Describing $\mathbb{Z}[x/2]$ 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.
 A: *

*We can define this rigorously as follows: consider the ring homomorphism $\phi: \mathbb Z[x] \longrightarrow \mathbb Q[x]$ defined by $x \mapsto \frac{x}{2}$, i.e. $f \mapsto f\left(\frac{x}{2}\right)$. The image is $\mathbb Z\left[\frac{x}{2}\right]$. This is generally how you adjoin elements from one ring to another - you adjoin a formal variable and evaluate that variable at your chosen element.


*Certainly! By the first isomorphism theorem, $\phi$ induces an isomorphism $\mathbb Z[x] / \ker(\phi) \longrightarrow \mathbb Z\left[\frac{x}{2}\right]$. However, you want to say that this is $\mathbb Z[x] / (2x - 1)$, which is not true. I'll prove in a moment that this is not the case, but first of all observe that $2x - 1 \notin \ker(\phi)$. Indeed, $\phi(2x - 1) = x - 1$.
What then is $\ker(\phi)$? Let's take some $f \in \ker(\phi)$ and write $f = \sum f_i x^i$, $f_i \in \mathbb Z$. Then
$$
0=\phi(f) = \sum f_i \left(\frac{x}{2}\right)^i = \sum \frac{f_i}{2^i} x^i.
$$
Remember that the image lives in $\mathbb Q[x]$, where $x$ is still just a formal variable. In particular, the set $\{1, x, x^2, \dots\}$ is linearly independent over $\mathbb Q$. Hence, as $\sum \frac{f_i}{2^i} x^i = 0 \in \mathbb Q[x]$, each coefficient $\frac{f_i}{2^i} = 0$. Then just multiply by $2^i$ and deduce that each $f_i = 0$ as well. Hence, $f = 0$ so $\phi$ is an isomorphism and $\mathbb Z[x] \cong \mathbb Z\left[\frac{x}{2}\right]$. By the way, this means in particular that $2$ is not a unit in $\mathbb Z\left[\frac{x}{2}\right]$ so the quotient $\mathbb Z[x]/(2x-1) \not\cong \mathbb Z\left[\frac{x}{2}\right]$.
This may seem a bit strange. After all, it seems like your intuition was that we are somehow inverting something. And this intuition makes sense, after all we are throwing a $\frac{1}{2}$ in there! However, this formal proof I just gave tells us that no additional relations arise when you replace $x$ with $\frac{x}{2}$. If it's easier to see it this way, let $y = \frac{x}{2}$. Then the inverse map simply sends $y \mapsto 2y$. I think it's easier to see why this map doesn't add any extra units.
3, 4, and 5 all follow from your knowledge of $\mathbb Z[x]$, as we have just shown that $\mathbb Z\left[\frac{x}{2}\right] \cong \mathbb Z[x]$. Let's go through them real quick.


*Yes, as $\mathbb Z[x]$ is a UFD.


*The isomorphism $Z[x] \longrightarrow \mathbb Z\left[\frac{x}{2}\right]$ restricts to an isomorphism on unit groups. The units of $\mathbb Z[x]$ are $\pm 1$, so the units of $\mathbb Z\left[\frac{x}{2}\right]$ are $\phi(1) = 1$ and $\phi(-1) = -1$.


*$\phi$ is a ring isomorphism so these ring theoretic concepts of primality and irreducibility are preserved by it. $f \in \mathbb Z[x]$ is prime iff $\phi(f) = f\left(\frac{x}{2}\right) \in \mathbb Z\left[\frac{x}{2}\right]$ is prime, and prime = irreducible here as we just showed this was a UFD.


*Yup, except for $a = 0$ this is an isomorphism for the same reason. Take $\mathbb Z[x] \longrightarrow \mathbb Q[x]$ via $x \mapsto ax$. For $a \neq 0$, the exact same proof works for computing the kernel of this map. This is because $a \neq 0$ is a unit in $\mathbb Q$, therefore not a zero divisor, so if $f_i a^i = 0$ iff $f_i = 0$.
