$k$-algebra of differential operators with coefficients in $k(t)$, does ring have any zero divisors? Let $A = k(t)\langle {d\over{dt}}\rangle$ be the $k$-algebra of differential operators with coefficients in $k(t)$. Thus, an element of $A$ is an expression$$F = f_n {{d^n}\over{dt^n}} + f_{n - 1} {{d^{n - 1}}\over{dt^{n - 1}}} + \ldots + f_1 {d\over{dt}} + f_0, \quad f_i \in k(t), \quad i = 1, \ldots, n,$$thought of as an operator $k(t) \to k(t)$. The sum $F + G$ of differential operators $F$ and $G$, is defined by taking the sum of corresponding coefficients $f_k$ and $g_k$, respectively, in front of ${{d^k}\over{dt^k}}$. Multiplication in the ring $k(t)\langle {d\over{dt}}\rangle$ is defined as the composition of differential operators (it is easy to see that such a composition is a differential operator again).
One has a ring imbedding $k(t) \hookrightarrow A= k(t)\langle {d\over{dt}}\rangle$, where a function $f \in k(t)$ is identified with the multiplication operator $m_f : g \mapsto f \cdot g$. Note that for $f \in k(t)$, we have$${d\over{dt}} \circ m_f - m_f \circ {d\over{dt}} = m_{f'}, \quad \text{where }f' := {{df}\over{dt}}.$$So, the elements $f$ and ${d\over{dt}}$ do not commute in $A$, in general.
Question. Does the ring $A$ have any zero divisors?
 A: First we will assume that $k$ has characteristic zero. 
$A$ acts linearly on the ring $k[[t]][t^{-1}]$ of formal Laurent series with coefficients in $k$ (where we allow only finitely many negative powers). This ring is an infinite dimensional vector space over $k$. Each differential operator, being linear in this action, has a kernel. This kernel is always finite dimensional for nonzero operators (this is the part that fails in positive characteristic). Thus the kernel of the product of two nonzero operators is finite dimensional, and in particular the product can never be zero.
We can now use this to prove the result in characteristic $p$. It is sufficient to prove this for the prime subfield $\mathbb{F}_p$ because taking the tensor product with the full field gives us the result.
Let $k=\mathbb{Q}$, and let $P$ be the subring of $k(t)$ consisting of all quotients of polynomials with integer coefficients where the denominator is monic. The result holds for $A_P$, the subring we get when we take coefficients in $P$ instead of $k(t)$. The advantage is that $P$ admits a surjective homomorphism onto $\mathbb{F}_p(t)$, which can be realized by taking the tensor product $\mathbb{F}_p\otimes P$ over $\mathbb{Z}$. Similarly we have $\mathbb{F}_p\otimes A_P\simeq A_{\mathbb{F}_p}$ as rings, and since $p$ is prime and the kernel consists of all multiples of $p$ we have that $A_{\mathbb{F}_p}$ remains a domain.
