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There is an absolutely unexpected (to me, anyway) construction of Mikhalev and Zolotykh Mikhalev--Umirbaev--Zolotykh of a Lie algebra $L$ in characteristic $p$ which is not free but whose universal enveloping $U(L)$ is a free Lie algebra. The definition is very simple: it has three generators $a,b,c$ and one relation $a=[b,c]+\mathrm{ad}(a)^p(c)$.
[The universal enveloping is obviously free since $\mathrm{ad}(a)^p=\mathrm{ad}(a^p)$, so the relation becomes $a=[b+a^p,c]$, and after applying the automorphism $a\mapsto a'=a, b\mapsto b'=b+a^p, c\mapsto c'=c$, we have $a'=[b',c']$, hence the universal enveloping algebra is obviously freely generated by $b'$ and $c'$. The proof of non-freeness of $L$ is much more subtle, using Fox derivatives and stuff like that.]
There is an absolutely unexpected (to me, anyway) construction of Mikhalev and Zolotykh of a Lie algebra $L$ in characteristic $p$ which is not free but whose universal enveloping $U(L)$ is a free Lie algebra. The definition is very simple: it has three generators $a,b,c$ and one relation $a=[b,c]+\mathrm{ad}(a)^p(c)$.
[The universal enveloping is obviously free since $\mathrm{ad}(a)^p=\mathrm{ad}(a^p)$, so the relation becomes $a=[b+a^p,c]$, and after applying the automorphism $a\mapsto a'=a, b\mapsto b'=b+a^p, c\mapsto c'=c$, we have $a'=[b',c']$, hence the universal enveloping algebra is obviously freely generated by $b'$ and $c'$. The proof of non-freeness of $L$ is much more subtle, using Fox derivatives and stuff like that.]