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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:

  1. $MR \neq 0$.

  2. $M$ 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

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

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

  3. 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.

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  • $\begingroup$ I can't talk for all but I think this question is waaaaaaaaay too long and there's a good chance many won't even take the time to read it, leave alone to sit down and try to solve it. I'd advice you to try to "cut it" into pieces and, perhaps and if needed, ask 2-3 much questions. $\endgroup$
    – DonAntonio
    Aug 6, 2013 at 12:20
  • $\begingroup$ @DonAntonio: In fact, the question begins with Theorem 1.2.5, assuming that we are all familiar with all of the definitions, so it's just a half page long. :( How should I make it look better? :( $\endgroup$
    – user49685
    Aug 6, 2013 at 14:30
  • $\begingroup$ Perhaps the proof presented by Herstein is incomplete. See the Russian version of this book. The proof for this theorem that appears in the Russian version is different. $\endgroup$
    – freshman
    Sep 6, 2017 at 19:34
  • $\begingroup$ Write this Некоммутативные кольца in Library Genesis. $\endgroup$
    – freshman
    Sep 6, 2017 at 19:39

1 Answer 1

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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 $J(A)=A\cap J(R)$.

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