# Polynomial without prime factorization is reducible

I need help proving the following: If $f(x) \in R[x]$ has no prime factorization and has degree at least 1, where $R$ is a UFD, then $f(x)$ is reducible.

This is a step in a larger proof, see below, that I can't work out (polynomial ring over UFD is UFD).

Proof for "if $R$ is UFD, then $R[x]$ is UFD":

We need to show if $f(x) \in R[x]$, $f(x)\neq 0$, then $f(x)$ has prime factorization. Suppose this is false, pick $f(x)$ such that it has no prime factorization and $deg(f)$ is as small as possible, then $deg(f) \ge 1$.

Define $c(f)$ to be the GCD of the coefficients of $f(x)$. Write $f(x) = c(f) f_1$, then $c(f_1(x))$ is a unit. Either $c(f)$ is a unit in $R$, or $c(f)$ has prime factorization. $f_1(x)$ cannot have prime factorization, in particular, $f_1(x)$ is reducible. (Note: the preceding claim I can not verify)

So $f_1(x) = g(x) h(x)$ such that $g(x)$ and $h(x)$ are not units. By another lemma, $c(f_1) = c(g) c(h)$, hence $c(g)$ and $c(h)$ are units. Hence, $deg(g)$ and $deg(h) \ge 1$. Thus $deg(g)$ and $deg(h) < deg(f_1) = deg(f)$. By the choice of $f(x)$, $g(x)$ and $h(x)$ have prime factorization, which is a contradiction. $\square$

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Please be more precise. What precisely do you mean by "has no prime factorization"? – Bill Dubuque Dec 6 '12 at 19:58
Sure. Originally, this came out of a step in the proof that "polynomial ring over UFD is UFD". So that we can't yet assume that $f(x)$ has a prime factorization. Is this true? – beeflavor Dec 6 '12 at 20:16
Have you already proved that irreducibles are prime? It would help greatly if you provided a link to the proof you are reading. – Bill Dubuque Dec 6 '12 at 20:21
I have edited my post to reflect this. Thank you. – beeflavor Dec 6 '12 at 20:45
Thanks for appending the proof. What is your definition of a prime element of $R[x]$? – Bill Dubuque Dec 6 '12 at 21:17

You can show existence in the same manner (but without contradiction) easily. Take $f(x), c(f)$ being the gcd of the coefficients, $f(x)=f_1(x)c(f)$. Therefore $f_1(x)$ is irreducible or factors into lower-degree polynomials. Factorization follows trivially.
To prove that $R[x]$ is a UFD you also have to show uniqueness.