Ring of integer-valued polynomials over rational integers is not Noetherian

Statement
The ring of integer-valued polynomials over rational integers is not a Noetherian ring.

Related facts

 * Ring of integer-valued polynomials over normal domain is normal
 * Ring of integer-valued polynomials over rational integers is not a UFD

Proof idea
The ring $$R$$ of integer-valued polynomials over rational integers is the same as the ring generated by binomial polynomials over integers, which is a free $$\mathbb{Z}$$-module over the binomial polynomials:

$$1,x, \binom{x}{2}, \binom{x}{3}, \dots$$.

We can show that the ideal of polynomials without constant term, which is the same as the intersection with $$R$$ of the ideal $$(x)$$ in $$\mathbb{Q}[x]$$, is not finitely generated.

Proof details: binomial polynomial for a prime is not in the ideal generated by smaller nonconstant binomials
To prove: If $$p$$ is prime, $$\binom{x}{p}$$ is not in the ideal generated by $$\binom{x}{r}$$ for $$1 \le r \le p-1$$.

Proof: The first thing we use about the ring of integer-valued polynomials is that any polynomial in this ring can be written uniquely as:

$$\sum_{i=0}^r a_i \binom{x}{i}$$

where $$a_i \in \mathbb{Z}$$, $$\binom{x}{0}$$ is the constant polynomial $$1$$, and the others are defined by:

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

Further, we use that the multiplication rule for two binomial coefficients, where $$r \le s$$, is:

$$\binom{x}{r} \binom{x}{s} = \sum_{i=0}^r \binom{s+i}{s}\binom{s}{r-i} \binom{x}{s+i}$$

Now, consider the binomial polynomial $$\binom{x}{p}$$ when $$p$$ is prime. We show that $$\binom{x}{p}$$ is not in the ideal generated by $$\binom{x}{r}$$, for $$1 \le r \le p-1$$. For this, consider first the product of a binomial coefficient $$\binom{x}{r}$$, with $$1 \le r \le p-1$$, with any integer-valued polynomial:

$$\left(\binom{x}{r}\right)\left(\sum_{i=1}^{n} a_i \binom{x}{i} \right)$$

The previous formulae yield that the coefficient of $$\binom{x}{p}$$ is given by:

$$\sum_{j=0}^r a_{p-j} \binom{p}{p-j} \binom{p-j}{r-j}$$

Notice that if $$j > 0$$, then $$\binom{p}{p-j}$$ is divisible by $$p$$, since $$0 < j \le r < p$$. If $$j = 0$$, then the term $$\binom{p-j}{r-j}$$ is divisible by $$p$$. In either case, each term in the summation is divisible by $$p$$, so the coefficient of $$\binom{x}{p}$$ in the product is a multiple of $$p$$.

Any element in the ideal generated by $$\binom{x}{r}$$ for $$1 \le r \le p - 1$$, is a sum of elements of the form above. Since the coefficient of $$\binom{x}{p}$$ is a multiple of $$p$$ in each case, we obtain that the coefficient of $$\binom{x}{p}$$ for any element in the ideal is a multiple of $$p$$. In particular, since the expression as a $$\mathbb{Z}$$-linear combination of binomial polynomials is unique, $$\binom{x}{p}$$ is not in the ideal generated by $$\binom{x}{r}$$, for $$1 \le r \le p - 1$$.

Proof detail: finishing it up
Define $$I_k$$ as the ideal generated by all the binomial polynomials $$\binom{x}{r}, 1 \le r \le k$$. We thus have an ascending chain of ideals:

$$I_1 \subseteq I_2 \subseteq \dots $$.

Moreover, we just proved that $$\binom{x}{p}$$ is not in $$I_{p-1}$$. Thus, $$I_{p-1}$$ is a proper ideal in $$I_p$$. Since there are infinitely many primes, there are infinitely many strict inclusions, and we thus have an infinite ascending chain of ideals that does not stabilize. This shows that $$R$$ is not Noetherian.

In fact, it shows that the ascending union of the $$I_k$$s, which is the ideal generated by all $$\binom{x}{r}$$, is not a finitely generated ideal. This ideal is the same as the ideal of polynomials without constant term, or the intersection with $$R$$ of the ideal $$(x)$$ in $$\mathbb{Q}[x]$$.