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I want to prove that: if $L$ is leibniz algebra and $J(L)$is jacobson radical of $L$ always: $$J(L)\subset L^2$$. I read this proof in (CLASSIFYING SEVERAL CLASSES OF LEIBNIZ ALGEBRAS Batten Ray) this prove is:

if $x$ is not in $L^2$ ,then we can find a complementary subspace ,$M$ ,of $x$ in $L$ that contains $L^2$and since $L^2\subset M$,$M$ is maximal ideal(?) of $L$ and $x$ is not in $M$. I dont know why $M$ is maximal ideal?

thanks for help

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  • $\begingroup$ I think every thing contains $L^2$ must be ideal.why $M$ is maximal? $\endgroup$
    – pink floyd
    Commented Nov 4, 2016 at 17:06
  • $\begingroup$ Note added 2017-10-25: someone has voted to close this question as "off-topic". Even if it is, what is the rationale for voting to close a question that was answered swiftly and satisfactorily? $\endgroup$
    – Yemon Choi
    Commented Oct 25, 2017 at 20:37

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Since it is a complementary subspace it has codimension one in $L$, and so is maximal. It is an ideal of $L$ because $L^2$ is inside $M$.

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