Polynomial factorisation for unique factor domain

Question

Suppose $R$ is a UFD and $f \in R[X]$ such that $\deg f > 0$ and $f$ has a root $\alpha \in R$. Show that $f = (X - \alpha) g$ for some $g \in R[X]$. (Suggestion: Write $f = a_0 + a_1 X + \dotsc + a_n X^n$, $g = b_0 + b_1 X + \dotsc + b_{n-1} X^{n-1}$ where $a_i \in R$, $b_i \in Q$ where $Q$ is the field of fractions of $R$. Expand $(X - \alpha)g$ and use that $\alpha \in R$ and $R$ is a ring to show that $b_{n-1},\dotsc,b_0 \in R$.)

This is the question I am attempting from my textbook. My initial thought was to use the division algorithm, but this does not seem to work. Can anyone tell me how to do this for a UFD?

Answer 1

You do not need an UFD. An integral domain is enough. Use the division algorithm in the quotient field and show that the resulting $g$ is contained in $R[X]$.

Some details:

We have $$a_0+a_1X + \dotsb a_nX^n = (X-\alpha)(b_0 + b_1X + \dotsb + b_{n-1}X^{n-1})$$

Considering the coefficients at $X^n$ shows $b_{n-1}=a_n \in R$. We proceed like this and conclude $b_{n-2} \in R$ by looking at $X^{n-1}$ and so on...

Answer 2

Evaluate $f(x) - f(α)$ (this is equal to $f(x)$, due to $α$ being a root of $f$).

A factor of $(x-α)$ can be taken out after some manipulation.

So you will now have $f(x)=(x-α)g(x)$

$g(x)$ will most likely look quite messy, but if you collect like terms and relabel these as your $b_i$'s you will be able to see that $g$ is in fact in $R[x]$ too, due to $R$ being a ring.