0
$\begingroup$

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. $\square$

  • 0
    Toward the bottom, you'll find the relevant proof to this question.2012-12-18

1 Answers 1

1

Your proof: firstly it doesn't even fully prove that the ring of polynomials over a UFD is a UFD. What you're doing is basically showing existence of prime factorization via a contradiction, which you don't even need incidentally (and your proof has various errors/typos that make it unclear/incorrect).

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.