(a) Prove R is an integral domain and R×=±1.
(b) Show that the irreducibles of R are the irreducibles of Z together with the irreducible polynomials of Q[x] with constant term ±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 Q[x] it is an integral domain. Note that the degree norm N has the property N(αβ)=N(α)+N(β) as in a typical polynomial ring, so that the only possible units are within Z, and as such the only units are ±1.
(b) Should p=αβ we have 0=N(p)=N(α)+N(β), so that α,β∈Z and thus by the primality of p we have one of the two is equal to ±1, i.e. is a unit. Letting p(x)=qnxn+...+q1x±1 be an irreducible element of Q[x] with constant term ±1, assuming p(x)=αβ for nonunit α,β∈R, by necessity α and β have constant term ±1 so that N(α),N(β)>0. But now α and β are nonunits in 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 ±1 else for a prime divisor p of the term we have a(x)=pa(x)p is factorization when a(x)≠±p. Assuming a(x)=b(x)c(x) for some nonunit b(x),c(x)∈Q[x], we have the constant term of b(x) is mn and necessarily the constant term of c(x) is nm so that b(x)=mnb′(x) and c(x)=nmc′(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×.
For prime p the ideal (p)⊆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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