Are the two notions of free $\mathbb{G}_a$-actions equivalent?

Question

Consider a finitely generated integral $\mathbb{C}$-domain $B$. An algebraic $\mathbb{G}_a$-action on $X:=\mathrm{Spec}(\mathcal{O}(X))$ is equivalent to a locally nilpotent $\mathbb{C}$-derivation $$\partial:\mathcal{O}(X)\to \mathcal{O}(X)$$ and the corresponding co-action map is $$\mathcal{O}(X)\to \mathcal{O}(X)[u],\quad f\mapsto\sum_{n=0}^\infty\frac{\partial^nf}{n!}u^n.$$

Here I see two definitions about freeness of the action:

• Fixed point free: there is no $\mathbb{C}$-point $x\in X$ such that $g.x=x$ for all $g\in\mathbb{G}_a$. This is from Algebraic theory of locally nilpotent derivations, section 1.3, page 19. An equivalent condition is that $\mathrm{im}(\partial)$ generates the unital ideal $\mathcal{O}(X)=(\partial \mathcal{O}(X))$, section 1.5.2, page 34.
• AG free: The morphism $\mathbb{G}_a\times X\to X\times X$ is a closed immersion, or equivalently the ring map is surjective $$\mathcal{O}(X)\otimes_{\mathbb{C}}\mathcal{O}(X)\to\mathcal{O}(X)[u],\quad f\otimes g\mapsto \sum_{n=0}^\infty\frac{\partial^nf}{n!}g\;u^n. $$ This is from Geometric invariant theory, Definition 0.8 (iv).

It is easy to see that $$\textrm{AG-free}\implies\textrm{Fixed point free}. $$ My question is whether the converse also holds. A counterexample is also welcome.

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Freudenburg, Gene, Algebraic theory of locally nilpotent derivations, Encyclopaedia of Mathematical Sciences 136. Invariant Theory and Algebraic Transformation Groups 7. Berlin: Springer (ISBN 978-3-662-55348-0/hbk; 978-3-662-55350-3/ebook). xxii, 319 p. (2017). ZBL1391.13001.

Mumford, D.; Fogarty, J.; Kirwan, F., Geometric invariant theory., Ergebnisse der Mathematik und ihrer Grenzgebiete. 2. Folge. 34. Berlin: Springer-Verlag. 320 p. (1994). ZBL0797.14004.

Answer 1

No, they are not. A nice example has been constructed by Winkelmann in "On free holomorphic $\mathbb C$-actions on $\mathbb C^n$ and homogeneous Stein manifolds", Math. Ann. 286 (1990), 593–612, Lemma 8.

In that example $B=\mathbb C[x_1,x_2,x_3,x_4]$. The action of $\mathbb G_a$ is given by $$t*(x_1,x_2,x_3,x_4)=(x_1,x_2+tx_1,x_3+tx_2+\frac12t^2x_1,x_4+t(x_2^2-2x_1x_3-1))$$ which corresponds to the derivation $$\partial=x_1\partial_{x_2}+x_2\partial_{x_3}+(x_2^2-2x_1x_3-1)\partial_{x_4}.$$ Since $\partial$ has no zeros, the $\mathbb G_a$-action has no fixed point. On the other side, the image of the morphism $\mu:\mathbb G_a\times X\to X\times X$ is not closed. In fact, consider the two points $p=(0,1,0,0)$ and $q=(0,-1,0,0)$ of $X$. Since $t*p=(0,1,t,0)$, the two points $p$ and $q$ do not lie in the same orbit. Hence $(p,q)$ does not lie in the image of $\mu$. On the other side, we have $$t*(a,1,0,0)=(a,ta+1,t+\frac12t^2a,0).$$ Putting $t=-2/a$ yields $$(-2/a)*(a,1,0,0)=(a,-1,0,0).$$ Thus $$(p,q)=\lim_{a\to0}\mu(-2/a,(a,1,0,0))$$ is in the closure of the image of $\mu$.