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Let $k$ be an algebraically closed field of characteristic $p>0$, and let $\mathfrak{g}$ be a restricted Lie algebra, with restricted enveloping algebra $u(\mathfrak{g})$. We can place a Hopf algebra structure on $u(\mathfrak{g})$ by requiring that all elements in $\mathfrak{g}\subset u(\mathfrak{g})$ be primitive. With this structure, might there be primitive elements in $u(\mathfrak{g})\setminus\mathfrak{g}$, or is it the case that the set of primitive elements in $u(\mathfrak{g})$ is precisely $\mathfrak{g}$?

I know that in characteristic $0$, it is true that the set of primitive elements in the universal enveloping algebra $U(\mathfrak{g})$ is precisely the Lie algebra $\mathfrak{g}$, but I'm not sure if this holds for positive characteristic.

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  • $\begingroup$ As a disclosure, I must mention that my characteristic $p$ knowledge is not the greatest. So, I will put a potential answer in the comments in the hope that someone with greater expertise can validate or invalidate it. Take the element (for $x \in \mathfrak{g}$) $$x^{p} = x \otimes \cdots \otimes x.$$ If $\Delta$ is the comultiplication, then $$\Delta(x^{p}) = \Delta(x)^{p} = (x \otimes 1 + 1 \otimes x)^{p} = \sum_{i=0}^{p} \binom{p}{i} (x^{i} \otimes x^{p-i}) = x^{p} \otimes 1 + 1 \otimes x^{p}.$$ $\endgroup$ – Siddharth Venkatesh Apr 24 '14 at 7:05
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    $\begingroup$ @SiddharthVenkatesh: Thanks for the comment! You're right to say that $p$th powers of primitive elements are primitive. However, in $u(\mathfrak{g})$ we have the relation $x^p=x^{[p]}\in\mathfrak{g}\subset u(\mathfrak{g})$, where $[p]$ is the $p$-restriction map of the restricted Lie algebra $\mathfrak{g}$. Thus, for all elements $x\in\mathfrak{g}$, we have $x^p\in\mathfrak{g}$. $\endgroup$ – Jared Apr 24 '14 at 16:48
  • $\begingroup$ Oh. Thanks for informing me. I really don't know much about what restricted meant for the universal enveloping algebra. So the enveloping algebra here has the additional relations $x^{[p]} = x^{p}.$ Is that the only new relation? $\endgroup$ – Siddharth Venkatesh Apr 24 '14 at 18:34
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    $\begingroup$ @SiddharthVenkatesh: Yes. The restricted enveloping algebra is the quotient of the universal enveloping algebra by the ideal generated by all elements of the form $x^p-x^{[p]}$ for $x\in\mathfrak{g}$. $\endgroup$ – Jared Apr 24 '14 at 18:49
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This is a very old question but it may be useful for somebody reading this post to know that the set of primitive elements in $u(\mathfrak{g})$ is precisely $\mathfrak{g}$.

A generalisation of this statement is proven for colour Lie $p$-superalgebras in Chapter 3, Theorem 2.11 of ``Infinite dimensional Lie superalgebras'' by Bahturin et al.

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