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 ring of integer-valued polynomials over $R$ is contained in the ring generated by binomial polynomials over $R$.

Definitions used

Ring of integer-valued polynomials

Further information: 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

Further information: 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 $0\leq i\leq n$.

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.