Question

Let R be a commutative ring with unity. If I is a prime ideal of R, prove that I[x] is a prime ideal of R[x].

Answer 1

Theorem 14.3:

Statement:An ideal I of a commutative ring R is prime, iff the quotient ring R/I is an integral domain.
Solution: To prove that I[r] is a prime ideal of R[r], by Theorem 14.3, it is equivalent to prove that R[x]/I[x] is an hence (by Theorem 14.3), R/I is an integral domain. By Theorem 16.1, this implies that (R/I)x] is an isomorphisms. Therefore, if we prove that integral domain. By assumption, I is a prime ideal of R and integral domain. Recall that being integral domain is preserved under an Ra/I (R/I[ then we will have shown that R[x]/I[x| is an prime ideal of R[x] Consider the map p : R[x] integral domain and, hence, that I[x] is a (R/I)z given by, for ana" + ..+ ao in R[x], ... + ao) 3 (а, + I)a" + ...+ (aо + 1) Ф(а,т" +

We will prove that p is an onto homomorphism with Kerp = I[x]. Onto? Let p(r) e (R/I)[x]. Then p(x) R/I. Choose coset representatives, an, . . . , ao: (An)r" +.. + Ao for Ao, . .. , An cosets in An Ao = an + I. = an + 1 Then (anх" + ... + ao) 3D p(х). Homomorphism: since cosets operations are defined by ring operations on the rep- resentatives, polynomial addition and multiplication is preserved Kernel? Suppose p(x) e I[x]. Then p(x) By definition, p(p(x)) 0R/I 0. Conversely, suppose p(p(x)) the coefficients in p(x) is sent to the zero element in R/I, hence to I. By definition of p, this implies that each of the coefficients in p(x) is in I. Therefore, p(x) e I[x]. By the First Isomorphism Theorem, we conclude that R[x]/I[x](R/I аnх" +: ..+аo with a; € I for i —D 0, .. ., п. (аn + I)x" ..+ (ao + I) — Iz^ + + I2n — ОRIII" + =0 for some p(x) e R[x]. Then each of - as required.