# Lie algebra: ideal

1. Sep 22, 2011

### Ted123

1. The problem statement, all variables and given/known data

Let $\mathfrak{g}$ be any lie algebra and $\mathfrak{h}$ be any ideal of $\mathfrak{g}$.

The canonical homomorphism $\pi : \mathfrak{g} \to \mathfrak{g/h}$ is defined $\pi (x) = x + \mathfrak{h}$ for all $x\in\mathfrak{g}$.

For any ideal $\mathfrak{f}$ of the quotient lie algebra $\mathfrak{g/h}$, consider the inverse image of $\mathfrak{f}$ in $\mathfrak{g}$ relative to $\pi$, that is: $$\pi ^{-1} (\mathfrak{f}) = \{X\in\mathfrak{g} : \pi (X)\in \mathfrak{f} \} .$$
Prove that $\pi ^{-1} (\mathfrak{f})$ is an ideal of the lie algebra $\mathfrak{g}$.

3. The attempt at a solution

See below

Last edited: Sep 22, 2011
2. Sep 22, 2011

### Ted123

This is my attempt:

Let $x\in\mathfrak{g}$ and $y\in\pi ^{-1}(\mathfrak{f})$.

We want to show that $[x,y]\in\pi ^{-1}(\mathfrak{f})$. That is, if $\pi ([x,y]) \in \mathfrak{f}$.

Now $\pi ([x,y]) = [\pi (x) , \pi (y)]$ since $\pi$ is a homomorphism.

And since $y\in \pi ^{-1}(\mathfrak{f})$, $\pi (y) \in \mathfrak{f}$.

But $[\pi (x) , \pi (y)] \in \mathfrak{f}$ for all $\pi (x) \in \mathfrak{g/h}$ and $\pi (y) \in \mathfrak{f}$ since $\mathfrak{f}$ is an ideal of $\mathfrak{g/h}$ and $\pi (x) = x+\mathfrak{h} \in \mathfrak{g/h}$.

Therefore $\pi ([x,y])\in\mathfrak{f}$ and $\pi ^{-1}(\mathfrak{f})$ is an ideal of $\mathfrak{g}$.

Can anyone spot any mistakes or anything I've done wrong?

Last edited: Sep 22, 2011