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][1]
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][2] (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:

https://mathoverflow.net/questions/80025/is-the-duflo-map-for-lie-algs-unique

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][3] 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. 

 


  [1]: http://en.wikipedia.org/wiki/Duflo_isomorphism
  [2]: http://en.wikipedia.org/wiki/Universal_enveloping_algebra
  [3]: http://en.wikipedia.org/wiki/Capelli's_identity