Let $\mathfrak{g}$ be positively graded Lie algebra over $\mathbb{Q}$, concentrated in even degrees. 



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> **Question:** If $\mathfrak{g}$ is not free, must there exist linearly independent elements $a,b\in\mathfrak{g}$ such that $[a,b]=0$?
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Zero-divisors in a graded Lie algebra
Consider the complex Lie algebra $\mathfrak{h}=\mathbb{C}[t,t^{-1}]\partial\_t$ with Lie bracket $$[t^m\partial\_t,t^n\partial\_t]=(n-m)t^{n+m-1}\partial\_t$$ ie the natural bracket that arises considering the natural action of $\mathfrak{h}$ on $\mathbb{C}[t,t^{-1}]$ by derivations. This is known as the [Witt algebra](http://en.wikipedia.org/wiki/Witt_algebra). If we give $t^n\partial\_t$ degree $n-1$ then this is a $\mathbb{Z}$-graded Lie algebra.


I claim that the $\mathbb{Q}$-Lie-subalgebra $\mathfrak{g}$ of $\mathfrak{h}$ spanned by the elements $t^n\partial\_t$ with $n\geq 3$ and odd does not contain linearly independent elements $a,b$ such that $[a,b]=0$. Indeed $\mathfrak{h}$ does not. 


Suppose $a=\sum\_{i\leq n}\lambda\_it^i\partial\_t$ and $b=\sum\_{j\leq n}\mu\_jt^j\partial\_t$ are linearly independent with $[a,b]=0$. We can assume WLOG that $\lambda\_n=1$. Suppose that $m\leq n$ is the largest number such that $\mu\_m\neq 0$. Then the degree ${n+m-2}$ piece of $[a,b]$ is simultaneously $\mu\_m (n-m)$ and $0$. Thus $m=n$ and by rescaling $b$ we may assume $\mu\_n=1$. 


Since $a\neq b$ we may now find $k$ maximal such that $\lambda\_k\neq \mu\_k$. Write $a=a\_1+a\_2$ and $b=b\_1+b\_2$ with $a\_1,b\_1$ a sum of homogenous terms of degree less than $k$ and $a\_2=b\_2$ a sum of homogenous terms at least $k$. Then $$0=[a,b]=[a\_1,b\_1]+[a\_2,b\_1]+[a\_1,b\_2]+[b\_2,a\_2]=[a\_1,b\_1]+[a\_2,b\_1-a\_1].$$ 


The leading term of $[a\_1,b\_1]$ has degree at most $(2k-2)$ and the leading term of $[a\_2,b\_1-a\_1]$ is $(\mu\_k-\lambda\_k)(n-k)t^{n+k-1}\partial\_t$ which has degree $n+k-2$ and so we obtain the desired contradiction. 


It is straightforward to verify that $\mathfrak{g}$ is not free.


