I trying to do problem $9.3.1$ in Dummut & Foote; I think they use some "abuse of notation" that I don't understand.
Let $R$ be an integral domain with quotient field $F$ and let $p(x)$ be a monic polynomial in $R[x]$. Assume $p(x) = a(x)b(x)$ where $a(x)$ and $b(x)$ are monic polynomials in $F[x]$ of smaller degree than $p(x)$. $\color{red}{\text{Prove that if a(x)} \not \in \text{R[x] then R is not a U.F.D.}}$ Deduce that $\mathbb{Z}[2\sqrt2]$ is not a U.F.D.
The part in red is what's confusing me. The reason why I think it's "abuse of notation" is because there is nothing to distinguish $a(x)$ from $b(x)$. So I could see an argument for each of the following interpretations, and I don't know which one is correct:
- Exactly one of $a(x), b(x)$ is not in $R[x]$
- At least one one of $a(x), b(x)$ is not in $R[x]$
- Precisely both of $a(x), b(x)$ are not in $R[x]$
Any help is appreciated.
Edit for Lord Shark the Unknown:
Proof that if $(x+a)(x+b) \in \mathbb{Z}[x]$ and $a, b \in \mathbb{Q}$ then $a, b \in \mathbb{Z}$.
Since $(x+a)(x+b) = x^2 + (a+b)x + ab$ we know $ab \in \mathbb{Z}$ and $a+b \in \mathbb{Z}$. Thus:
$$a^2+2ab+b^2 \in \mathbb{Z}$$
$$\implies a^2-2ab +b^2\in \mathbb{Z}$$
$$\implies (a-b)^2 \in \mathbb{Z}$$
$$\implies a-b \in \mathbb{Z}$$
$$\implies (a+b)(a-b) \in \mathbb{Z}$$
$$\implies a^2-b^2 \in \mathbb{Z}$$
Now we can also get that $a^2+b^2 \in \mathbb{Z}$ using the same trick we used in the first steps. So $(a^2+b^2)-(a^2-b^2) \in \mathbb{Z}$,
$$\implies 2a^2 \in \mathbb{Z}$$
$$\implies a^2 \in \mathbb{Z}$$
$$\implies a \in \mathbb{Z}$$
And by a similar argument, $b \in \mathbb{Z}$.