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 $R$?

  • $\begingroup$ 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? $\endgroup$ – trisct Sep 12 at 10:41
  • $\begingroup$ Yes, that is absolutely fine! $\endgroup$ – астон вілла олоф мэллбэрг Sep 12 at 12:21

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