Reading through Serre's book on Galois cohomology, I encounted the following problem:  

Let $n$ be an integer. Consider families of integers $c(i,j,k)$ with $i,j,k \in [1,n]$ which are alternating in $(i,j).$ Show that for every $n \geq 3,$ there exists such a family with the following property:  

(\*) - If the elements $x\_1,\ldots,x\_n$ of a Lie algebra of characteristic $p$ satisfy the relations $$[x\_i,x\_j] = \sum\_k c(i,j,k) x\_k,$$ then $x\_i =0$ for all $i.$ 


For $n=3,$ I managed to work this through by hand. What I did was to let $[x\_1,x\_2] = x\_2, [x\_2,x\_3] = x\_3$ and $[x\_3,x\_1] = x\_1.$ Then using the Jacobi relations, we see that  

$$[x\_1,[x\_2,x\_3]]+[x\_3,[x\_1,x\_2]]+[x\_2,[x\_3,x\_1]] = [x\_1,x\_3] + [x\_3,x\_2]+[x\_2,x\_1] = -x\_1-x\_3-x\_2 = 0.$$
We then have a non-zero relation between the $x\_i$'s. One then shows that this forces $x\_1=x\_2=x\_3=0$ (assuming I have not made a mistake) 


I have not worked through this for higher $n$'s. However, I am somewhat troubled with my answer since I did not use the hypothesis that $k$ has characteristic $p>0.$ Could someone verify that the answer depends on the characteristic, and if so, indicate in some way (or post a complete answer, it is up to you) in what way it enters?



Serre Galois Cohomology exercise on Lie algebras
As you do, set $[x\_1,x\_2] = x\_2$, and $[x\_3,x\_1] = x\_1$, and $[x\_2,x\_n] = x\_n$ for $n\ge 3$ (and what you like for other brackets, provided the bracket is alternating). 


As you have observed, these conditions (for $n\le 3$) force $x\_1=x\_2=x\_3=0$ in any Lie ring, and the remaining $[x\_2,x\_n] = x\_n$ force $x\_n=0$. 


