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Thursday, May 16, 2013

Polynomial Factorization Practice (9.3.4)

Dummit and Foote Abstract Algebra, section 9.3, exercise 4:

MathJax TeX Test Page Let R=Z+xQ[x]Q[x].

(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,βZ 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,βR, 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:RZ/pZ by qnxn+...+q1x+zz+pZ. For irreducible qQ[x] with constant term ±1, we have q is prime and thus R/(q) is contained in the integral domain Q/(q) and is therefore an integral domain itself, proving q is prime.

(c) Note that any vxR for vQ is not irreducible, since vx=(2)(v2x) is a valid factorization. Now, if x=π1...πn for irreducible πi, then 1=N(π1)+...+N(πn) so that exactly one πk is of the form qx+z and for ik we have πiZ. By our remark above z0, but now the product π1...πn 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)={¯a+bx | aZ,bQ0b<1}. Since clearly , we must show the reverse: For arbitrary a(x)R, we have a(x)=qnxn+...+q1x+z=(qnxn1+...q2x)x+(w+q1)x+z=(qnxn1+...q2x+w)x+q1x+z for some wZ and 0q1<1, so that ¯a(x){¯a+bx | aZ,bQ0b<1}. From here, the isomorphism from R/(x) onto Z+(Q/Z)x is evident, where Q/Z is the ring extension of Q/Z as additive groups by standard multiplication. 

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