Example 50.19.1. Here is an example where we do not have a trace map on de Rham complexes. For example, consider the $\mathbf{C}$-algebra $B = \mathbf{C}[x, y]$ with action of $G = \{ \pm 1\} $ given by $x \mapsto -x$ and $y \mapsto -y$. The invariants $A = B^ G$ form a normal domain of finite type over $\mathbf{C}$ generated by $x^2, xy, y^2$. We claim that for the inclusion $A \subset B$ there is no reasonable trace map $\Omega _{B/\mathbf{C}} \to \Omega _{A/\mathbf{C}}$ on $1$-forms. Namely, consider the element $\omega = x \text{d} y \in \Omega _{B/\mathbf{C}}$. Since $\omega $ is invariant under the action of $G$ if a “reasonable” trace map exists, then $2\omega $ should be in the image of $\Omega _{A/\mathbf{C}} \to \Omega _{B/\mathbf{C}}$. This is not the case: there is no way to write $2\omega $ as a linear combination of $\text{d}(x^2)$, $\text{d}(xy)$, and $\text{d}(y^2)$ even with coefficients in $B$. This example contradicts the main theorem in [Zannier].

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