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 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 ?

Notations

U(g) - 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 ?

Capelli determinant = Duflo ( determinant) - was it known ?

Remark

For the case of gl_n there are so-called Capelli generators 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. Polynomial algebra of n-generators.

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