Field of fractions of PID R with infinitely many max ideals not a f.g. R-algebra

Question

Let $R$ be a principal ideal domain, with field of fractions $F$. Show that if $R$ has infinitely many maximal ideals then $F$ is not finitely generated as an $R$-algebra.

Answer 1

Suppose on contrary that, if there exist a surjective ring homomorphism $\phi : R[x_{1},...,x_{n}] \rightarrow F$. Call $\phi(x_{i}) = \frac{a_{i}}{b_{i}}$ where $i = 1,...,n$ and $a_{i}, b_{i}(\neq 0) \in R$. So the image of $\phi$ can contain only those terms whose denominator contains terms of the form $b_{1}^{m_{1}}...b_{n}^{m_{n}}$, where $m_{i} \geq 0$, so such expression involves only finitely many irreducible divisors. Since we are given that there are infinitely many maximal ideal(generated by single element which is irreducible/prime). Therefore if you choose a irreducible element different from those set of divisors, that will not be in the image, which leads a contradiction.