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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?

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  • $\begingroup$ I guess Serre quantified $p$. Is it before or after "there exists"? As in your previous post, a page number would be welcome. $\endgroup$
    – YCor
    Commented Mar 1, 2018 at 13:58
  • $\begingroup$ Sorry, pg. 37. No, Serre did not, as I can see, quantify $p.$ I do not think it is a mistranslation issue, since the same statement appears in the french edition on pg. 34. $\endgroup$
    – user44591
    Commented Mar 1, 2018 at 14:01
  • $\begingroup$ OK. Reading the itemization in Serre's exercise (p34 of French edition), it seems pretty clear that he means: that we have to construct $c$ working for all $p$. So actually the question is equivalent to showing that for every $n$, there exists $c\in{\mathbf{Z}}^{\{1,\dots,n\}^3}$ alternating in the first 2 coordinates, such that the Lie $\mathbf{Z}$-algebra presentation $\langle x_1,\dots,x_n|[x_i,x_j]=\sum c(i,j,k)x_k,\forall i,j\rangle$ yields a Lie algebra $V$ such that $V/pV$ is zero for every prime $p$. It's even better if we get $V=0$, of course. $\endgroup$
    – YCor
    Commented Mar 1, 2018 at 14:07
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    $\begingroup$ @YCor Unless I made a mistake above, I think the case $n=3$ should be done. But maybe I made a mistake, and of course, I did not use any assumption on the characteristic (and maybe that is not needed?) $\endgroup$
    – user44591
    Commented Mar 1, 2018 at 14:20
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    $\begingroup$ I'll post a cw answer so that you can accept an answer (after editing it if you wish) and the question is considered settled by the system. $\endgroup$
    – YCor
    Commented Mar 1, 2018 at 22:17

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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 $x_n=0$.

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