If $x$ is an algebraical element over $F$, then $F(x) = F[x]$

Question

Let $K / F$ be an extension of fields and let $F[X]$ be the ring of polynomials $\sum_{i = 0}^n a_i X^i$, where $n \in \mathbb{N}$ and $a_i \in F$ for all $i = 1 , \ldots , n$. The inclusion $F[x] \subset F(x)$ is trivial for all $x \in K$, being $$ F(X) = \left\{\frac{p(X)}{q(X)} : p(X) , q(X) \in F[X] \mbox{ and } q(X) \not \equiv 0\right\}\mbox{,} $$ but also $F(x) \subset F[x]$ if $x$ is an algebraic element over $F$. How can I prove that? In other words, how can I show that there exists $\frac{p(x)}{q(x)} \in F(x)$ such that it does not belong to $F[x]$ if $x$ is trascendent over $F$?

Answer 1

If $x$ is algebraic over $F$, then $F[x]$ is a finite dimensional vector space over $F$: indeed let $f$ be the minimal polynomial for $x$; if you have a polynomial $g$, then $g=fq+r$, with the degree of $r$ less than the degree of $f$ and $g(x)=r(x)$. Hence, if the degree of $f$ is $n$, $F[x]$ is spanned by $\{1,x,\dots,x^{n-1}\}$.

Let now $g$ be a polynomial with $\deg g<\deg f$, with $g(x)\ne0$. Then, as $f$ is irreducible, $\gcd(f,g)=1$, so by Bézout there exist polynomials $p$ and $q$ with $1=pf+qg$. This implies $$ q(x)g(x)=1 $$ so $q(x)\in F[x]$ is the inverse of $g(x)$.

If instead $x$ is transcendental over $F$, then $\frac{1}{x}\notin F[x]$.