(a) Prove $R$ is an integral domain and $R^\times = \pm 1$.
(b) Show that the irreducibles of $R$ are the irreducibles of $\mathbb{Z}$ together with the irreducible polynomials of $\mathbb{Q}[x]$ with constant term $\pm 1$. Prove these irreducibles are prime.
(c) Show $x$ cannot be written as the product of irreducibles in $R$ (in particular, $x$ is not irreducible) and conclude that $R$ is not a UFD.
(d) Show $x$ is not prime in $R$ and describe $R/(x)$.
Proof: (a) Since this set is clearly closed under subtraction and multiplication, it is a subring, and by its containment in $\mathbb{Q}[x]$ it is an integral domain. Note that the degree norm $N$ has the property $N(\alpha \beta)=N(\alpha)+N(\beta)$ as in a typical polynomial ring, so that the only possible units are within $\mathbb{Z}$, and as such the only units are $\pm 1$.
(b) Should $p=\alpha \beta$ we have $0=N(p)=N(\alpha)+N(\beta)$, so that $\alpha,\beta∈\mathbb{Z}$ and thus by the primality of $p$ we have one of the two is equal to $\pm 1$, i.e. is a unit. Letting $p(x)=q_nx^n + ... + q_1x \pm 1$ be an irreducible element of $\mathbb{Q}[x]$ with constant term $\pm 1$, assuming $p(x)=\alpha \beta$ for nonunit $\alpha,\beta∈R$, by necessity $\alpha$ and $\beta$ have constant term $\pm 1$ so that $N(\alpha),N(\beta) > 0$. But now $\alpha$ and $\beta$ are nonunits in $\mathbb{Q}[x]$, violating the irreducibility of $p(x)$. Finally, assuming $a(x)$ is an irreducible element of $R$ by necessity we have its constant term is $\pm 1$ else for a prime divisor $p$ of the term we have $a(x)=p \dfrac{a(x)}{p}$ is factorization when $a(x)≠\pm p$. Assuming $a(x)=b(x)c(x)$ for some nonunit $b(x),c(x)∈\mathbb{Q}[x]$, we have the constant term of $b(x)$ is $\dfrac{m}{n}$ and necessarily the constant term of $c(x)$ is $\dfrac{n}{m}$ so that $b(x)=\dfrac{m}{n}b'(x)$ and $c(x)=\dfrac{n}{m}c'(x)$ for $b'(x),c'(x)∈R$ and now $a(x)=b'(x)c'(x)$ is a valid factorization of $a(x)$ in $R$ by $0 < N(b(x))=N(b'(x))$ and $0 < N(c(x))=N(c'(x))$ so that $b'(x),c'(x)∉R^\times$.
For prime $p$ the ideal $(p) \subseteq R$ is the kernel of the surjective homomorphism $φ : R → \mathbb{Z}/p\mathbb{Z}$ by $q_nx^n + ... + q_1x + z \mapsto z + p\mathbb{Z}$. For irreducible $q∈\mathbb{Q}[x]$ with constant term $\pm 1$, we have $q$ is prime and thus $R/(q)$ is contained in the integral domain $\mathbb{Q}/(q)$ and is therefore an integral domain itself, proving $q$ is prime.
(c) Note that any $vx∈R$ for $v∈\mathbb{Q}$ is not irreducible, since $vx=(2)(\dfrac{v}{2}x)$ is a valid factorization. Now, if $x=\pi_1...\pi_n$ for irreducible $\pi_i$, then $1=N(\pi_1)+...+N(\pi_n)$ so that exactly one $\pi_k$ is of the form $qx+z$ and for $i≠k$ we have $\pi_i∈\mathbb{Z}$. By our remark above $z≠0$, but now the product $\pi_1...\pi_n$ has a nonzero constant term, a contradiction.
(d) If $x$ were prime in $R$, it would be irreducible, a contradiction by the above. We now claim that $R/(x)=\{\overline{a+bx}~|~a∈\mathbb{Z},b∈\mathbb{Q} \land 0≤b<1\}$. Since clearly $\supseteq$, we must show the reverse: For arbitrary $a(x)∈R$, we have $$a(x)=q_nx^n + ... + q_1x+z=$$$$(q_nx^{n-1}+...q_2x)x+(w+q_1')x+z=(q_nx^{n-1}+...q_2x+w)x+q_1'x+z$$ for some $w∈\mathbb{Z}$ and $0≤q_1'<1$, so that $\overline{a(x)}∈\{\overline{a+bx}~|~a∈\mathbb{Z},b∈\mathbb{Q} \land 0≤b<1\}$. From here, the isomorphism from $R/(x)$ onto $\mathbb{Z}+(\mathbb{Q}/\mathbb{Z})x$ is evident, where $\mathbb{Q}/\mathbb{Z}$ is the ring extension of $\mathbb{Q}/\mathbb{Z}$ as additive groups by standard multiplication.$~\square$
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