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.