Degree of $F(x)$ over $F(y)$ where $y=f(x)/g(x)$ is a rational function.

Question

A problem states:

Let $x$ be a transcendental element over field $F$, and $y = f(x)/g(x)$ be a rational function, with relatively prime polynomials $f,g \in F[t]$. Let $n = \max(\deg f,\deg g)$. Suppose $n \ge 1$. Prove that $[F(x) : F(y)] = n$.

By taking $h(t)=yg(t)-f(t)\in F(y)[t]$, I see that $h(x)=0$, thus $[F(x) : F(y)] \le n$. But I have trouble proving the reverse inequality. Can any one help me?

Answer 1

The point is that $h(t,y)=g(t)y-f(t)$ is an irreducible polynomial in the polynomial ring $F[t,y]$. Therefore it is irreducible over $F(y)[t]$ and its degree is exactly $n$, and so adjoining a zero gives a degree $n$ extension of $F(y)$.