Show that submodules of $L$ are precisely the ideals of $L$

Question

Let $L$ be a Lie algebra. We can consider $L$ into an $L-$module via the adjoint representation $$ad: L \rightarrow gl(L) , \space \space (ad \space x)y = [x,y]$$

I need to prove that submodules of $L$ are precisely the ideals of $L$.

So, let $I$ be an ideal of $L$. I need to show that for every $x \in L$ and for every $k \in I ,$ we have $ [x,k] \in I$.

Now, I am using the following statement - if $x \in L$ and $y \in I \cap J$ , where $I$ and $J$ are ideals of $L$, then $[x,y] \in I \cap J$.

Here I am considering $J=L$. Is this enough or am I missing something?

"I need to show that for every $x \in L$ and for every $k \in I ,$ we have $ [x,k] \in I$." This is just the definition that $I$ is an ideal. But you are assuming this already. — Dietrich Burde, Mar 12 '17 at 17:28

score 3 · Accepted Answer · answered Mar 12 '17 at 17:24

3

There is nothing to prove since it holds by definition. If $I$ is a submodule, then $x.I={\rm ad}(x)(I)\subseteq I$ for all $x\in L$, which says that $[x,I]\subseteq I$ for all $x\in L$, i.e., that $I$ is a Lie algebra ideal by definition.

answered Mar 12 '17 at 17:24

Dietrich Burde

130,978

Show that submodules of $L$ are precisely the ideals of $L$

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