10 Locally Nilpotent Radical

Definition

Let $V$ be a class of algebras, we say $R\subseteq V$ is called radical class iff

• $R$ is closed under homomorphisms
• for any $A\in V$, there exists $R(A)⊲A$ with $R(A)\in R$ and $I⊲A,I\in R⟹I\subseteq R(A)$
• $R(A/R(A))=0$.

Definition

An algebra $A$ is called locally nilpotent if any finitely generated subalgebra of $A$ is nilpotent.

In this section, our main goal is to prove the following theorem.

The property of local nilpotency is a radical in the class of Jordan algebra.

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Intersection Characterization

Definition

If $R$ is a radical class, every algebra $A$ contains a unique maximal ideal belonging to $R$, denoted by $R(A)$, which is constructed as the sum of all $\mathcal{R}$-ideals in $A$

$$R(A)=\sum \{I⊲A\mid I\in \mathcal{R}\}.$$

We say an algebra $A$ is $R$-semisimple if $R(A)=0$.

Theorem

Let $R$ be a radical class. Then for any algebra $A$, the radical $R(A)$ coincides with the intersection of all ideals $P$ such that the quotient $A/P$ is $R$-semisimple

$$R(A)=\cap \{P⊲A\mid R(A/P)=0\}.$$

Furthermore, if the radical class is special (e.g., related to prime algebras, like the prime radical or locally nilpotent radical), this intersection can be restricted to prime ideals

$$R(A)=\cap \{P⊲A\mid P\text{ is prime and }R(A/P)=0\}.$$

a Criterion for Radical Classes

Proposition

Let $R\subseteq V$ be a radical class, and $A\in V$.

Then if $I⊲A$ and $I,A/I\in R$, then $A\in R$. $(\ast )$

\begin{proof} Define $\overline{A}=A/R(A)$, then $A/R(A)\cong A/I/R(A)/I$. Since $I\subseteq R(A)$, we have $A/R(A)$ is a homomorphic image of $A/I$. Since $A/R(A)\in R$ and $A/R(A)⊲A/R(A)$, by (ii) we have $A/R(A)\subseteq R(A/R(A))=0$, there is $A/R(A)=0$ and so $A=R(A)\in R$. \end{proof}

Lemma

Let $V$ be a class of algebras and $R$ be a subclass $R\subseteq V$ satisfying condition $(\ast )$ of ^86874b, i.e., for any $A\in V$ and $I⊲A$, $A/I\in R⟹A\in R$. Then if $A\in V$ and $I_1,I_2⊲A$ such that $I_1,I_2\in R$, then $I_1+I_2\in R$.

\begin{proof} Note that $I_1+I_2/I_1\cong I_2/I_1\cap I_2\in R$, and $I_1\in R$. Then we have $I_1+I_2\in R$. \end{proof}

Remark. We can generalize it to any finite sum of ideals.

Definition

We say that the class of algebras $R$ is locally defined if the following condition is satisfied:

$$A\in R⟺\text{for each finitely generated subalgebra }B\subseteq A,B\in R.$$

Proposition

Let $V$ be a class of algebras, and let $R\subseteq V$ be a subclass that is:

• closed under homomorphism,
• locally defined, and
• satisfies $(\ast )$ in ^86874b.

Then $R$ is a radical class.

\begin{proof} Let $A\in V$, and let $R(A)=\sum \{I⊲A:I\in R\}$. We need to show that $R(A)\in R$ and $R(A/R(A))=0$. Let $a_1,\cdots ,a_k\in R(A)$ with $a_i\in I_i\in R$ and $I_i⊲A$. By generalization of ^86874b $(\ast )$, $I_1+\cdots +I_k\in R$. It follows that $\mathrm{alg}⟨a_1,\cdots ,a_n⟩\in R$. Since $R$ is locally defined, $R(A)\in R$.

Now if $\overline{I}⊲A/R(A)$, then $\overline{I}\in R$. Let $I⊲A$ with $R(A)\subseteq I$.

Then $I/R(A)=\overline{I}\in R$ and $R(A)\in R$, by $(\ast )$ there is $I\in R$, $I\subseteq R(A)$ and so $\overline{I}=0$. \end{proof}

Nilpotency in Finitely Generated Algebras

Lemma

Let $J$ be a Jordan algebra $J=Fv+I$ with $J^2\subseteq I$. If $I$ is nilpotent, then $J$ is also nilpotent.

Remark. Any subspace of $J$ which contain $J^2$ is an ideal.

\begin{proof} If $J=Fv+I$, then we have the following inclusion

$$U(J{)}^{3m}\subseteq (U_J(I){)}^m+(U_J(I{)}^m)U(J)$$

by the proof of ^j7zg8g, where $U_J(I)={\mathrm{alg}}_{U(J)}⟨R_a:a\in I⟩$.

Define $\pi :U(J)\to \mathrm{Mult}(J)$. Then we have $R(J{)}^{3m}\subseteq (R_J(I){)}^m+R_J(I{)}^{m}R(J)$. Since $I$ is nilpotent, $R_J(I)$ is nilpotent. Thus we have $n\in \mathbb{N}$ with $R_J(I{)}^n=0$ and so $R(J)$ is nilpotent. Therefore, $J$ is nilpotent. \end{proof}

Lemma

Let $J$ be a finitely generated Jordan algebra, and let $J^2$ be a nilpotent ideal. The $J$ is nilpotent.

\begin{proof} Assume that $J=\mathrm{alg}⟨a_1,\cdots ,a_k⟩$ with $a_i\in J$. Define $I_0=J^2$, $I_1=J^2+F{a}_1$, $\cdots$, $I_k=J^2+F{a}_k$. These $I_0,\cdots ,I_k$ are ideals. By ^345566, $I_1$ is nilpotent.

Since $I_{i+1}^2\subseteq I_i$, repeat the argument above and we have $I_k=J$ nilpotent. \end{proof}

Lemma

If $J$ is finitely generated, then $J^2$ is finitely generated as well.

\begin{proof} WLOG suppose $J$ is a finitely generated free Jordan algebra. Consider $I=(J^2{)}^3$, which is an ideal of $J$. Define $\overline{J}=J/I$, then $({\overline{J}}^2{)}^3=0$ and so $\overline{J}$ is nilpotent by ^gjzrak. There exists $m$ such that $(\overline{J}{)}^m=0$, which follows that $J^m\subseteq I\subseteq (J^2{)}^2$.

Let $S=\{v\in J:v\text{ is product of generators of }J\text{ with }2⩽\mathrm{deg}(v)⩽m\}$. Then $S$ is a finite set because $J$ is finitely generated. We claim that $J^2=\mathrm{alg}⟨S⟩$.

Firstly, $\mathrm{alg}⟨S⟩\subseteq J^2$. To show that $J^2\subseteq \mathrm{alg}⟨S⟩$, take $v\in J^2$ monomial. If $\mathrm{deg}v<m$, then $v\in S$; if $\mathrm{deg}v⩾m$, then by $J^m\subseteq (J^2{)}^2$, $v=\sum v_i{u}_i$ for some $v_i,u_i\in J^2$. As $\mathrm{deg}{u}_i+\mathrm{deg}{v}_i=\mathrm{deg}v$ and $2⩽\mathrm{deg}{v}_i,u_i<m$ (otherwise repeat this procedure), we deduces that $u_i,v_i\in S$ and so $v\in \mathrm{alg}⟨S⟩$. Hence $J^2\subseteq \mathrm{alg}⟨S⟩$ and we prove the claim.

Since $J=F(J)/I$ and $F(J{)}^2$ is finitely generated, $J^2=F(J{)}^2/I$ is finitely generated. \end{proof}

Corollary

If $J$ is finitely generated, then for any $m⩾1$, $J^{(m)}$ is also finitely generated, where $J^{(1)}=J^2$ and $J^{(i+1)}=(J^{(i)}{)}^2$ for any $i⩾2$.

Exercise. Show that if $J$ is finitely generated, then $J^n$ is finitely generated.

While nilpotency always implies solvability (see ^8bdmmg), the converse is not true in general (see ^tlwx2b). However, if we impose the condition of being finitely generated, the equivalence holds.

Lemma

Any finitely generated solvable Jordan algebra is nilpotent.

\begin{proof} Let $J$ be a solvable finitely generated Jordan algebra. We do induction on index of solvability.

If $J^{(1)}=J^2=0$, we have done.

Suppose $J^{(m)}=0$ and any Jordan algebra $A$ with $A^{(k)}=0$ and $k<m$ is nilpotent. By ^31zeou, $J^2$ is finitely generated. Since $(J^2{)}^{(m-1)}=J^{(m)}=0$, $J^2$ is nilpotent.

By ^gjzrak, $J$ is nilpotent. \end{proof}

Zhevlakov’s Theorem

Lemma

Let $J$ be a Jordan algebra. If there exists $I⊲J$ such that $I$ and $J/I$ are locally nilpotent, then $J$ is locally nilpotent.

\begin{proof} We may assume that $J$ is finitely generated. Let $\overline{J}=J/I$ be finitely generated and nilpotent. Thus there exists $m$ such that $(\overline{J}{)}^m=0$ and $J^m\subseteq I$.

Take $n$ with $2^n⩾m$, then $J^{(n)}\subseteq J^{2^n}\subseteq J^m\subseteq I$ is finitely generated by ^644g46. Since $I$ is nilpotent, $J^{(n)}$ is nilpotent and so is solvable. There exists $k$ such that $(J^{(n)}{)}^{(k)}=J^{(n+k)}=0$ and so $J$ is solvable. Then $J$ is nilpotent by ^8bdmmg. \end{proof}

Zhevlakov theorem

The property of local nilpotency is a radical in the class of Jordan algebra.

Exercise. Let $A$ be a finitely generated associative algebra, then $A^{+}$ is a finitely generated Jordan algebra.

Exercise. Let $A$ be associative with $A^{+}$ nilpotent, then $A$ is nilpotent as well.

Remark. Golod-Shafarevich constructed associative finitely generated algebra $A$ with $A=\mathrm{Nil}(A)$ and $\mathrm{LocNil}(A)=0$. Then $J=A^{+}$ is nil, finitely generated, but not nilpotent.

Jordan algebra which is solvable but not nilpotent

Let $V$ be a vector space over $F$, and let $\{v_i:i⩾1\}$ be a set of basis. Define Grassman algebra $G(V)=\mathrm{alg}⟨v_i:i⩾1,v_i{v}_j=v_j{v}_i⟩$. Define $J=V\oplus G(V)$. Denote basis of $V$ in $G(V)$ as $\overline{v_i}$. $V\circ V=G\circ G=0$, $v_i\circ w=w\circ v_i=w\overline{v_i}$ for some $w\in G$ and $v_i\in V$.

• $J$ is a Jordan algebra: check linearized Jordan identity.
• $J$ is solvable: $J^2\subseteq G$ and $G^2=0$ yield $(J^2{)}^2=0$.
• $J$ is not nilpotent: $(\cdots (({\overline{v}}_1\cdot v_2)\cdot v_3)\cdots )\cdot v_k=\overline{v_1}\cdots \overline{v_k}\ne 0$ in $G$.

Exercise. Show that if $\mathrm{Lie}⟨x,y⟩$ is a free Lie algebra with two generators, $[\mathrm{Lie}⟨x,y⟩,\mathrm{Lie}⟨x,y⟩]$ is not finitely generated. (Hint: consider $[\cdots [[x,y],y],\cdots ,y]$.)