# Contracting principal ideals in univariate polynomial rings to subrings generated by a monomial

Let $S=\mathbb{Z}[x^{\pm1}]$ be the ring of Laurent polynomials with integer coefficients. Fix $n>1$, and let $R = \mathbb{Z}[x^{\pm n}]$, so that $S$ is an integral ring extension of $R$. For an arbitrary polynomial $f(x)\in S$, is the contraction of the principal ideal $f(x)S$ to $R$ always principal, and if so how do you compute a generator?

If $f(x)$ is irreducible, the principal ideal $f(x)S$ is prime and of height one, so its contraction is also prime of height one (Going Down), hence principal. This should go through for products of distinct irreducibles as well, but for powers this isn't so clear to me how to procede.

To find a generator, one is tempted to consider the product $f_n(x)$ of $f(\zeta x)$ over all $n$th roots of unity, which is indeed in the contracted ideal, but need not generate. Is $f_n(x)$ always a power of the generator?

I have found a proof that the contraction is principal, and this proof generalizes to Laurent polynomials in $d$ variables. There $\mathbb{Z}[x^{\pm n}]$ is replaced by the integral group ring generated by a finite-index subgroup of $\mathbb{Z}^d$. I have not yet resolved the last part of the question, but I think it should not be hard.
This all turns out to be a special case of a very general theorem. Let $R$ and $S$ be unique factorization domains, $R$ contained in $S$, and $S$ integral over $R$. Then every principal ideal in $S$ contracts to a principal ideal in $R$. (Although this seems like it should be easily found in the literature I haven't been able to locate it). The proof is to observe that you have Going Down for prime ideals, and that a prime ideal here is principal iff and only if it has height one. By Going Down, prime ideals of height one contract to prime ideals of height one