Why is this $ \mathbb{G}_{a} $ bundle trivial

Question

Please tell me why the following example of a principal $ \mathbb{G}_{a} $-bundle over an affine ring is trivial. Let $ \{x_{1},x_{2},x_{3}\} $ a basis of $ \mathbf{V}^{\ast} $, $ c_{1}(t),c_{2}(t) $ two non-zero, linearly independent, additive polynomials such that the only common root is zero, and $ \beta: \mathbb{G}_{a} \to \operatorname{GL}(\mathbf{V}) $ be the representation whose co-action is the following one: \begin{align*} \beta^{\sharp}(x_{1}) &= x_{1} \\ \beta^{\sharp}(x_{2}) &= x_{2} \\ \beta^{\sharp}(x_{3}) &= x_{3}+c_{2}(t)x_{2}+c_{1}(t)x_{1}. \end{align*} The stabilizer of any closed point $ y \in D(x_{1}x_{2}) $ is equal to $ 0 $. Therefore $ D(x_{1}x_{2}) $ is a principal $ \mathbb{G}_{a} $-bundle over its image in $ \operatorname{Spec}(k[x_{1},x_{2},x_{3}]^{\mathbb{G}_{a}}) $. By a theorem of Demazure, Gabriel, and others it is a trivial $ \mathbb{G}_{a} $-bundle.

If a variety $ X $ is a trivial $ \mathbb{G}_{a} $-bundle over $ Y $, then there is clearly a separable $ \mathbb{G}_{a} $-equivariant map from $ X $ to $ \mathbb{G}_{a} $. If there is a separable $ \mathbb{G}_{a} $-equivariant map $ \phi: X \to \mathbb{G}_{a} $, then let $ Y $ be the fibre over $ 0 $. There is an isomorphism $ \psi: \mathbb{G}_{a} \times Y \to X $ which sends $ (t,y) $ to $ t \ast y $. The inverse to $ \psi $ is the map which sends $ x $ to $ \left(\phi(x), \left(-\phi(x)\right) \ast x\right) $. This shows that a variety $ X $ is a trivial $ \mathbb{G}_{a} $-bundle over $ X//\mathbb{G}_{a} $ if and only if there is a separable, $ \mathbb{G}_{a} $-equivariant morphism $ \phi: X \to \mathbb{G}_{a} $.

If $ D(x_{1}x_{2}) $ is a trivial $ \mathbb{G}_{a} $-bundle over $ \operatorname{Spec}(k[x_{1},x_{2},x_{3}]_{x_{1}x_{2}}^{\mathbb{G}_{a}}) \cong \operatorname{Spec}(k[x_{1},x_{2}]_{x_{1}x_{2}}) $, then there is a $ \mathbb{G}_{a} $-equivariant morphism $ \phi: D(x_{1}x_{2}) \to \mathbb{G}_{a} $.

If $ \Delta_{\mathbb{G}_{a}}: \mathbb{G}_{a} \times \mathbb{G}_{a} \to \mathbb{G}_{a} $ is the multiplication morphism of $ \mathbb{G}_{a} $, then the existence of the morphism $ \phi $ means that \begin{equation*} \Delta_{\mathbb{G}_{a}} \circ (\operatorname{id}_{\mathbb{G}_{a}}, \phi) = \phi \circ \beta \end{equation*} A consequence of this is that if $ \phi^{\sharp}(t) = g(X)/(x_{1}x_{2})^{e} $, then \begin{align*} \beta^{\sharp}(g(X)/(x_{1}x_{2})^{e}) &= \beta^{\sharp} \circ \phi^{\sharp}(t) \\ &= (\operatorname{id}_{k[t]} \otimes \phi^{\sharp}) \circ \Delta^{\sharp}(t) \\ &= (\operatorname{id}_{k[t]} \otimes \phi^{\sharp})(t \otimes 1 + 1 \otimes t) \\ &= t+g(X)/(x_{1}x_{2})^{e} \end{align*} Assume that $ g(X) = \sum_{j=0}^{d} x_{3}^{j} g_{j}(x_{1},x_{2}) $. Since \begin{align*} \beta^{\sharp}(g(X)-g_{0}(x_{1},x_{2})) &= \beta^{\sharp}(g(X))-g_{0}(x_{1},x_{2}) \\ &= g(X)+t(x_{1}x_{2})^{e}-g_{0}(x_{1},x_{2}) \\ &= g(X)-g_{0}(x_{1},x_{2})+t(x_{1}x_{2})^{e} \end{align*} the pair of polynomials $ (g(X)-g_{0}(x_{1},x_{2}),(x_{1}x_{2})^{e}) $ has the same property as the pair $ (g(X),(x_{1}x_{2})^{e}) $. As a result, we may assume for the next part that $ x_{3} $ divides $ g(X) $. Because $ c_{1}(t),c_{2}(t) $ are additive polynomials, the following identities hold: \begin{multline*} \beta^{\sharp}\left( x_{3}+c_{2}(-g(X)/(x_{1}x_{2})^{e})x_{2}+c_{1}(-g(X)/(x_{1}x_{2})^{e})x_{1} \right) \\ = \beta^{\sharp}(x_{3})+c_{2}(\beta^{\sharp}(-g(X)/(x_{1}x_{2})^{e}))x_{2}+c_{1}(\beta^{\sharp}(-g(X)/(x_{1}x_{2})^{e}))x_{1} \\ = x_{3}+c_{2}(t)x_{2}+c_{1}(t)x_{1}+c_{2}(-g(X)/(x_{1}x_{2})^{e}-t)x_{2}+c_{1}(-g(X)/(x_{1}x_{2})^{e}-t)x_{1} \\ = x_{3}+(c_{2}(t)-c_{2}(t))x_{2}+(c_{1}(t)-c_{1}(t))x_{1}+c_{2}(-g(X)/(x_{1}x_{2})^{e})x_{2}+c_{1}(-g(X)/(x_{1}x_{2})^{e})x_{1} \\ =x_{3}+c_{2}(-g(X)/(x_{1}x_{2})^{e})x_{2}+c_{1}(-g(X)/(x_{1}x_{2})^{e})x_{1}. \end{multline*} This shows that if $ \beta^{\sharp}(g(X)) = g(X)+t(x_{1}x_{2})^{e} $, then $ x_{3}+c_{2}(-g(X)/(x_{1}x_{2})^{e})x_{2}+c_{1}(-g(X)/(x_{1}x_{2})^{e})x_{1} $ is an invariant rational function. However, by our assumption that $ x_{3} $ divides $ g(X) $, we know that \begin{align*} x_{3} & \mid \left(x_{3}+c_{2}(-g(X)/(x_{1}x_{2})^{e})x_{2}+c_{1}(-g(X)/(x_{1}x_{2})^{e})x_{1}\right)\\ & \in k[x_{1},x_{2},x_{3}]_{x_{1}x_{2}}^{\mathbb{G}_{a}} \\ & = k[x_{1},x_{2}]_{x_{1}x_{2}}. \end{align*} This is a contradiction. What is the error here?

Answer 1

I believe that the problem with the argument is in the statement that $ D(x_{1}x_{2}) $ is a principal $ \mathbb{G}_{a} $-bundle over $ \operatorname{Spec}(k[x_{1},x_{2}]_{x_{1}x_{2}}) $. In order for $ D(x_{1}x_{2}) $ to be a principal $ \mathbb{G}_{a} $-bundle, it must be locally trivial. If I am correct, this $ \mathbb{G}_{a} $-space is nowhere trivial unless one of $ c_{1}(t) $ or $ c_{2}(t) $ is equal to zero and the other is equal to $ t $. Namely, there is no open sub-variety of $ D(x_{1}x_{2}) $ which is a trivial $ \mathbb{G}_{a} $-bundle over its image in $ \operatorname{Spec}(k[x_{1},x_{2}]_{x_{1}x_{2}}) $. Therefore, the theorem does not apply.