11 Prime Radical

Prime and Semi-prime Algebras

Definition

We call algebra $A$ prime if for any $I,J⊲A$ such that $I\circ J=0$, either $I=0$ or $J=0$.

An ideal $P⊲A$ is called prime if $A/P$ is prime.

Definition

Let $A$ be an algebra. Denote the prime radical of $A$ by $\mathrm{rad}(A)={\cap }_{P⊲A,P\text{ prime}}P$.

Definition

We call algebra $A$ semi-prime if $I⊲A$ and $I^2=0$, then $I=0$.

An ideal $P⊲A$ is called semi-prime if $A/P$ is semi-prime.

Characterization via m-sequences

Definition

A sequence of elements $\{a_1,\cdots ,a_i,\cdots \}\subseteq A$ is called $m$-sequence if $a_{i+1}\in ({\mathrm{id}}_A(a_i){)}^2$. We say that element $a\in A$ annihilates $m$-sequence if any $m$-sequence started with $a$ contains zero, i.e. any $m$-sequence which contains $a$ also contains $0$.

Lemma

Let $a\in A$ be an element which annihilates all $m$-sequence, then $a$ belongs to all semi-prime ideals of $A$.

\begin{proof} Assume that $a$ is an element which annihilates all $m$-sequence, and there exists semi-prime ideal $P$ such that $a\notin P$. If $(\mathrm{id}⟨a⟩{)}^2\subseteq P$, then by $P$ semi-prime, there is $\mathrm{id}⟨a⟩\subseteq P$ and so $a\in P$, which is impossible.

Hence $(\mathrm{id}⟨a⟩{)}^2⊈P$. Then there exists $a_1\in (\mathrm{id}⟨a⟩{)}^2\setminus P$. Repeat this procedure, then we build $\{a_1,a_2,\cdots ,\}$, which is a $m$-sequence and does not contain $0$, leading to a contradiction. \end{proof}

Theorem

$\mathrm{rad}(A)=\{a\in A:a\text{ annihilates }m\text{-sequence}\}$.

\begin{proof} Let $M=\{a\in A:a\text{ annihilates }m\text{-sequence}\}$. By ^4a83a1, $M\subseteq \mathrm{rad}(A)$.

To show that $\mathrm{rad}(A)\subseteq M$, it is enough to prove that if $a\notin M$, then there exists prime ideal $P$ such that $a\notin P$. As $a\notin M$, there exists a $m$-sequence $m:=\{a,a_1,\cdots ,a_n,\cdots \}$ with $a_i\ne 0$. Define $\mathcal{M}=\{I⊲A:m\cap I=\varnothing \}$. Notice that $\mathcal{M}$ is a inductive set, as for any $I_1\subseteq \cdots \subseteq I_n\subseteq \cdots$, $I={\cup }_{i\in I}{I}_i\in \mathcal{M}$. Thus $\mathcal{M}$ contains a maximal ideal $P$ with $a\notin P$.

Next we show that $P$ is prime. Let $I,J⊲A$, $I,J\subseteq P$ and suppose $I,J⊈P$. WLOG we may suppose that $P⊊I$ and $P⊊J$ (if necessary substitute $I\to I+P$ and $J\to J+P$). Then maximality of $P$ implies that there exists $a_i\in I$ and $a_j\in J$ with $a_i,a_j\in m$.

Suppose $j>i$. Since $a_{i+1}\in (\mathrm{id}⟨a_i⟩{)}^2\subseteq I$ and then $a_{j+1}\in (\mathrm{id}⟨a_j⟩{)}^2\subseteq IJ\subseteq P$, leading to a contradiction. Therefore, $P$ is a prime ideal and so the proof is complete. \end{proof}

Structure of Semi-prime Algebras

Corollary

$\mathrm{rad}(A)$ is the minimal semi-prime ideal. Furthermore, if $A$ is semi-prime, then $\mathrm{rad}(A)=0$.

Exercise. Show that $\mathrm{rad}(\mathrm{rad}(A/\mathrm{rad}A))=0$.

\begin{proof} By ^4a83a1 and ^34q5nm, we know $\mathrm{rad}(A)$ is contained in all semi-prime ideals. Now we prove that $A/\mathrm{rad}(A)$ is semi-prime. For any ideal $I⊲A$ such that $I^2\subseteq \mathrm{rad}(A)$, there is $I^2\subseteq P$ for all prime ideals $P$ and so $I\subseteq P$. It follows that $I\subseteq \mathrm{rad}(A)$ and so $\mathrm{rad}(A)$ is semi-prime.

It remains to show that if $A$ is semi-prime, then $\mathrm{rad}(A)=0$. Otherwise, assume that there exists non-zero $a\in \mathrm{rad}(A)$. Take $a_0=a\ne 0$ and $I_0=\mathrm{id}(a_0)$. Since $A$ is semi-prime, $I_0\ne 0$ yields that $I_0^2\ne 0$ and so there exists non-zero $a_1\in (\mathrm{id}(a_0){)}^2$. Repeat this procedure, then $a$ does not annihilate $m$-sequence, which is impossible. Therefore, $\mathrm{rad}(A)=0$. \end{proof}

Proposition

An algebra $A$ is semi-prime iff $A$ is subdirect product of prime algebras.

\begin{proof} Assume that $A$ is semi-prime. Then by ^da3su8, $\mathrm{rad}(A)={\cap }_{P⊲A,P\text{ prime}}P=0$. Define $\varphi :A\to {\prod }_{P⊲A,P\text{ prime}}A/P$, then $\ker \varphi =0$ and so $\varphi$ is injective. Since $P$ is prime, the algebra $A/P$ is prime. Therefore, $A$ is a subdirect product of prime algebras.

Assume that $A$ is a subdirect product of prime algebras. Hence $A\subseteq \prod B_{\alpha }$, where $B_{\alpha }$ are prime and so semi-prime. If there exists $I⊲A$ such that $I^2=0$, then $I_{\alpha }^2=0$ and so $I_{\alpha }=0$. It deduces that $I=0$. Now we finish the proof. \end{proof}

Theorem

If $V$ is a homogeneous variety of algebras, with the property that locally nilpotent algebras is radical class. Define

$$\mathrm{LocNilp}(A)=\{I⊲A:I\text{ is locally nilpotent}\},$$

then

$$\mathrm{LocNilp}(A)=\cap \{P⊲A:P\text{ prime},\mathrm{LocNilp}(A/P)=0\}.$$

Furthermore, we have $\mathrm{rad}(A)⊊\mathrm{LocNilp}(A)⊊\mathrm{Nil}(A)$.