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:

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Corollary 6

Let R be a UFD, let F be its field of fractions and let $\displaystyle p(x) \in R[x] $.

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].

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The proof reads as follows:

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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].

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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

Re: Polynomial Rings, UFDs and Fields of Fractions

1. In the proof, D&F showed that $\displaystyle p(x)$ reducible in $\displaystyle F[x]$ if and only if $\displaystyle p(x)$ reducible in $\displaystyle R[x]$. Now $\displaystyle (P\iff Q) \iff (\neg P \iff \neg Q)$ which recovers the original statement of the theorem.

2. Suppose the gcd of the coefficients of $\displaystyle p(x)$ is 1 and $\displaystyle p(x)=a(x)b(x)$ for some $\displaystyle a(x),b(x)\in R[x];a(x),b(x)\not=p(x)$ (so $\displaystyle p(x)$ is reducible in $\displaystyle R[x]$). We want to show that $\displaystyle a(x),b(x)$ are non-constant.

Suppose otherwise. Then $\displaystyle p(x)=a(x)b(x)$ where $\displaystyle a(x)$ or $\displaystyle b(x)$ is constant (WLOG, take $\displaystyle a(x)=a$ to be constant). Then $\displaystyle p(x)=a\cdot b(x)$. Since the gcd of the coefficients of $\displaystyle p(x)$ is 1, we necessarily have $\displaystyle a=1$. Thus $\displaystyle p(x)=b(x)$ which is a contradiction.

Re: Polynomial Rings, UFDs and Fields of Fractions

Thanks for the help!

Just a point of clarification:

You write: "Suppose the gcd of the coefficients of p(x) is 1 and p(x)=a(x)b(x) for some $\displaystyle a(x),b(x)\in R[x] $; $\displaystyle a(x),b(x)\not=p(x) $ (so p(x) is reducible in R[x]). "

Can you show explicitly why this follows: "The gcd of the coefficients of p(x) is 1 and p(x) = a(x)b(x) for some $\displaystyle a(x), b(x) \in R[x] \Longrightarrow a(x), b(x) \ne p(x) $

Couldn't it be the case that a(x) = p(x) and b(x) = 1? What has thwe gcd of the co-efficients of p(x) got to do with this?

Peter

Re: Polynomial Rings, UFDs and Fields of Fractions

Quote:

Originally Posted by

**Bernhard** Couldn't it be the case that a(x) = p(x) and b(x) = 1? What has thwe gcd of the co-efficients of p(x) got to do with this?

That can certainly happen, I'm not saying every factorisation is of the form I stated. However, if $\displaystyle p(x)$ is reducible in $\displaystyle R[x]$, there is __some__ factorisation such that $\displaystyle a(x),b(x)\not = p(x)$.

The gcd of the coefficients of p(x) being 1 is a separate assumption. So we're actually assuming the gcd bit and that $\displaystyle p(x)$ is reducible in $\displaystyle R[x]$.

Re: Polynomial Rings, UFDs and Fields of Fractions

Thanks again Gusbob

But how do we know that there is even **some** factorization where $\displaystyle a(x)b(x) = p(x) $ and $\displaystyle a(x), b(x) \ne p(x) $.

How do we know that it is not the case that the **ONLY** factorization is a(x) = p(x) and b(x) = 1?

Can you clarify?

Peter

Re: Polynomial Rings, UFDs and Fields of Fractions

If the only factorisation is 1 and p(x), then p(x) is irreducible in R[x]. We wanted to assume it is reducible, so you need to assume some other factorisation exists.