Proof in field of quotients (form of elements)

Question

Let $R$ be an unique factorization domain. Being an integral domain, it has a field of quotients $F$. We can consider $R[x]$ to be a subring of $F[x]$.

Given any polynomial $f(x)\in F[x]$, then $f(x)=(f_0(x)/a)$, where $f_0(x)\in R[x]$ and where $a\in R$.

I don't understand why $a\in R$, in the proof of the theorem that Every integral domain can be imbedded in a field, we used $a/b$, where $a,b$ where in the integral domain. So why in this example above we have $a/b$ where $a\in R[x]$ and $b\in R$?

Answer 1

$F[x]$ is the set of polynomials of $x$ with coefficients in $F$, the field of quotients of $R$. You may be thinking of $F(x)$, the field of rational polynomials (field of quotients of $R[x]$.

Answer 2

The point is that we can choose a common denominator for all the fractional coefficients (e.g. the product or lcm of the denominators), which allows us to rewrite it in said form, just like every polynomial in $\,\Bbb Q[x]\,$ can be written as the quotient of a polynomial with integer coeff's and integer denominator.