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why if $C$ is a subalgebra of $A$ and $B$ is an ideal of $A$ contained in $F(C)$, then $B$ is contained in $F(A)$?

$F(A)$ is frattini subalgebra of $A$.can I use this for Leibniz algebras?

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Suppose that B is not contained in F(A). Then there is a maximal subalgebra M of A such that A = B + M. But now C = B + M \cap C = F(C) + M \cap C = M \cap C. It follows that C is contained in M, whence B is contained in M, a contradiction.

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    :How should I use the B ideal? – 2017-02-21
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    Sorry, I don't understand the question. What do you want to use it for? – 2017-02-21
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    :can B is subalgebra not ideal? – 2017-02-21
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    No, the proof doesn't work then, as B + M is not necessarily a subalgebra, and so we cannot assert that A = B + M. – 2017-02-21