TITLE: In which commutative algebras does any derivation possess a flow?
QUESTION [9 upvotes]: Definitions
Suppose $A$ is a commutative algebra over $\mathbb{R}$ with unity. $\mathbb{R}$-linear map $\xi\colon A\to A$ is a derivation of $A$ iff $\xi(ab)=a\xi(b)+\xi(a)b$ for any $a,b\in A$. If $\gamma\colon \mathbb{R}\to A$, $a\in A$, then we say, that $a=\frac{\partial}{\partial t}| _ {t=\tau} \gamma(t)$ iff $h(a) =\frac{\partial}{\partial t}|_{t=\tau} h(\gamma(t))$ for any $\mathbb{R}$-linear map $h\colon A\to\mathbb{R}$. 
Suppose $\xi$ is a derivation of $A$. Then $\Phi\colon A\times\mathbb{R}\to A$ is it's flow iff $\Phi(a,0)=a$ for any $a\in A$ and
$$\frac{\partial}{\partial t} \Phi(a,t) = \xi \Phi(a,t).\tag{1}$$
The question
1. I like algebras $A$, such that any derivation of $A$ possesses a flow. Is there any simple sufficient condition for them?
2. Is there any simple condition for an algebra and it's derivation, from which it follows, that this derivation possesses a flow?
Examples
1. Algebra $C^\infty(M)$ of smooth functions on a closed manifold $M$ --- yes (if I haven't made a mistake), any derivation possesses a flow. This, I believe, can be checked using Picard-Lindelof theorem.
1'. Algebra $C^\infty(M)$ of smooth functions on a non-compact manifold without boundary --- no (see example 2), but a derivation possesses a flow if it preserves some function $H\in A$, such that for any $c\in\mathbb{R}$ subspace $\{x\mid H(x) < c \}$ of the topological space $M$ is compact.
2. Algebra $C^\infty((0,1))$ of smooth functions on an interval --- no, $\frac{\partial}{\partial x}$ does not possess a flow.
3. Algebra $C^\infty([0,1])$ of smooth functions on a segment --- yes, any derivation possesses a flow, but it's not always unique (for example, for $\frac{\partial}{\partial x}$ it is not). In order to prove this, one can consider an embedding of $[0,1]$ to some closed manifold $N$ and prolong any function from $[0,1]$ to $N$. Then use example 1.
4. Algebra $C^\infty(\mathbb{R})$ --- no, because it is isomorphic to the algebra from example 2.
5. Algebra $\mathbb{R}[x]$ --- no. In order to prove this one can consider derivation $\xi=x^2\frac{\partial}{\partial x}$ and manually solve equation (1) for $a=x$. Any solution locally should be of the form $\frac{x}{1+xt}$. It is not in $\mathbb{R}[x]$.
6. Algebra $\mathbb{R}[x,y]/(x^2+y^2-1)$ --- no. In order to prove this take $\xi = y (x\frac{\partial}{\partial y} - y \frac{\partial}{\partial x})=\sin(\varphi)\frac{\partial}{\partial \varphi}$ and solve equation (1) manually (locally) in polar coordinates (take, for example, $a=y$). Check that the answer is not a polynomial.
7.(from Greg Muller's answer) Algebra $A$ is finite-dimensional --- yes, every derivation possesses the flow.
This question was already posted here on math.stackexchange.com, but it has received no answers even with a bounty. Any help is appreciated, both in the theme of question and in improving its wording.

REPLY [3 votes]: If you want to work in the algebraic category you should think about an algebraic notion of flow. If $M$ is a smooth manifold then a flow is an $\mathbb{R}$-action $\mathbb{R} \times M \to M$; dualizing this map gives a coaction $C^{\infty}(M) \to C^{\infty}(\mathbb{R}) \otimes C^{\infty}(M)$ where by $\otimes$ I mean a suitably completed tensor product. (Note that the group structure on $\mathbb{R}$ gives $C^{\infty}(\mathbb{R})$ a Hopf algebra structure, at least with respect to a suitably completed tensor product as above, so we can talk about comodules over it.)
So the most algebraic notion of flow would be a "polynomial flow," namely a coaction $A \to k[t] \otimes_k A$, where $A$ is a $k$-algebra and we are thinking of $k[t]$ in its incarnation as the ring of functions on the additive group scheme $\mathbb{G}_a$ over $k$. In fact in this language a derivation is precisely a coaction $A \to k[t]/t^2 \otimes_k A$ and the problem of integrating this to a flow is a lifting problem. 
A sufficient condition for a polynomial flow to exist is that the original derivation $D$ be nilpotent, but you knew that already. A slightly more interesting sufficient condition is that it be locally nilpotent in the sense that for every $a \in A$ there is some $n$ such that $D^n a = 0$. For example this is true of the derivation $\frac{\partial}{\partial x}$ on $k[x]$. I think this condition is also necessary when $A$ is an integral domain.
There are variations on this theme, e.g. we can talk about formal flows $A \to k[[t]] \otimes_k A$ but these always exist in characteristic zero (formally exponentiate) so this is in some sense uninteresting. Replacing $k[[t]]$ with other variations of $k[t]$ give other notions of flow. 
For example, to talk about flows on smooth manifolds in this algebraic language we should really talk about smooth algebras. Smooth algebras admit a smooth tensor product (this is the suitably completed tensor product I wanted) with respect to which $C^{\infty}(M) \otimes C^{\infty}(N)$ really is just $C^{\infty}(M \times N)$, and in particular $C^{\infty}(\mathbb{R})$ really is a Hopf algebra in the category of smooth algebras.
Now a flow is a coaction $A \to C^{\infty}(\mathbb{R}) \otimes A$, and these exist for smooth compact manifolds but also for other smooth algebras, e.g. any embedding of a finite-dimensional real commutative algebra $A$ into a real matrix algebra gives it a smooth algebra structure and all derivations on such things exponentiate to smooth flows.