Is Gauss' Lemma the Key to Non-UFDs in Polynomial Rings?

[Math Amateur]

Exercise 1, Section 9.3 in Dummit and Foote, Abstract Algebra, reads as follows:

Let R be an integral domain with quotient field F and let \(\displaystyle p(x) \in R[x] \) be monic. Suppose p(x) factors non-trivially as a product of monic polynomials in F[x], say \(\displaystyle p(x) = a(x)b(x) \), and that \(\displaystyle a(x) \notin R[x] \). Prove that R is not a unique factorization domain. Deduce that \(\displaystyle \mathbb{Z}[2\sqrt{2}\) is not a unique factorization domain.
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I found a proof on Project Crazy Project which reads as follows:

Suppose to the contrary that R is a unique factorization domain. By Gauss’ Lemma, there exist \(\displaystyle r, s \in F \) such that \(\displaystyle ra(x), sb(x) \in R[x] \)and \(\displaystyle (ra(x))(sb(x)) = p(x) \). Since p(x), a(x) and b(x) are monic, comparing leading terms we see that rs = 1. Moreover, since a(x) is monic and \(\displaystyle ra(x) \in R[x] \), we have \(\displaystyle r \in R \). Similarly, \(\displaystyle s \in R \), and thus \(\displaystyle r \in R \)is a unit. But then \(\displaystyle a(x) \in R \), a contradiction. So R cannot be a unique factorization domain.

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Question (1)

Consider the following statement in the Project Crazy Project Proof:

"Moreover, since a(x) is monic and \(\displaystyle ra(x) \in R[x] \), we have \(\displaystyle r \in R \)."

My reasoning regarding this statement is as follows:

Consider $$a(x) = x^n + a_{n-1}x^{n-1} + ... ... + a_1x + a_0$$ ... ... ... (a)

Then $$ra(x) = rx^n + ra_{n-1}x^{n-1} + ... ... + ra_1x + ra_0$$ ... ... ... (b)

In equation (b) r is the coefficient of $$x^n$$ and coefficients of polynomials in R[x] must belong to R

Thus $$r \in R$$

Is this reasoning correct? Can someone please indicate any errors or confirm the correctness.

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Question (2) The last part of the Project Crazy Project (PCP) reads as follows:

Moreover, since a(x) is monic and \(\displaystyle ra(x) \in R[x] \), we have \(\displaystyle r \in R \) Similarly, \(\displaystyle s \in R \), and thus \(\displaystyle r \in R \)is a unit. But then \(\displaystyle a(x) \in R \), a contradiction. So R cannot be a unique factorization domain.Why does r and s being units allow us to conclude that $$a(x) \in R[x]$$? (I am assuming that the author of PCP has made an error in writing R in this expression and that he should have written R[x]

I would be very appreciative of some help.

Peter

[Please note that this set of questions is also posted on MHF]

 

[Euge]

Since $r$ is a unit and $ra(x)\in R[x]$, then the product $r^{-1}(ra(x))\in R[x]$, that is, $a(x)\in R[x]$.