Let $k$ be a field of characteristic $p>0$, and let $(\mathfrak{g},[\cdot,\cdot],(\cdot)^{[p]})$ be a finite-dimensional restricted Lie algebra over $k$. Let $u(\mathfrak{g})$ be the restricted enveloping algebra defined as


where $T(\mathfrak{g})$ is the tensor algebra and $I$ is the ideal generated by elements of the form $[a,b]-a\otimes b-b\otimes a$ and $a^{[p]}-\underbrace{a\otimes\ldots\otimes a}_{p-\text{times}}$.

I'd like to show that $u(\mathfrak{g})$ is a local $k$-algebra, with unique maximal ideal $\mathfrak{m}$ satisfying $a^{[p]}=0$ for all $a\in\mathfrak{m}$. My idea is to define $\mathfrak{m}$ to be the ideal generated by all elements $a\in\mathfrak{g}$ such that $a^{[p]}=0$, but it is not clear to me that this is the unique maximal ideal, and that every element $a\in\mathfrak{m}$ has trivial restriction, $a^{[p]}=0$. Is what I'm trying to prove true? If so, can you help me prove it?


No, this is false. A local algebra $A,m$ has a unique simple module $A/m$. But restricted Lie algebras can have lots of different simple modules, e.g. restricted $\mathfrak{sl}_2(k)$ has $p$ simple modules, of dimensions $1,2,\ldots,p-1$ which are obtained by reducing the usual simples in characteristic zero mod $p$.

Further, the ideal generated by elements with $p$th power zero may contain elements whose $p$th power is not zero. Take $u(\mathfrak{sl}_2(k))$ again, which contains $e,f$ with $e^{[p]}=f^{[p]}=0$. Then $ef-fe=h$ and $h^{[p]}=h$.


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