If $g$ is Lie algebra over field char(k)=0,
then the following facts are well-known:


1) S(g) and U(g) are isomorphic as $g$-modules. (Symmetrization map S(g)->U(g) gives isomorphism).


2) S(g)^g and ZU(g)=U(g)^g are isomorphic as commutative algebras. (The [Duflo map](http://en.wikipedia.org/wiki/Duflo_isomorphism)
defines is isomorphism which is combination of symmetrization map with some intricate corrections by terms of smaller degree). 


Both facts are based on the symmetrization.
If we consider $g$ over field char(k)$\ne$0, there is NO symmetrization map.
So I wonder the following:


**Question** are the facts above true for Lie algebras over char(k)$\ne$0 ?




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Notations


U(g) - [universal enveloping algebra](http://en.wikipedia.org/wiki/Universal_enveloping_algebra) (non-commutative associative algebra defined by relations $[x\_i, x\_j] = \sum\_k c\_{ij}^k x\_k$, for any linear bases $x\_k$ of $g$.


S(g) - symmetric algebra of $g$ (defined as $k[x\_1...x\_n]$ for any bases $x\_k$ of $g$.


S(g)^g, U(g)^g means subspaces of g-invariants (g act by zero).


Symmetrization map is defined as $S(x\_1..x\_k) = 1/k! \sum\_{\sigma} \prod\_l {x\_{\sigma(l)} $.


Duflo map is not so easy to write so let me just mention some MO questions:


[Is the Duflo map for Lie algs. unique ?](https://mathoverflow.net/questions/80025/is-the-duflo-map-for-lie-algs-unique)


[Capelli determinant = Duflo ( determinant) - was it known ?](https://mathoverflow.net/questions/92348/capelli-determinant-duflo-determinant-was-it-known) 




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Remark


For the case of gl\_n there are so-called [Capelli generators](http://en.wikipedia.org/wiki/Capelli's_identity) of the center of U(g),
which can be defined over any char. So I think that the center of U(gl) is the same as 
for char(k)=0 the result is true. (It is NOT the same as in char=0, as I mistakenly wrote first, since there should be generators corresponding to a^p (which are S(g)^g for any a)).


I think the same is true for other classical semi-simples - there are analogous Capelli like formulas. 



Are S(g) and U(g) isomorphic as g-modules for g Lie algebra over F_p ? Are S(g)^g and U(g)^g isomorphic as com.algs ? 
Question 2 has a negative answer. Indeed, let $L =sl(2,k)$ where $k$ is an algebraically closed field of characteristic $p>3$ and let {$e,h,f$} be the standard basis of $L$. Then it is well-known (and was first proved by Rudakov-Shafarevich in the late 60s) that the centre $Z(L)$ of $U(L)$ is generated by $E=e^p$, $F=f^p$, $H=h^p-h$ and a Casimir element $C$ subject to the relation
$$4EF+H^2=\prod\_{i\in{\mathbb F}\_p}(C-i^2)=C^p-2C^{(p+1)/2}+C.$$ Furthermore, the maximal spectrum of $Z(L)$ is a hypersurface in ${\mathbb A}^4$ given by the above equation. On the other hand, it follows from a result of Brown-Goodearl that the Azumaya locus of $U(L)$ coincides with the smooth locus of ${\rm Specm\ }Z(L)$. So the modular representation theory of $sl(2,k)$ yields that ${\rm Specm\ }Z(L)$
has exactly $(p-1)/2$ singular points (one can also see this directly by differentiating the above equation).


On the other hand, it is also known (and not difficult to see directly) that $S(L)^L$ is generated by
$X=e^p$, $Y=f^p$, $Z=h^p$ and $\Omega=4ef+h^2$ subject to the relation $4XY+Z^2=\Omega^p$. Moreover, ${\rm Specm\ }S(L)^L$ is a hypersurface in ${\mathbb A}^4$ given by the above equation. This hypersurface has a unique singular point at the origin.
Since $(p-1)/2>1$ under our assumptions on $p$, we see that the $k$-algebras $Z(L)$ and $S(L)^L$ cannot be isomorphic.


As far as I know the answer to Question 1 is unknown, in general, although there do exist finite dimensional Lie algebras $L$ for which $L\subset U\_1(L)$ is NOT a direct summand of $U(L)$ (if $U(L)$ is isomorphic to $S(L)$ as $ad(L)$-modules, then $U(L)$ must contain an isomorphic copy of $L$ as a direct summand, but that copy doesn't have to lie in the first component $U\_1(L)$ of the canonical filtration of $U(L)$). 


If we require the stronger condition that an isomrphism $S(L)\rightarrow U(L)$ sends $S^1(L)$ onto $L\subset U\_1(L)$ (which is obviously true for the symmetrisation map), then the answer to Question 1 is NO. For that condition to hold the Lie algebra $L$ must admit at least one $[p]$-th power map, i.e. must be ${\it restrictable}$ (this observation is due to Michel Duflo). The majority of finite dimensional simple Lie algebras are non-restrictable, and the smallest example is the $3$-dimensional simple Lie algebra over an algebraically closed field of characteristic $2$.


