Roberts’ counterexample to finite generation
Phenomenon
A nonlinear action of the additive group \(\mathbb G_a\) on \(\mathbb A^7\) has a non-finitely generated ring of invariants (Roberts 1990). In contrast to Nagata’s counterexample, the group is one-dimensional and the action is not linear.
Setup
Let \(k\) be a field of characteristic zero and set \[ B=k[x_1,x_2,x_3,y_1,y_2,y_3,z]. \] For \(t\in\mathbb G_a\), define \[ \begin{aligned} t\cdot(x_1,x_2,x_3,y_1,y_2,y_3,z) =\bigl(&x_1,x_2,x_3,\\ &y_1+tx_1^3,\ y_2+tx_2^3,\ y_3+tx_3^3,\\ &z+t(x_1x_2x_3)^2\bigr). \end{aligned} \] Because the \(x_i\) remain fixed, composing two such transformations adds their parameters. Thus this formula defines an algebraic group action.
Its corresponding derivation is \[ D=x_1^3\frac{\partial}{\partial y_1} +x_2^3\frac{\partial}{\partial y_2} +x_3^3\frac{\partial}{\partial y_3} +(x_1x_2x_3)^2\frac{\partial}{\partial z}. \] It lowers total degree in \(y_1,y_2,y_3,z\), so it is locally nilpotent: every polynomial is killed by some iterate of \(D\). In characteristic zero, invariance is equivalent to being killed by this derivation. Roberts’ theorem states that \[ A:=B^{\mathbb G_a}=\ker D \quad\text{is not finitely generated over }k. \] See Roberts (1990) for the original construction and Dufresne and Kraft (2015, sect. 9) for this formulation.
Geometrically, the action translates each fibre of the projection to the three \(x\)-coordinates in a direction depending polynomially on the base point. Away from the origin of that base, the direction is nonzero. Over the origin, the whole fibre is fixed: the fixed-point locus is the affine four-space defined by \(x_1=x_2=x_3=0\).
Construction
The following polynomials are invariant: \[ x_i,\qquad u_{ij}=x_i^3y_j-x_j^3y_i,\qquad b_i=x_i z-x_j^2x_k^2y_i, \] where \(\{i,j,k\}=\{1,2,3\}\) in the last expression. In each case, applying \(D\) produces two terms that cancel.
The proof below uses Roberts’ existence lemma (Roberts 1990, Lemma 3; Freudenburg 2001, sect. 3.3): for every \(n\geq0\), there is an invariant \[ \beta_n=x_1z^n+\text{terms of lower }z\text{-degree}. \] The lemma is used without proof.
Proof of non-finite generation
Set \(I=(x_1,x_2,x_3)B\). First, every invariant is constant modulo \(I\). To see this, take \(f\in A\) and write its part of total \(x\)-degree zero as \(f_0(y_1,y_2,y_3,z)\). The part of \(Df=0\) of \(x\)-degree three gives \[ \sum_{i=1}^3x_i^3\frac{\partial f_0}{\partial y_i}=0. \] Hence \(f_0\) is independent of the \(y_i\). Next compare the coefficient of \(x_1^2x_2^2x_3^2\) in \(Df\). Contributions from the \(y_i\)-derivatives are divisible by some \(x_i^3\), so this coefficient is precisely \(\partial f_0/\partial z\). It also vanishes, and characteristic zero implies \(f_0\in k\).
Now suppose \(A\) had finitely many algebra generators. Subtracting their constants gives generators \(g_1,\ldots,g_r\in I\). Modulo \(I^2\), every polynomial in these generators is a constant plus a linear combination of their images: all products of two generators vanish.
Consequently, the image of \(A\) in \(B/I^2\) would be finite-dimensional over \(k\). But the images of the \(\beta_n\) are linearly independent. In any finite relation, the term with largest index has a unique highest power of \(z\), with nonzero coefficient \(x_1\) modulo \(I^2\). This is a contradiction.
The argument shows that the \(\beta_n\) remain linearly independent modulo \(I^2\), where products of invariants lying in \(I\) vanish.
Outro
In characteristic zero, Weitzenböck’s theorem gives finite generation for every linear \(\mathbb G_a\)-action (Freudenburg 2001, sect. 1). Roberts’ example therefore cannot be made linear by a polynomial change of coordinates.