A$ is an ideal of $R$
then $J(A)=A\cap J(R)$.
Proof.
If $a\in A\cap J(R)$
then as an element of $J(R)$,
$a$ is right-quasi-regular.
Its quasi-inverse
$a'=-a-aa'$
is thus in $A$ since $A$ is an ideal of $R$.
In short,
$A\cap J(R)$ is a quasi-regular ideal of $A$
so must be contained in $J(A)$
by Theorem 1.2.3.
Suppose now that $\rho$ is a maximal regular right
ideal of $R$ and let $\rho_A=A\cap \rho$.
If $A\nsubseteq \rho$ the maximality of
$\rho$ forces $A+\rho=R$ therefore
$$R/\rho \cong \frac{A+\rho}{\rho}\cong \frac{A}{A\cap \rho}=A/\rho_A.$$
Since $R/\rho$ is irreducible we get that
$\rho_A$ is a maximal right ideal of $A$.
Since $\rho$ is regular $x-bx\in \rho$ for some
$b\in R$;
$b=a+r$ with $a\in A$,
$r\in \rho$.
Hence $\rho \ni x-bx=x-(a+r)x=x-ax-rx$
giving us that $x-ax\in \rho$.
In particular we see that $\rho_A$
is regular in $A$.
Therefore $J(A)\subseteq \rho_A$
for all maximal regular right ideals $\rho$ of $R$
which do not contain $A$
and certainly also for those which do.
In other words,
$J(A)\subseteq \cap \rho_A=(\cap \rho)\cap A=J(R)\cap A$.
The two opposite containing relations give
us the desired result $J(A)=A\cap J(R)$.
