In GCD in a subring is GCD in a bigger ring? it is noted that $\gcd(2, x) = 1$ in $\mathbb{Z}[x]$ (since $\mathbb{Z}[x]$ is a UFD and the prime/irreducible factorizations of $2$ and $x$ share no primes), but $\gcd(2, x) = 2$ in $\mathbb{Z}[x/2]$. I am struggling to understand $\mathbb{Z}[x/2]$.

- What is a rigorous definition of $\mathbb{Z}[x/2]$?
- Is it isomorphic to a quotient of $\mathbb{Z}[x]$? This won't imply UFD since $\mathbb{Z}[x]/(x^2 + 5) \cong \mathbb{Z}[\sqrt{-5}]$.
- Is it a UFD?
- What are the units?
- What are the primes and irreducibles?
- Does all of this generalize to $\mathbb{Z}[ax]$ with $a \in \mathbb{Q}$?

Here are some of my scattered ideas, which may contain incorrect statements (no need to read this). The polynomials in $\mathbb{Z}[x/2]$ ought to look like $$ \sum_{i = 0}^n a_i \left( \frac x 2 \right)^i = \sum_{i = 0}^n \frac{a_i}{2^i} x^i \in \mathbb{Q}[x]$$ For (1.), my first idea is to conceptualize $\mathbb{Z}[x/2]$ as some sort of monoid ring just like $\mathbb{Z}[x]$ is the monoid ring $\mathbb{N}[\mathbb{Z}]$. But now it looks like $\mathbb{Z}[x/2]$ might be "equal" to $(\mathbb{Z}[1/2])[x]$. Or maybe we can localize $\mathbb{Z}[x]$ at some multiplicative set like $(2\mathbb{Z})[x]$.

On (2.), $\mathbb{Z}[x]/(2x - 1) \cong \mathbb{Z}[1/2]$ ($(2x - 1)$ is the kernel of evaluation at $1/2$: if $f \in \mathbb{Z}[x]$ satisfies $f(1/2) = 0$, then the polynomial $2x - 1$ has coprime coefficients and divides the integer polynomial $f$ in $\mathbb{Q}[x]$ and thereby in $\mathbb{Z}[x]$ by Gauss's lemma), but I have few ideas on how to influence the formal symbol $x$ with a quotient -- I believe Laurent polynomials over a field $F$ come from $F[[x]][y]/(xy - 1)$, which certainly influences $x$ by adding $\frac 1 x$.

For (3.) if we could show Noetherian integral domain then it would suffice to prove that irreducible implies prime. $\mathbb{Z}$ is Noetherian, so $\mathbb{Z}[x_1, \dots, x_n]$ and $\mathbb{Z}[[x_1, \dots, x_n]]$ are Noetherian as well as any quotients of these. Integral domain should be true because the lead coefficients of two nonzero polynomials in $\mathbb{Z}[x/2]$ cannot multiply to zero.

For (4.), there is a lovely classification of the units of a polynomial ring. If the same applies here, the units should be the units of $\mathbb{Z}$, namely $\pm 1$. For (5.), primes and irreducibles will be the same assuming $\mathbb{Z}[x/2]$ is a UFD. The irreducibles in $\mathbb{Z}[x]$ are precisely the integer primes and the polynomials with coprime coefficients that remain irreducible in $\mathbb{Q}[x]$ (and associates of these). So maybe the primes in $\mathbb{Z}[x/2]$ are similar modulo something about $2$.

For (6.), my guess is this all generalizes to $\mathbb{Z}[x/p]$ for primes $p \in \mathbb{Z}$ but not any further than that.