Write $\operatorname{Int}(\mathbb{Z})$ for the set of polynomials $p$ mapping integers to integers, i.e., $p(\mathbb{Z}) \subseteq \mathbb{Z}$. I have shown that $\operatorname{Int}(\mathbb{Z})$ is a subring of $\mathbb{Q}[X]$. I am now asked to prove that $\operatorname{Int}(\mathbb{Z})$ is not isomorphic to $\mathbb{Z}[X]$.
By dabbling around online I found that $\operatorname{Int}(\mathbb{Z})$ is non-Noetherian, whilst $\mathbb{Z}$ as a PID is Noetherian and so $\mathbb{Z}[X]$ is Noetherian, hence the two rings in question cannot be isomorphic.
However, I would like to have a more elementary proof. We have $\mathbb{Z}[X] \subsetneq \operatorname{Int}(\mathbb{Z})$ since the latter contains polynomials such as $\frac{1}{2} X^2 + \frac{1}{2} X$. I thought I could get a contradiction by considering the 'extra elements', but since both rings are infinite and a ring automorphism needs not be the identity I don't know how to proceed. What are some other properties of ring isomorphisms that I can consider to reach a contradiction?
Best Answer
$\mathbb{Z}[x]$ is generated by a single element, namely $x$, so you can show that $\text{Int}(\mathbb{Z})$ is not generated by a single element, or equivalently that no homomorphism $\mathbb{Z}[x] \to \text{Int}(\mathbb{Z})$ can be surjective.
In fact $\text{Int}(\mathbb{Z})$ is not even finitely generated. The reason is that any finite set of polynomials in $\text{Int}(\mathbb{Z})$ generates a subring of $\mathbb{Q}[x]$ with the property that the denominators of all coefficients that appear are divisible by finitely many primes. But $\text{Int}(\mathbb{Z})$ contains polynomials ${x \choose n}$ (which form a $\mathbb{Z}$-linear basis) with coefficients whose denominators are divisible by arbitrarily large primes.