Question

6.6.7. Show that R = Z + xQ[x] does not satisfy the ascending chain condition for principal ideals. Show that irreducibles in R are prime.

From Goodman's "Algebra: Abstract and Concrete"

Answer 1

66 To Show: R = 2 + XQ [x] does not satisfy the ascending chain condition for principal ideals. To show that IR is not a PID we consider the Ideal P= x Q[x] CR We'll show that p is not principal. In fact we'll show that P is not finitely generated Suppose to the contrary, and let p = fi fr ... fx) for some kai Each fi has constant term zeno let qui be the Coefficient by x for each fi, isisk

Then any element of P say I has the coefficient by x a - Za + Zast... +2 gifit 92 f2 t... tgx fx member of the set & Q because Q is not a finitely generated group hus p is not finitely generated ! and therefore is not principal. Thus R is not a. PID and doesn't satisfy the ascending chain condition for principal ideals. To show:- irreducibles in R = 2 + a Q [x] are prime. Suppose first fixy is of the form tp for some prime per 94 frx) could be written as a produet fra)= a(w)6XIEIR the degrees of the terms would need to add to o. Hence this. would give a factorization of the prime p into a product of integeri. I since primes in 2 are irreducible acx) or 6(x) is a unit Hence text is irreducible in R . . irreducibles in k are tp Where p is a prime.