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Let $\mathbb Q[x]$ be the ring of polynomials in the variable $x$ with rational coefficients. Let $S$ be the subring of $\mathbb Q[x]$ consisting of all those polynomials $f(x)$ with the property that $f(α)$ is an integer whenever α is an integer. Then
(a) Given a positive integer $n$ and distinct integers $ a_1, . . . , a_n$ and any integers $b_1, . . . , b_n $, there exists a polynomial $f(x)$ in $S$ such that $f(a_i)=b_i$ for all $i\in\{1,2,3,...,n\}$;
(b) The ring $S$ is finitely generated as a ring over the ring $\mathbb Z$ of integers.

I tried the above question by first taking $f(x)$ a polynomial in $\mathbb Q[x]$ which satisfies the above conditions. Then I got $n$ equations, which had a solution, so I got that there exists such a polynomial in $\mathbb Q[x]$, but I couldn't prove that it belongs to $S$. I have no idea how to proceed with second part. Any help will be appreciated. Thank you.

user26857
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Derwal Meena
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    Is your question to prove (a) and (b)? Also, how do you define a finitely generated ring? Do you mean that $S$ is a finitely generated $\Bbb{Z}$-algebra? – soggycornflakes Jan 05 '24 at 14:32
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    Don't try to prove too much. There are infinitely many polynomials in $\mathbb Q[x]$ that pass through each of the points $(a_i,b_i)$...After all, you could just take any integer $a_{n+1}$ distinct from the other $a_i$ and force the polynomial through $(a_{n+1},B)$ for any integer $B$ you want. There is no need to prove that all of these are in $S$. Indeed, they don't have to be...$f(x)=\frac x2$ passes through $(0,0)$ and $(2,1)$ but it is not in $S$. – lulu Jan 05 '24 at 14:35
  • Related. – Jyrki Lahtonen Jan 05 '24 at 15:05
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    I'm not sure how you reconcile 2. with this https://math.stackexchange.com/questions/408219 – user26857 Jan 05 '24 at 15:38
  • @soggycornflakes ,Yes I want to prove (a) and (b).I think finitely generated here means as finitely generated $Z$ algebra. – Derwal Meena Jan 05 '24 at 15:44

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Part A: we construct a polynomial that's a linear combination of polynomials that act like indicator functions

Consider the following polynomials.

  • $B_1(x) = x$ sends $1$ to $1$ and fixes the integers.
  • $B_2(x) = \frac{1}{2}x(x-1)$ sends $2$ to $1$ and fixes the integers.
  • $B_3(x) = \frac{1}{6}x(x-1)(x-2)$ sends $3$ to $1$ and fixes the integers.

As proof that the values of $B_n$ are integers, we note that they are binomial coefficients when evaluated at $x \le n$ and $0$ when $0 \le x < n$.

In general, let $B_i(x)$ be $\frac{1}{i!}\prod_{0 \le k < i} (x-k)$.

Additionally, let $B_0(x) = 1$ and $B_{-n}(x) = B_n(-x)$.

Armed with this, let's define:

$F_{ab}$ as a function which sends $a$ to $0$ and $b$ to $1$, and is undefined when $a = b$.

$F_{ab}(x)$ is $B_{b-a}(x-a)$.

Now let's define $Q_{(\vec{a}, i)}$ which sends $a_i$ to $1$ and everything else to $0$.

$$ Q_{(\vec{a}, i)} = \prod_{j \neq i} F_{\vec{a}_j\vec{a}_i}(x) $$

So now let's define $P_{(\vec{a},\vec{b})}$ which sends $\vec{a}$ to $\vec{b}$.

$$ P(\vec{a}, \vec{b}) = \sum_{i} \vec{b}_i Q_{(\vec{a}, i)} $$

Part B: We show that the sequence $B_1, B_2, \cdots$ cannot be part of any finitely generated subring.

First, we have the degree of a polynomial $p(x)$, which is defined standardly.

Next, we have what I will nonstandardly call the codegree.

Let the codegree of a rational number $p/q$ be the number of distinct prime factors of $q$ when written in simplest form.

The codegree of an integer is defined to be zero.

Let the codegree of a polynomial with rational coefficients $p(x)$ be maximum codegree of any of its coefficients.

I claim that any finitely generated subring of $\mathbb{Q}[x]$ will have a universal bound on its codegree.

Let $T$ be a finitely generated subring of $\mathbb{Q}[x]$ generated by the finite set $G$.

Let $u$ be the sum of all codegrees of all coefficients of all elements of $G$. This is a conservative bound; sharper bounds are probably possible.

$u$ bounds the codegree of any element of $T$, since prime factors cannot appear in any coefficient of any polynomial of $S$ unless that prime factor was present in one of the coefficients of one of the generators in $G$.

Now, let's suppose $S$, the ring of polynomials with rational coefficients that fix the integers setwise, is generated by some finite set of polynomials with rational coefficients $H$. Note that all $B_i$ must be in $S$.

Let $v$ be the sum of the codegrees of all coefficients of elements of $H$.

Let $r$ be the $v+1$st prime.

The polynomial $B_r$ contains $\frac{1}{r!}$ as a coefficient. Therefore $B_r$ has codegree greater than or equal to $v+1$.

Therefore $B_r$ is not in $S$, which is a contradiction.

Greg Nisbet
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