Describing $\mathbb{Z}[x/2]$

1. 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.

2. 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.

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

4. 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 $\phi(-1) = -1$.

5. $\phi$ 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.

6. 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 $f_i = 0$.