Where does the algebraic closure enter into Block's Theorem?

Question

When applying Block's Theorem on the structure of differentiably simple rings to Lie algebras most authors require an algebraically closed field, but I can see no reference to algebraic closure in Block's paper. What extra does the algebraic closure give? I apologise for asking such a basic (and perhaps easy) question but I've talked to a number of other algebraists and asked this on another forum without getting any answer.

Answer 1

Block's theorem does not require the base field $k$ to be algebraically closed but one has to be careful when $k$ is imperfect. Then $k$ will admit field extensions of the form $K=k(a)$ with $a\not\in k$ and $a^p\in k$. Note that $K$ will become a truncated polynomial ring in one variable over a field extension $L=k(b)$ with $b^p=a^p$. If $p>2$ then the $k$-Lie algebra $\mathfrak{g}=\mathfrak{sl}_2\otimes_k K$ is simple of dimension $3p$ (but not absolutely simple) and ${\rm Der}_k(K)$ is a nontrivial $k$-form of the Witt algebra $W(1;\underline{1})$ (such twisted forms can only exist over imperfect fields). By Block's theorem, the derivation algebra of $\mathfrak{g}$ is the semidirect product ${\rm Der}_k(K)\ltimes {\rm ad}\,\mathfrak{g}$ of dimension $4p$ over $k$ but it is not isomorphic to ${\rm Id}\otimes_k W(1;\underline{1})\ltimes ({\rm ad}\,\mathfrak{sl}_2)\otimes_k k[X]/(X^p)$ over $k$. However, such an isomorphism will exist over $L$.