Equivalent conditions of prime ideals in Lie algebra

Question

I am studding on Lie algebra and prime ideals, but some problems rise up to me. please look on this and try to help me.

Definition: An ideal $P$ of $L$ is called prime if $[H, K] \subseteq P$ with $H, K$ ideals of $L$ implies $H \subseteq P$ or $K \subseteq P$.

THEOREM : Let $P$ it be an ideal of L. Then the following conditions are equivatent}:

i) $P$ is prime.

ii) If $[a, H]\subseteq P$ for $a \in L$ and an ideal $H$ of $L$, then either $a\in P$ or $H\subseteq P.$

iii) If $[a, <b^{L}>]\subseteq P$ for $a, b\in L$, then either $a\in P$ or $b\in P.$

PROOF. $\mathrm{i}$) $\Rightarrow \mathrm{i}\mathrm{i}\mathrm{i}$). For each $a\in L,$

$$ <a^{L}>=\sum_{i=0}^{\infty} V_{i}, $$ where $V_{0}=(a)$ and $V_{i} [(a),\underline{{L],\ldots,}_{i-times}}L]$. If $[a,\ <b^{L}>]\subseteq P$, we assert that $$ $$ $[V_{i},\ <b^{L}>]\subseteq P$ for all $i\geq 0$. In fact, it is true for $i=0$. Let $i\geq 1$ and assume that the assertion is true for $i-1$. Then $$ [V_{i},\ <b^{L}>]=[[V_{i-}{}_{1}L],\ <b^{L}>] $$ $$ \subseteq[[V_{i-1},\ <b^{L}>],\ L]+[V_{i-1},\ [L,\ <b^{L}>]] $$ $$ \subseteq[P,\ L]+[V_{i-1},\ <b^{L}>]\subseteq P_{:} $$ Thus we have the assertion.

It follows that $$ [<a^{L}>,<b^{L}>]\subseteq P. $$ Since $P$ is prime, either $<a^{L}>\subseteq P$ or $<b^{L}>\subseteq P$ and so $a\in P$ or $b\in P.$

$\mathrm{i}\mathrm{i}\mathrm{i})\Rightarrow \mathrm{i}\mathrm{i})$ . Let $a\in L\backslash P$ and let $H$ be an ideal of $L$ such that $[a,\ H]\subseteq P$. For any $b\in H, [a,\ <b^{L}>]\subseteq P$ since the ideal $<b^{L}>$ is contained in $H$. As $a\not\in P,$ iii) implies $b\in P$. Hence $H\subseteq P.$

$\mathrm{i}\mathrm{i})\Rightarrow \mathrm{i})$ . Let $H, K$ be ideals of $L$ such that $[H,\ K]\subseteq P$ and $H\not\subset P$. Since $[a,\ K]\subseteq P$ for any $a\in H\backslash P$, we have $K\subseteq P$ by ii). Therefore $P$ is prime.

My questions( I need more explanation about this points which have color) :-

1) It follows that $[<a^{L}>,<b^{L}>]\subseteq P.$ (why this true) ??

2) Since $P$ is prime, either $<a^{L}>\subseteq P$ or $<b^{L}>\subseteq P$ and so $a\in P$ or $b\in P.$ (why this true) ??

3) Since $[a,\ K]\subseteq P$ for any $a\in H\backslash P$ (why this true) ??

Answer 1

The source is Naoki Kawamoto, On Prime Ideals of Lie Algebras (Hiroshima Math J. 4 (1974), 679--684), Theorem 1.

To quote the paper and to correct your typesetting: $$\langle a^L \rangle = \sum_{i=0}^\infty V_i \qquad \text{where} \qquad V_0 = (a) \qquad \text{and} \qquad V_i = [\ldots[(a),\underbrace{L],\ldots,}_i\ L]. \tag{1}$$

That is, $V_1 = [(a),L], \quad V_2 = [[(a),L],L], \quad V_3=[[[(a),L],L],L],\quad$ etc.

1) To show $[\langle a^L \rangle,\langle b^L \rangle] \subseteq P$ it suffices to show $[V_i, \langle b^L \rangle] \subseteq P$ for each $i \geqslant 0$. This sufficiency follows from $(1)$. Showing $[V_i, \langle b^L \rangle] \subseteq P$ is what is done in the proof. To summarize the argument: the case $i=0$ is true by assumption, and the rest follows by induction and the Jacobi identity.

2) Since $P$ is prime, either $\langle a^L\rangle \subseteq P$ or $\langle b^L \rangle \subseteq P$ (this is the definition of prime). If $\langle a^L \rangle \subseteq P$ then we have $a \in \langle a^L \rangle \subseteq P$, so $a \in P$. In the same way, if $\langle b^L \rangle \subseteq P$ then $b \in P$. So $a \in P$ or $b \in P$.

3) $[a,K] \subseteq P$ for any $a \in H\setminus P$ is true by the first assumption on $H$ and $K$ made in the preceding sentence, that is $[H,K] \subseteq P$. Indeed, for $a \in H\setminus P$ we have $a\in H$ and hence $[a,K] \subseteq [H,K] \subseteq P$.

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To answer the questions in the comment:

4. What is the meaning of $(a)$?

   $L$ is a Lie algebra over a field $\Phi$. For any element $a$ of $L$, $\langle a^L \rangle$ is defined to be the smallest ideal of $L$ containing $a$. An ideal $I$ is a $\Phi$-vector subspace of $L$ satisfying $[I,L] \subset L$. So what is $\langle a^L \rangle$ as a $\Phi$-vector space? Of course it must contain $a$, and all its scalar multiples, i.e. the $1$-dimensional $\Phi$-vector subspace $(a) := \Phi a \subset \langle a^L \rangle$. But if it contains $(a)$, then (by definition of ideal) it must also contain $[(a),L]$, and hence $[[(a),L],L]$, etc. Because it is a $\Phi$-vector space it must contain their sum. This leads to $(1)$.

5. How you use the Jacobi identity in the first one?

   It is used to go from line 1 to line 2 in the display equation: $[[V_{i-1},L], \langle b^L \rangle]$ is a set consisting of $[[x,y],z]$ such that $x \in V_{i-1}, y \in L, z \in \langle b^L \rangle$. By the Jacobi identity $(I)$ and antisymmetry $(II)$ we have: $$\begin{align*}[[x,y],z] &= -[[y,z],x] -[[z,x],y] \tag{I} \\ & = [[x,z],y] + [x,[y,z]] \tag{II} \\ &\in [[V_{i-1},\langle b^L \rangle],L] + [V_{i-1},[L,\langle b^L \rangle]].\end{align*}$$ Hence $[[V_{i-1},L], \langle b^L \rangle] \subseteq [[V_{i-1},\langle b^L \rangle],L] + [V_{i-1},[L,\langle b^L \rangle]]$.

6. How can we prove that if $\langle b^L \rangle$ is contained in $H$ then $[a,\langle b^L \rangle] \subseteq [a,H]$?

   This is by definition, or the meaning of the notation: $[a,\langle b^L \rangle]$ is a set consisting of $[a,y]$ with $y \in \langle b^L \rangle$ and $[a,H]$ is a set consisting of $[a,z]$ with $z \in H$. Since $y \in \langle b^L \rangle \subseteq H$, we have $y \in H$, so $[a,y] \in [a,H]$, hence $[a,\langle b^L \rangle] \subseteq [a,H]$

7. How can we prove that if $a \in H$ then $[a,K] \subseteq [H,K]$?

   Same/similar reason: $[H,K]$ is a set consisting of $[x,y]$ with $x \in H$ and $y \in K$, and $[a,K]$ is a set consisting of $[a,z]$ with $z \in K$. Since $a \in H$, we have $[a,z] \in [H,K]$, so $[a,K] \subseteq [H,K]$.