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- Jun 22, 2012

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On page 304, part way through the proof, D&F write:

"Assume d is not a unit (in R) and write d as a product of irreducibles in R, say [TEX] d = p_1p_2 ... p_n [/TEX] . Since [TEX] p_1 [/TEX] is irreducible in R, the ideal [TEX] (p_1) [/TEX] is prime (cf Proposition 12, Section 8.3) so by Proposition 2 above the ideal [TEX] p_1R[x] [/TEX] is prime in R[x] and [TEX] (R/p_1R)[x] [/TEX] is an integral domain. ..."

My problems with the D&F statement above are as follows:

(1) I cannot see why the ideal [TEX] (p_1) [/TEX] is a prime ideal. Certainly Proposition 12 states that "In a UFD a non-zero element is prime if and only if it is irreducible" so this means [TEX] p_1 [/TEX] is prime since we were given that it was irreducible. But does that make the principal ideal [TEX] (p_1) [/TEX] a prime ideal? I am not sure! Can anyone show rigorously that [TEX] (p_1) [/TEX] a prime ideal?

(2) Despite reading Proposition 12 in Section 8.3 I cannot see why the ideal [TEX] p_1R[x] [/TEX] is prime in R[x] and [TEX] (R/p_1R)[x] [/TEX] is an integral domain. ...". (Indeed, I am unsure that [TEX] p_1R[x] [/TEX] is an ideal!) Can anyone show explicitly and rigorously why this is true?

I would really appreciate clarification of the above matters.

Peter

Note: Proposition 2 referred to above states the following:Note: Proposition 2 referred to above states the following:

Let I be an ideal of the ring R and let (I) = I[x] denote the ideal of R[x] generated by I (the set of polynomials with co-efficients in I). Then

\(\displaystyle R[x]/(I) \cong (R/I)[x]\)

In particular, if I is a prime ideal of R then (I) is a prime ideal of R[x]