# The center of nilpotent Lie algebra

Let $L$ be a non-abelian nilpotent Lie algebra and $Z(L)$ be its center. Is it possible for $Z(L)$ to be a maximal ideal in $L$?

An attempt: Since $L$ is not abelian then there exists $x\in L, x\notin Z(L)$, now $\langle x\rangle_F + Z(L)$ is an ideal of $L$, by maximality of $Z(L)$ this ideal is $L$. Thus $dim\ L/Z(L)=1$ and the nilpotency class of $L$ is $2$ i.e $[L,[L,L]]=0$ if this implies that $Z(L)=[L,L]$ then we are done. since for a nilpotent Lie algebra $L$ we must have $dim\ L/[L,L]>1$.

• what about L abelian? – Tim kinsella Feb 16 '16 at 8:00
• sorry , i forgot to exclude abelian case! – Ronald Feb 16 '16 at 8:00
• By centralizer you mean center? – Mariano Suárez-Álvarez Feb 16 '16 at 8:22
• Yes .. I changed it – Ronald Feb 16 '16 at 8:24
• Why is what you say is an ideal an ideal? – Mariano Suárez-Álvarez Feb 16 '16 at 8:38

Suppose such an $L$. Then $[L,L]+Z(L)$ is an ideal (why is $[L,L]+Z$ not all of $L$? because $[[L,L],L]$ must be strictly smaller than $[L,L]$, since $L$ is nilpotent.. rt?). Since $Z$ is maximal, $[L,L]\subset Z$, but then $Z$ must have codimension one if it is to be maximal. But then $L$ is abelian. contradiction.
• Sorry, could you please explain why if $[L,L]\subset Z$ then $L$ is abelian? – Ronald Feb 16 '16 at 8:41
• Suppose $[L,L]\subset Z$. Every linear subspace containing $[L,L]$ is an ideal, so if $Z$ is going to be maximal, it cannot be properly contained in a proper subspace of $L$. So $dim(Z) = dim(L)-1$. So $L= Z\oplus \mathbb{R}x$ where $x\notin Z$. Now $[\mathbb{R}x, Z] = [\mathbb{R}x, \mathbb{R}x]= [Z,Z]=0$, so $[L,L]=0$. – Tim kinsella Feb 16 '16 at 8:45