- Thread starter
- #1

- Jun 22, 2012

- 2,918

**[This item has also been simultaneously posted on MHF]**

Polynomial Rings, UFDs and Fields of Fractions

Polynomial Rings, UFDs and Fields of Fractions

In Dummit and Foote Section 9.3 Polynomial Rings that are Unique Factorization Domains, Corollary 6, reads as follows:

================================================== ==========================

Corollary 6

Let R be a UFD, let F be its field of fractions and let
.

Suppose the gcd the of the coefficients of p(x) is 1.

Then p(x) is irreducible in R[x] if and only if it is irreducible in F[x].

In particular, if p(x) is a monic polynomial that is irreducible in R[x], then p(x) is irreducible in F[x].

================================================== ===========================

The proof reads as follows:

================================================== ===========================

Proof:

By Gauss' Lemma, if p(x) is reducible in F[x] then p(x) is reducible in R[x].

Conversely, the assumption that gcd the of the coefficients of p(x) is 1 implies that if it is reducible in R[x], then p(x) = a(x)b(x) where neither a(x) nor b(x) are constant polynomials in R[x]. This same factorization shows that p(x) reducible in F[x].

================================================== ===========================

My problems requiring clarification are as follows:

Problem 1: The Corollary talks in terms of irreducibility while the proof talks in terms of reducibility. Why is this? How is the statement of the Corollary and the proof reconciled? Can someone give a clear clarification?

Problem 2: The proof reads "the assumption that gcd the of the coefficients of p(x) is 1 implies that if it is reducible in R[x], then p(x) = a(x)b(x) where neither a(x) nor b(x) are constant polynomials in R[x]". Can someone please show rigorously and explicitly how this follows.

Peter

================================================== ==========================

Corollary 6

Let R be a UFD, let F be its field of fractions and let

Suppose the gcd the of the coefficients of p(x) is 1.

Then p(x) is irreducible in R[x] if and only if it is irreducible in F[x].

In particular, if p(x) is a monic polynomial that is irreducible in R[x], then p(x) is irreducible in F[x].

================================================== ===========================

The proof reads as follows:

================================================== ===========================

Proof:

By Gauss' Lemma, if p(x) is reducible in F[x] then p(x) is reducible in R[x].

Conversely, the assumption that gcd the of the coefficients of p(x) is 1 implies that if it is reducible in R[x], then p(x) = a(x)b(x) where neither a(x) nor b(x) are constant polynomials in R[x]. This same factorization shows that p(x) reducible in F[x].

================================================== ===========================

My problems requiring clarification are as follows:

Problem 1: The Corollary talks in terms of irreducibility while the proof talks in terms of reducibility. Why is this? How is the statement of the Corollary and the proof reconciled? Can someone give a clear clarification?

Problem 2: The proof reads "the assumption that gcd the of the coefficients of p(x) is 1 implies that if it is reducible in R[x], then p(x) = a(x)b(x) where neither a(x) nor b(x) are constant polynomials in R[x]". Can someone please show rigorously and explicitly how this follows.

Peter

Last edited: