For a positive integer $n$, let $\mathcal{P}$ be the set of primes not dividing $n$ and $\mathbb{Z}_{\mathcal{P}} = \mathbb{Z}\left[\left\{p^{-1}:p\in\mathcal{P}\right\}\right]$ be the localization of $\mathbb{Z}$ at $\mathcal{P}$. The primes in $\mathbb{Z}_{\mathcal{P}}$ then are exactly those dividing $n$. Are there any ways to find or characterize the square-free elements of this ring?
For example, in the integers $n$ is square-free if and only if $n = \prod_{p^e\vert\vert n}p^{1/e}$, or if and only if $\prod_{p^e\vert\vert n}p^{e-1}=1$.
I want to find conditions for $\prod_{p^e\vert\vert n}p^e(p-1)$ to be square-free in $\mathbb{Z}_{\mathcal{P}}$, because this is exactly the condition for $n$ to be "cyclic" (i.e. coprime to its totient $\varphi(n)$).
Any answers at all are appreciated.
Every nonzero element $x\in\mathbb{Z}_\mathcal{P}$ can be uniquely factorized in the form $x=u\prod_{i=1}^mp_i^{d_i}$, where $u\in\mathbb{Z}_{\mathcal{P}}$ is a unit, $p_1,\dots,p_m$ are the prime factors of $n$, and $d_1,\dots,d_m\in\mathbb{N}$. Explicitly, representing $x$ as a rational number in lowest terms, $p_i^{d_i}$ is the largest power of $p_i$ dividing the numerator of $x$, and $u=x/\prod p_i^{d_i}$ is a unit because all the prime factors of its numerator are in $\mathcal{P}$. The element $x$ is then squarefree iff $d_i<2$ for all $i$. Concretely, this means that when you represent $x$ as a rational number in lowest terms, its numerator has no prime factor of $n$ as a repeated factor.
In particular, $\prod_{p^e\vert\vert n}p^e(p-1)$ is square-free in $\mathbb{Z}_\mathcal{P}$ iff $n$ is square-free in $\mathbb{Z}$ (i.e., all of the $e$s are $1$) and whenever $p$ and $q$ are two primes dividing $n$, $q\not\mid p-1$ (because this means that the $p-1$ factors don't contribute any additional factors of different primes dividing $n$).