1
$\begingroup$

I am currently going through a proof of Ado's theorem. I am stuck in one step. Suppose $\mathfrak{g}$ is a solvable Lie algebra which is not nilpotent. Then one can show that there is an ideal $\mathfrak{a}$ of codimension 1 which contain $nil\mathfrak{g}$. Then we have a complementary vector subspace $\mathfrak{h}$ so that $\mathfrak{g}=\mathfrak{a} \oplus \mathfrak{h}$ as vector space. Then the nilradical $nil\mathfrak{g} = nil\mathfrak{a}$. Why is this true? Clearly, since $nil\mathfrak{g} \subset \mathfrak{a}$ we have $nil\mathfrak{g} \subset nil\mathfrak{a}$. How can I get the other inclusion?

$\endgroup$
0
$\begingroup$

The quotient ${\cal g}/nil(g)$ is a reductive algebra. Since ${\cal g}$ is solvable, this reductive algebra is commutative. This implies that every vector space which contains $nil({\cal g})$ is an ideal of ${\cal g}$. We deduce that $nil({\cal a})$ is a nilpotent ideal of ${\cal g}$ since $nil({\cal g})$ is the maximal nilpotent ideal of ${\cal g}$, we deduce that $nil({\cal a})\subset nil({\cal g})$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.