# Let $x$ be transcendental over $F$. Let $y=f(x)/g(x)$ be a rational function. Prove $[F(x):F(y)]=\max(\deg f,\deg g)$

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

My attempt: By replacing $$y$$ with $$\frac{1}{y}$$ if necessary we assume $$\deg g\geq\deg f$$. By the Euclid algorithm we assume $$\deg g>\deg f$$. Then $$\deg g=\max(\deg f,\deg g)$$, let $$n=\deg g$$. Let $$R(t)$$ be this polynomial $$R(t)=yg(t)-f(t)=\frac{f(x)}{g(x)}g(t)-f(t)$$ Then $$\deg R=\deg g=n$$ because $$n=\deg g>\deg f$$. The coefficients of $$R(t)$$ lie in $$F(y)$$ and hence $$R(t)\in F(y)[t]$$. Clearly $$R(x)=0$$.

Now I only have to prove $$R(t)$$ is actually the minimal polynomial of $$x$$ over $$F(y)$$. I believe the next step is to obtain a contradiction that $$R(t)$$ being reducible can lead to $$f,g$$ being not coprime, but I can't figure out how to do that.

P.S. A hint is preferred over a full answer.

Hint : Since $$x$$ is transcendental over $$F$$, we get that $$y$$ is also transcendental. In particular, $$F[y]$$ is a UFD, therefore to show that $$R$$ is irreducible in $$F(y)[t]$$, it is enough to show that it is irreducible in $$F[y][t]$$ by Gauss' lemma.
But then $$F[y][t] = F[t][y]$$, so you can act like $$y$$ is a variable and we can take the $$y$$ degrees during comparison of equality in this integral domain. What is the $$y$$ degree of $$R(t)$$?
To finish, what is the $$t$$ degree of $$R$$?
• Got it. If $R(t)$ has a factorization in $F[y,t]$ then one of the factors must have degree $0$ in $y$. Divide $R(t)$ by this factor one yields a polynomial, meaning this factor is common factor of $f,g$. But they are coprime hence the factor must be a constant. Am I right? – trisct Sep 12 at 10:41