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Here is an example of $G$ for which $k[\mathfrak{g}]$ is not a free $A$-module, where $A=k[\mathfrak{g}]^G$. A=k[\mathfrak{g}]^G$(I hope there is no mistake in what follows). Let$G=\mathrm{PGL}_p$where$p\ge 5$, a simple algebraic group of adjoint type with$\mathfrak{g}=\mathfrak{pgl}_p$. The Chevalley Restriction Theorem holds for$G$as follows, e.g., from the Springer-Steinberg paper (see Ch. II, 3.17). Then we have that$A\cong k[\mathfrak{t}]^W$where$\mathfrak{t}$is the image of the diagonal matrices in$\mathfrak g$. Of course,$W\cong\mathfrak{S}_p$and$\mathfrak{t}$is the quotient of the natural$\mathfrak{S}_p$-module$V$with basis${e_1,\ldots, e_p}$, permuted by$\mathfrak{S}_p$, by its trivial submodule$k(e_1+\cdots+e_p)$. By a result of Gregor Kemper, for$p\ge 5$the ring$S(V^*)^{\mathfrak{S}_p}\cong k[\mathfrak{t}]^W$is not Cohen-Macaulay (see Corollary 2.8 and Example 2.9 in Kemper's paper published in J. Algebra, Vol. 215 (1999), 33--351). In particular,$k[\mathfrak{g}]^W$is not a polynomial algebra, which resolves in the negative Problem 3.18 in the Springer-Steinberg paper. Since$\mathfrak g$is a restricted Lie algebra and$\mathfrak t$is its toral Cartan subalgebra of dimension$p-1$, it follows from a result I proved that there exist homogeneous$f_0,\ldots, f_{p-2}\in A$with${\rm deg}\ f_i=p^{p-1}-p^i$such that$x^{[p]^{p-1}}=\sum_{i=0}^{p-2}f_i(x)x^{[p]^i}$for all$x\in\mathfrak{g}$. Furthermore, the zero locus of the$f_i$'s equals the nullcone$\mathcal N$of$\mathfrak g$. Since$\mathcal N$is irreducible of codimension$p-1$it follows that {$f_0,\ldots, f_{p-2}$} a regular sequence in$k[\mathfrak{g}]$and$k[\mathfrak{g}]$is a free module over$A_0:= k[f_0,\ldots, f_{p-2}]$. Since$A\cong k[\mathfrak{t}]^W$has Krull dimension$p-1$, the$f_i$'s form a homogeneous sytem of parameters for$A$. If$k[\mathfrak{g}]$is free over$A$then$A$is finitely generated and projective over$A_0$. But then$A$is free over$A_0$implying that$A$is Cohen-Macaulay. This contradiction shows that Kostant's freeness theorem fails in our case. 2 added 49 characters in body Here is an example of$G$for which$k[\mathfrak{g}]$is not a free$A$-module, where$A=k[\mathfrak{g}]^G$. Let$G=\mathrm{PGL}_p$where$p\ge 5$, a simple algebraic group of adjoint type with$\mathfrak{g}=\mathfrak{pgl}_p$. The Chevalley Restriction Theorem holds for$G$as follows, e.g., from the Springer-Steinberg paper (see Ch. II, 3.17). Then we have that$A\cong k[\mathfrak{t}]^W$where$\mathfrak{t}$is the image of the diagonal matrices in$\mathfrak g$. Of course,$W\cong\mathfrak{S}_p$and$\mathfrak{t}$is the quotient of the natural$\mathfrak{S}_p$-module$V$with basis${e_1,\ldots, e_p}$, permuted by$\mathfrak{S}_p$, by its trivial submodule$k(e_1+\cdots+e_p)$. By a result of Gregor Kemper, for$p\ge 5$the ring$S(V^*)^{\mathfrak{S}_p}\cong k[\mathfrak{t}]^W$is not Cohen-Macaulay (see Corollary 2.8 and Example 2.9 in Kemper's paper published in J. Algebra, Vol. 215 (1999), 33--351). In particular,$k[\mathfrak{g}]^W$is not a polynomial algebra, which resolves in the negative Problem 3.18 in the Springer-Steinberg paper. Let Since$x\in\mathfrak{gl}_p$\mathfrak g$ is a restricted Lie algebra and write $I$ for the identity matrix \mathfrak t$is its toral Cartan subalgebra of order dimension$p$. Let p-1$, it follows from a result I proved that there exist homogeneous $f_1(x),\ldots, f_{p-1}(x)$ be the coefficients of f_0,\ldots, f_{p-2}\in A$with$t^{p-1}, {\rm deg}\ ldots, t$in$\chi_x(t):=|tI-x|$. Since for all$\mu\in k$we have that $$|tI-(x+\mu I)|^p=|(t-\mu)^p-x^p|,$$ it is easy to see f_i=p^{p-1}-p^i$ such that $f_i(x+\mu I)=f_i(x)$ x^{[p]^{p-1}}=\sum_{i=0}^{p-2}f_i(x)x^{[p]^i}$for all$i$. So x\in\mathfrak{g}$. Furthermore, the $f_i$'s can be regarded as elements zero locus of $A$. Since $A\cong k[\mathfrak{t}]^W$ has Krull dimension $p-1$, the $f_i$'s form a homogeneous sytem equals the nullcone $\mathcal N$ of parameters for $A$. \mathfrak g$. Since the nilpotent cone of$\mathfrak{g}$\mathcal N$ is irreducible of codimension $p-1$ and coincides with the zero locus of the $f_i$'s it follows that {$f_0,\ldots, f_{p-2}$} a regular sequence in $k[\mathfrak{g}]$ and $k[\mathfrak{g}]$ is a free module over $A_0:= k[f_1,\ldotsk[f_0,\ldots, f_{p-1}]$f_{p-2}]$. Since$A\cong k[\mathfrak{t}]^W$has Krull dimension$p-1$, the$f_i$'s form a homogeneous sytem of parameters for$A$. If$k[\mathfrak{g}]$is free over$A$then$A$is finitely generated and projective over$A_0$. But then$A$is free over$A_0$implying that$A$is Cohen-Macaulay. This contradiction shows that Kostant's freeness theorem fails in our case. 1 Here is an example of$G$for which$k[\mathfrak{g}]$is not a free$A$-module, where$A=k[\mathfrak{g}]^G$. Let$G=\mathrm{PGL}_p$where$p\ge 5$, a simple algebraic group of adjoint type with$\mathfrak{g}=\mathfrak{pgl}_p$. The Chevalley Restriction Theorem holds for$G$as follows, e.g., from the Springer-Steinberg paper (see Ch. II, 3.17). Then we have that$A\cong k[\mathfrak{t}]^W$where$\mathfrak{t}$is the image of the diagonal matrices in$\mathfrak g$. Of course,$W\cong\mathfrak{S}_p$and$\mathfrak{t}$is the quotient of the natural$\mathfrak{S}_p$-module$V$with basis${e_1,\ldots, e_p}$, permuted by$\mathfrak{S}_p$, by its trivial submodule$k(e_1+\cdots+e_p)$. By a result of Gregor Kemper, for$p\ge 5$the ring$S(V^*)^{\mathfrak{S}_p}\cong k[\mathfrak{t}]^W$is not Cohen-Macaulay (see Corollary 2.8 and Example 2.9 in Kemper's paper published in J. Algebra, Vol. 215 (1999), 33--351). In particular,$k[\mathfrak{g}]^W$is not a polynomial algebra, which resolves in the negative Problem 3.18 in the Springer-Steinberg paper. Let$x\in\mathfrak{gl}_p$and write$I$for the identity matrix of order$p$. Let$f_1(x),\ldots, f_{p-1}(x)$be the coefficients of$t^{p-1}, \ldots, t$in$\chi_x(t):=|tI-x|$. Since for all$\mu\in k$we have that $$|tI-(x+\mu I)|^p=|(t-\mu)^p-x^p|,$$ it is easy to see that$f_i(x+\mu I)=f_i(x)$for all$i$. So the$f_i$'s can be regarded as elements of$A$. Since$A\cong k[\mathfrak{t}]^W$has Krull dimension$p-1$, the$f_i$'s form a homogeneous sytem of parameters for$A$. Since the nilpotent cone of$\mathfrak{g}$is irreducible of codimension$p-1$and coincides with the zero locus of the$f_i$'s it follows that$k[\mathfrak{g}]$is a free module over$A_0:= k[f_1,\ldots, f_{p-1}]$. If$k[\mathfrak{g}]$is free over$A$then$A$is finitely generated and projective over$A_0$. But then$A$is free over$A_0$implying that$A\$ is Cohen-Macaulay. This contradiction shows that Kostant's freeness theorem fails in our case.