Let $A$ be a commutative ring and $f\in A[x]$ a split monic. When $f$ is separable with roots $\mathrm Z(f)= \{ a_1,\dots ,a_k \}$, the Chinese remainder theorem (CRT) ensures that evaluation is an $A$-algebra isomorphism $A[x]/\langle f\rangle\cong \mathsf{Set}(\mathrm Z(f),A)$. In this case, Lagrange interpolation furnishes a "low degree" inverse, i.e formulas that produce minimal degree representatives of equivalence classes in the quotient.
Now assume $f\in A[x]$ is split with the same roots, but no longer separable. Assuming the differences of the roots are invertible $a_i-a_j\in A^\times$, CRT ensures that taking jets is an $A$-algebra isomorphism $\tfrac {A[x]}{\langle f\rangle}\cong \tfrac {A[x]}{\langle (x-a_1)^{m_1}\rangle}\times \cdots \times \tfrac {A[x]}{\langle (x-a_k)^{m_k}\rangle}$, where $m_i$ is the multiplicity of the root $a_i\in A$. Here by "jet" I mean the divided difference expansion, where $\tfrac 1{k!}\tfrac{\mathrm d^k}{\mathrm dx ^k}f$ is replaced with $f[\underbrace{x,\dots ,x} _k]$, to avoid assuming invertibility of integers in $A$.
One can concoct an inverse using powers and derivatives of the Lagrange basis, but these expressions have high degree - they do not give the minimal degree representatives of classes in the quotient.
On the other hand, the "abstract" inverse of the CRT isomorphism uses modular inverses and I don't see how to get a general formula here. (Maybe this is the way, i.e via computing gcd with powers of the $x-a_i$, I just don't understand if it is.)
Is there a general "jet interpolation" formula that gives a minimal degree polynomial with prescribed jets? What is it?
