Let $R$ be a unique factorization domain, let $F$ be the field of fractions of $R$ and let $f(x)\in R[x]$. I want to show that if $f(x)$ is irreducible in $F[x]$ then $f(x)$ is irreducible in $R[x]$.
Suppose that $f$ is reducible in $R[x]$. Then $f(x)=a(x)b(x)$ for $a(x),b(x)\in R[x]$ where $\deg a(x)$ and $\deg b(x)$ are smaller than the $\deg f(x)$.
If the only way to factor $f(x)$ is so that one of $a(x)$ is a constant in $R[x]$, then wouldn't $f(x)$ be irreducible in $F[x]$ since $a(x)$ would be a unit in $F[x]$? What am I missing here?
There was a corollary to Gauss's Lemma proved in the text that says:
Let $R$ be a unique factorization domain, let $F$ be its field of fractions and let $p(x)\in R[x]$. Suppose that the greatest common divisor of the coefficients of $p(x)$ is 1. Then $p(x)$ is irreducible in $R[x]$ if and only if it is irreducible in $F[x]$.
The proof of this corollary is easy with the extra condition that the greatest common divisor of the coefficients is 1. Is it still true if the greatest common divisor of the coefficients is larger than 1?
Best Answer
The statement you have at the beginning of the question is false. Consider $f(x) = 2x$. Since $2$ is not a unit in $\mathbb Z$, it follows that $f(x)$ is reducible in $\mathbb Z[x]$. On the other hand, $f(x)$ is irreducible in $\mathbb Q[x]$. In fact, $\mathbb Q[x] / (2x) = \mathbb Q[x] / (x) \cong \mathbb Q$.