Do derivations over a commutative ring form a ring?

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Given a commutative ring $R$ and a derivation $D : R \rightarrow R$, we might consider a ring built over "differential operators" built over this operator, with the elements of the ring being spans of the operators $D,D^2,D^3,\dots$ over the ring $R$. For example, if $R$ is the ring of smooth functions $\mathbb{R} \rightarrow \mathbb{R}$, an example of such an element of this ring would be $sin(x) \frac{d^2}{dx^2} + 5\frac{d}{dx}$. Multiplication in this ring is just composition of operators, and addition is just operator addition.

  1. Does this construction actually give a ring in general?
  2. If this is a ring, what are its properties? (Specifically, I'd like to know if $R$ is reduced, and $D$ is a derivation on $R$, is the ring of differential operators over $R$ also reduced? But that might be better as another question)
  3. What is such a ring typically called in the literature?
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No, in general the set of derivations over a commutative ring do not form themselves a ring under addition and composition. The problem is that the composition of two derivations need not be itself a derivation.