Ring of integer-valued polynomials is contained in ring generated by binomial polynomials

Statement
Suppose $$R$$ is an integral domain of characteristic zero. Then, the fact about::ring of integer-valued polynomials over $$R$$ is contained in the fact about::ring generated by binomial polynomials over $$R$$.

Ring of integer-valued polynomials
Let $$R$$ be an integral domain and let $$K$$ be its field of fractions. The ring of integer-valued polynomials over $$R$$ is defined as the subring of $$K[x]$$ comprising those polynomials $$f(x) \in K[x]$$ such that $$f(a) \in R$$ for all $$a \in R$$.

Ring generated by binomial polynomials
Let $$R$$ be a commutative unital ring of characteristic zero. Let $$L$$ be the localization of $$R$$ at the multiplicatively closed subset of nonzero integers. The ring generated by binomial polynomials over $$R$$ is the subring of $$L[x]$$ comprising $$R$$-linear combinations of the binomial polynomials:

$$\binom{x}{r} = \frac{x(x-1)\dots (x - r + 1)}{r!}$$

for $$r$$ a nonnegative integer (for $$r = 0$$, the corresponding binomial polynomial is defined as the constant polynomial $$1$$).

Related facts

 * Equivalence of definition of ring generated by binomial polynomials: The proof used for this theorem actually shows something more general: if $$R$$ is a ring of characteristic zero, the ring generated by binomial polynomials over $$R$$ is precisely the ring $$\operatorname{Int}(\mathbb{Z},R)$$: the ring of those polynomials that send elements of $$\mathbb{Z}$$ to within $$R$$.

Proof
Given: A ring $$R$$ with field of fractions $$K[x]$$. A polynomial $$f \in K[x]$$ such that $$f(a) \in R$$ for any $$a \in R$$.

To prove: $$f(x) = \sum_{i=0}^n a_i \binom{x}{i}$$ where $$a_i \in R$$ and $$n$$ is the degree of $$f$$.

Proof: First, note that the binomial polynomials form a basis for $$K[x]$$, so $$f$$ can be written uniquely as a $$K$$-linear combination of elements of $$K[x]$$.

Further, the largest $$\binom{x}{n}$$ for which the coefficient in $$f$$ is nonzero, is the same $$n$$ as the degree of $$f$$.

Thus, we have:

$$f(x) = \sum_{i=0}^n a_i \binom{x}{i}$$.

Here, $$a_i \in K$$.

We now need to prove $$a_i \in R$$ each $$0 \le i \le n$$.

We do this by inducting on $$i$$:


 * $$a_0 = 0$$: We have $$a_0 = f(0)$$, and $$f(0) \in R$$ since $$0 \in R$$.
 * Suppose $$a_0, a_1, \dots, a_{i-1}$$ are all in $$R$$. We now show that $$a_i \in R$$: Consider $$f(i)$$. Since $$i \in R$$, $$f(i) \in R$$ by assumption. But we have:

$$f(i) = \sum_{j=0}^n a_j \binom{i}{j}$$.

Note that for $$j > i$$, $$\binom{i}{j} = 0$$, by definition, so we have:

$$f(i) = a_i + \sum_{j=0}^{i-1} a_j\binom{i}{j}$$.

This rearranges to yield:

$$a_i = f(i) - \sum_{j=0}^{i-1} a_j\binom{i}{j}$$.

Since $$f(i) \in R$$, the right side is in $$R$$, so $$a_i \in R$$.

This completes the proof.