Annihilator and Jacobson Radical

Annihilator

Definition

Let $R$ be a ring, and let $M$ be a left $R$-module. For any $x\in M$, define

$${\mathrm{Ann}}_R(x)=\{r\in R:rx=0\}$$

as the *annihilator *of $x$ in $R$. Notice that ${\mathrm{Ann}}_R(x)$ is a left ideal of $R$.

For a set $X\subseteq M$, define ${\mathrm{Ann}}_R(X)={\cap }_{x\in X}{\mathrm{Ann}}_{R}x$. If $N$ is a $R$-submodule of $M$, then ${\mathrm{Ann}}_R(N)$ is a $2$-sided ideal of $R$.

Lemma

Let $R$ be a ring with identity $1$. Let $S$ be a left simple $R$-module. Then $S$ is a quotient of the left regular module ${}_{R}R$.

\begin{proof} Let $x\in S$ be an nonzero element of $S$. Consider cyclic module $0\ne Rx\subseteq S$, then $Rx=S$. Define

$$\phi :{}_{R}R\to S,r\mapsto rx$$

and note that $\phi$ is an epimorphism. Hence $S\simeq R/{\mathrm{Ann}}_R(x)$ is a quotient of the left regular module ${}_{R}R$. \end{proof}

Jacobson Radical

Definition

Let $R$ be a ring. The Jacobson radical $J(R)$ of $R$ is the intersection of all left maximal ideals of $R$. As a left $R$-module, $J(R)=\mathrm{Rad}({}_{R}R)$.

Intersection of Annihilators of Simples

Theorem

The following holds for the Jacobson radical of a ring $R$.

• $J(R)$ is the intersection of all the annihilator ideals of simple (left) $R$-modules. Consequently, $J(R)$ is a $2$-sided ideal of $R$.
• $a\in J(R)$ iff $1-xa$ has a left inverse for all $x\in R$, i.e., there exists $b\in R$ such that $b(1-xa)=1$.
• $J(R)$ is the largest one among the ideals of $I$ satisfying the following condition: if $a\in I$, then $1-a$ is a unit of $R$.
• $J(R)$ coincides with the intersection of all maximal right ideals of $R$.

\begin{proof} i) Let $S$ be a simple $R$-module. As $S\simeq R/{\mathrm{Ann}}_R(x)$ for any $x\in S$ by ^09efb3, ${\mathrm{Ann}}_R(x)$ is a maximal left ideal of $R$. It deduces that $J(R)\subseteq {\mathrm{Ann}}_R(x)$ and so $J(R)\subseteq {\mathrm{Ann}}_R(S)$.

Conversely, we aim to show that, for any $x\in R$, if $x\in {\mathrm{Ann}}_R(S)$ for each simple $R$-module $S$, then $x$ is contained in all maximal left ideals. For a maximal left ideal $I$, $R/I$ is a simple $R$-module. Since $x\in {\mathrm{Ann}}_R(R/I)$, we know $x\cdot (1+I)\in I$ and so $x\in I$. Now we finish the proof.

ii) "→" If $a\in J(R)$ and $1-xa$ has no left inverse for some $x\in R$, then $R(1-xa)\ne R$. Hence there is a maximal left-ideal $M$ of $R$ with $1-xa\in M$. Then $a\in M$ yields $1\in M$, which is a contradiction.

"←" If $a\notin J(R)$, then $a\notin M$ for some maximal left ideal $M$ of $R$. Then $Ra+M=R$ and $xa+b=1$ for some $x\in R$ and $b\in M$. So $1-xa\in M$ and $1-xa$ has no left inverse.

iii) Let $a\in J(R)$. Then there exists $b$ such that $b(1-a)=1$ and then $b=1+ba$ has a left inverse, i.e. there exists $c$ such that $cb=1$. It deduces that $c=cb(1-a)=1-a$ and so $1-a$ is a unit.

Let $I$ be a $2$-sided ideal with the property ”$a\in I$ then $1-a$ is a unit of $R$“. If $I⊈J(R)$, then there exists a maximal ideal $M$ such that $I⊈M$ and $I+M=R$. So $a^{\prime }+b=1$ for some $a^{\prime }\in I$ and $b\in M$. As $1-a^{\prime }\in M$, $1-a^{\prime }$ is not a unit, which is a contradiction.

iv) Suppose $K(R)$ is the intersection of all maximal right ideals of $R$, then by the argument of iii), we can prove that $K(R)$ is the largest one among the ideals of $I$ satisfying the following condition: if $a\in I$, then $1-a$ is a unit of $R$. Then by iii) $K(R)=J(R)$ and so we finish the proof. \end{proof}

Nilpotency of $J(A)$ in Artinian Rings

Definition

• An element $a\in R$ is called nilpotent if $a^n=0$ for some $n\in {\mathbb{Z}}_{+}$. If so, $(1-a)(1+a+\cdots +a^{n-1})=1$ and $1-a$ is a unit.
• An left ideal $I$ of $R$ is called a nil left ideal if every element of $I$ is nilpotent.
• $I$ is called nilpotent if $I^n=0$ for some $n$. Clearly, nilpotent ideals are nil.

Theorem

If $I$ is a nil left ideal, then $I\subseteq J(R)$.

\begin{proof} Let $a\in I$. For any $x\in R$, $xa\in I$. Since $I$ is nil, $1-xa$ is a unit and so $a\in J(R)$ by ^p95eh4. \end{proof}

Theorem

If $A$ is Artinian, then $J(A)$ is nilpotent.

\begin{proof} Consider $J(A)\supseteq J(A{)}^2\supseteq \cdots$. Since $A$ is Artinian, there exists $n$ such that $J(A{)}^n=J(A{)}^{n+1}=\cdots$. Assume that $N=J(A{)}^n\ne 0$.

Consider the set of left ideals $\{I:I\subseteq A,NI\ne 0\}$. It is a nonempty set since $N^2\ne 0$. Let $I_0$ be the minimal member of this set. There exists $a\in I_0$ such that $Na\ne 0$. Since $Na\subseteq I_0$ and $N(Na)=Na\ne 0$, by the minimality of $I_0$, we know $Na=I_0$. In particular, there exists $b\in N$ such that $ba=a$, i.e. $(1-b)a=0$. Recall that $b\in N\subseteq J(A)$, then $1-b$ is a unit, leading to a contradiction. \end{proof}

Corollary

Let $A$ be an Artinian ring, then $J(A)$ is the maximal nilpotent left ideal of $A$.

\begin{proof} By ^112eab and ^a43f1d. \end{proof}

Corollary

Let $A$ be an Artinian ring. Then $J(A)$ is the unique left ideal $U$ of $A$ with the property that $U$ is nilpotnent and $A/U$ is semisimple.

\begin{proof} By ^r8oe32 and ^8uxtcw. \end{proof}

Theorem

Let $I$ be a $2$-sided ideal of $R$ with $I\subseteq J(R)$. Then $J(R/I)=J(R)/I$.

\begin{proof} By ^9eaa9e. \end{proof}

Theorem

Let $A$ be an Artinian. Then $A$ is semisimple iff $J(A)=0$.

\begin{proof} "←" Assume that $\{M_{\lambda },\lambda \in \Lambda \}$ are all maximal left ideals of $A$. We consider the set of left ideals that are expressed a finite intersection of some $M_{\lambda }$. By Artinian, this set has minimal element, say $L=M_{{\lambda }_1}\cap \cdots \cap M_{{\lambda }_r}$. If $L\ne J(A)$, then there exists $M_{\lambda }$ such that $M_{\lambda }\cap L⊊L$, which contradicts to the minimality of $L$. So

$$L=J(A)=M_{{\lambda }_1}\cap \cdots \cap M_{{\lambda }_r}.$$

Define a map $A\to A/M_{{\lambda }_1}\oplus \cdots \oplus A/M_{{\lambda }_r}$, which is a monomorphism and so ${}_{A}A$ is semisimple. (This part can also be proved by ^r8oe32.)

"→" Assume ${}_{A}A$ is semisimple, then $A=S_1\oplus \cdots \oplus S_r$ with $S_i$ simple. Since $J(A)\subseteq {\mathrm{Ann}}_R(S_i)$, $J(A)A=0$. It deduces that $J(A)=0$. \end{proof}