Question

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.

Answer 1

If you really mean to get away with treating the algebra as an algebraic object -- no topology is given on this vector space -- then you don't even have the smooth compact manifold example. The given definition of $a=\gamma'(\tau)$ is rarely satisfied. There are too many linear maps from $A$ to $\mathbb R$. A function $\gamma:\mathbb R\to A$ cannot have a derivative in your sense at $\tau$ unless there is some $\epsilon$ such that the vectors $\gamma(t)$ for $|t-\tau|<\epsilon$ span a finite-dimensional vector subspace of $A$.

On the other hand, it is true that a smooth compact manifold (with its (sheaf of) smooth functions) can be reconstructed from the purely algebraic object which is the algebra of global functions, and also that the only derivations of that algebra are the ones arising from tangent vector fields. I don't how to describe (without introducing a topology on the algebra, or something like that) the relation between the derivations and the corresponding homomorphisms from the additive group of real numbers into the automorphism group of the algebra.

Answer 2

My guess is this is the kind of algebra you don't care about (since they aren't subrings of real-valued functions), but algebras of the form $\mathbb{R}[x]/x^n$ will have your property. When looking for a flow corresponding to a derivation $\xi$, consider the formal power series $$ \sum_i \frac{t^i\xi^i}{i!} $$ The problem, of course, is showing that this power series converges in the algebra of appropriate endomorphisms. However, a lazy trick for guaranteeing its convergence is to hope that $\xi^n=0$ for some $n$. This happens for every derivation in the ring $\mathbb{R}[x]/x^n$, or more generally when the nilradical is maximal.

Edit: As is pointed out in the comments, this is a correct claim but the wrong reason. The real reason flows exponentiate in these rings, as well as all finite dimensional ones, is that a derivation is given by a matrix (after some choice of basis), and all matrices can be exponentiated (that is, the above series converges).