Correspondence Theorem for Lie algebras

Question

I am trying to prove the following:

Let $L$ be a Lie algebra and $I$ an Ideal of $L$. There exists a bijection between the Ideals of the quotient algebra $L/I$ and the Ideals of $L$, that contain $I$.

• Let $J$ be an Ideal of $L$, which contains $I$ $\Rightarrow$ $J/I$ is an Ideal of $L/I$
• Let $K$ be an Ideal of $L/I$ $\Rightarrow$ $J:=\{z\in L \mid z+I \in K\}$ is an Ideal of $L$, which contains $I$

I already know that this holds true for vector spaces (as it is true for modules). Knowing that, how can I show that this also holds true for Lie algebras?

Answer 1

I think this well-know proof holds for Lie algebras.

Let $J$ be an ideal of $L$ that contain $I$. Consider the map $$\begin{array}\ \phi:& L & \rightarrow & L/I \\ & J &\mapsto & J/I\end{array}$$ Note that $J/I$ is an ideal of $L/I$ since for $X+I \in J/I$ and $Y+I\in L/I$ we have $[X+I,Y+I]=[X,Y]+I$, where $[X,Y]$ is in $J$ because it is an ideal of $L$.

Now, let $K$ be an ideal of $L/I$ and consider the map $$\begin{array}\ \psi:& L/I & \rightarrow & L \\ & K &\mapsto & \{ X\in L\ |\ X+I \in K\}\end{array}$$

We need to check that this set is in fact an ideal of $L$, but this is easy. For $X\in \psi(K)$ and $Y\in L$ we have $[X,Y]\in L$ and $[X,Y] +I=[X+I,Y+I] \in K$, because $K$ is an ideal.

• $\psi(\phi(J))=\psi(J/I) = \{X\in L\ |\ X+I \in J/I\}=J$.
• $\phi(\psi(K))=\phi(\{Y\in L\ |\ Y+I \in K\})=\{Y+I\in L/I \ |\ Y+I\in K\} = K$.

This show the bijection.