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

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 $1\leq r\leq p-1$.

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 $r\leq s$, 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 $1\leq r\leq p-1$. For this, consider first the product of a binomial coefficient ${\binom {x}{r}}$, with $1\leq r\leq p-1$, 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 $0<j\leq r<p$. 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 $1\leq r\leq p-1$, 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 $1\leq r\leq p-1$.

Proof detail: finishing it up

Define $I_{k}$ as the ideal generated by all the binomial polynomials ${\binom {x}{r}},1\leq r\leq k$. 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 $\mathbb {Q} [x]$.