Eisenbud 1.21a $G(n):= F(n)-F(n-1)$, given $G(n)$ is integer-valued, then so is $F(n)$.

Question

This question is based on a subproblem from 1.21 (a) from Eisenbud's commutative algebra

Given $F: \mathbb Z \to \mathbb Z$, we then define $G(n):= F(n)-F(n-1)$. Is it then true that:

$G(n)$ is an integer-valued polynomials (with rational coefficients) $\implies F(n)$ is an integer-valued polynomial (with rational coefficients).

Therefore, suppose $$G(n):= q_0 + q_1 n + \dots q_k n^k $$ We get that: $$ F(1)-F(0) = G(0)=q_0 \in \mathbb Z$$ $$ F(2)-F(1) = G(1) \in \mathbb Z$$ $$ F(3)-F(2) = G(2) \in \mathbb Z$$ $$\vdots$$ $$ F(n)-F(n-1) = G(n) \in \mathbb Z$$ Okay great, so we deduce that this function differs an integer value from its previous value, which is great, but I am missing a step or argument here.

Any ideas?

Note: I have used the hint, and was able to prove that any such integer-valued polynomial has a basis of $G_k(n)= \binom{n}{k} \in \mathbb Q[n]$. I do not see how this can be used.

Answer 1

$$F(n)-F(0)=G(1)+G(2)+... +G(n)$$ $$=nq_0+\frac{n(n+1)}{2}q_1+...$$ and is therefore a rational polynomial in $n$ by Faulhaber's formula.

It is integer valued since $F(0)$ and all $G(i)$ are.