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Given an Lie Algebra $L$ (of finite dimension and over an algebraically closed field with zero characteristic) and an ideal $I$, is it truth that

$rad\left(\dfrac{L}{I}\right)= \pi(rad(L))$,

where $\pi$ is the projection? What happens if $L$ isn't finite dimensional? And if the field has positive characteristic?

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2 Answers 2

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A counter-example for the infinite dimensional case:

Let $V$ be a vector space with a basis $\{e_i\mid i\in \mathbb{Z}_+\}$ and $V_p=\langle e_i\mid i=1,\dots,p\rangle$. Consider $$\mathfrak{t}_\infty:= \{ x \in \mathfrak{gl}(V)\mid x(V_i)\subset V_i \} .$$ as a lie subalgebra of $\mathfrak{gl}(V)$.

Let $E_{ij}$ be the linear transformations on $V$ such that $E_{ij}(e_j)=e_i$ and $E_{ij}(e_p)=0$ if $p\neq j$. Then, we have the formula $$[E_{ij},E_{kl}]=\delta_j^kE_{il}-\delta_l^iE_{kj}.$$

Convince yourself that $$\mathfrak{h}:=\{x \in t_\infty \mid x(V) \subset \langle e_i\mid i>1\rangle \}$$ is an ideal of $\mathfrak{t}_\infty$.

So, we have that $$rad( \mathfrak{t}_\infty / \mathfrak{h})=$$

$$\langle E_{1j} \mid j>1\rangle / \mathfrak{h}.$$

Suppose, by absurd, that $$\pi ( rad(\mathfrak{t}_\infty) ) = $$

$$rad( \mathfrak{t}_\infty / \mathfrak{h}) =$$

$$\langle E_{1j}\mid j>1\rangle / \mathfrak{h}.$$ This implies that $$\langle E_{1j}\mid j>1\rangle + \mathfrak{h} = $$

$$rad(\mathfrak{t}_\infty) + \mathfrak{h}.$$

Therefore, $$\langle E_{1j}\mid j>1\rangle + \mathfrak{h} \ni E_{12}=$$

$$x + y \in rad(\mathfrak{t}_\infty) + \mathfrak{h}$$

for some $x \in rad(\mathfrak{t}_\infty)$ and $y \in \mathfrak{h}$. Then

$$rad(\mathfrak{t}_\infty) \ni $$

$$[E_{11},x]=$$

$$[E_{11},E_{12}]-[E_{11},y]=$$

$$E_{12}+0=E_{12}.$$

This, again by the formula given above, implies that

$$rad(\mathfrak{t}_\infty)$$

contains

$$t_\infty ':=\{x \in \mathfrak{t}_\infty\mid x(V_1)=0\textrm{ and } x(V_i)\subset V_{i-1}, i>1\}$$ that is a non-solvable ideal of $\mathfrak{t}_\infty$. A contradiction.

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I am not sure about infinite dimensions and/or positive characteristic, but the answer is Yes in the finite-dimensional, zero characteristic case.

In your situation we have an exact sequence of Lie algebras $$ 0 \longrightarrow I \longrightarrow L \stackrel{\pi}{\longrightarrow} L/I \longrightarrow 0 $$ Let $r_L < L$ be the radical of $L$. Then $\pi(r_L)$ is a solvable ideal of $R/I$ and hence it is contained in its radical $r_{L/I}$, so that $$ \pi(r_L) < r_{L/I} $$

My original answer had an error, which Kevin Ventullo pointed out in a comment below. He also kindly fixed it in his second comment, which I include here for completeness, but the credit is due solely to him.

The Levi factor of $L$, which is semisimple and isomorphic to $L/r_L$, surjects onto $(L/I)/\pi(r_L)$, whence $(L/I)/\pi(r_L)$ has to be semisimple as well, whence we get the other inclusion $$ r_{L/I} < \pi(r_L) $$

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  • $\begingroup$ I'm confused. $\pi^{-1}(r_{L/I})$ contains $I$, so it need not be solvable. $\endgroup$ Apr 15, 2011 at 1:25
  • $\begingroup$ Here's a fix for the second half: if $\pi(r_L)$ was not all of $r_{L/I}$, we could quotient $L$ and $L/I$ by $r_L$ and $\pi(r_L)$ respectively, and get a surjective map from a semisimple Lie algebra to a non-semisimple Lie algebra, which is impossible. $\endgroup$ Apr 15, 2011 at 1:37
  • $\begingroup$ Thanks for the comment. Yes, I made an error in the second part (although not sure exactly where!) and your second comment fixes it. I will edit the question accordingly. $\endgroup$ Apr 15, 2011 at 2:09

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