Degree of the extension $k(x)/k(q(x))$ for a rational function $q\in k(x)$ Let $k$ be an algebraically closed field, and $q\in k(x)$ a nonzero rational function, expressible as $q(x)=r(x)/s(x)$ for $r$ and $s$ coprime, and $d=\deg r\ge\deg s$. Then will $[k(x):k(q(x))]=d$? It is easy to see that $d$ is an upper bound, by using the polynomial
$$
r(T)-q(x)s(T)\in k(x)[T]
$$ which vanishes for $T=x$, but I don't see why this polynomial must be irreducible. Presumably, the proof of this will at some point involve Gauss's lemma, applied, say, to
$$
 s(x) r(T)-r(x)s(T)\in k[x,T],
$$ but I can't see how to make the details of the argument work.

Set $L:=k(x)$, and its subfield $K:=k(q)$, where $q=r/s$ for coprime polynomials $r,s\in k[x]$. The claim is that $L\cong K[t]/(r(t)-qs(t))$, and it is enough to show that $m:=r(t)-qs(t)$ is an irreducible polynomial in $K[t]$. (This is where you are making a mistake: we don't want to ask about this polynomial in $L[t]$.)
Now, $m$ is an irreducible element of $k(t)[q]$ (it is linear in $q$), and is primitive in $k[t][q]$, so by Gauss's Lemma it is irreducible in $k[q,t]$. Then Gauss's Lemma once more implies that it is irreducible in $k(q)[t]=K[t]$.
Thus $L\cong K[t]/(m)$, and so $L/K$ has degree $\max\{\deg(r),\deg(s)\}$.
