MHB How Do You Determine the Degree of the Field Extension K/L in POTW #173?

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Here is this week's POTW:

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Let $x$ be a variable and $K = F(x)$ the field of rational functions in $x$ over a field $F$. Let $L = F\left(\frac{f(x)}{g(x)}\right)$, where $f(x),\, g(x)\in F[x]$ are relatively prime and $\frac{f}{g}\in K\setminus F$. Show that $K/L$ is a finite extension and evaluate $[K:L]$ in terms of $f$ and $g$.
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No one answered this week's problem. You can read my solution below.

Spoiler

Let $r = \frac{f(x)}{g(x)}$. Since $K = L(x)$ and $x$ is a root of the polynomial $q(t) := f(t) - rg(t)$, which is a polynomial over $L$, then $[K:L]$ is finite. Since $r\notin F$, $\deg q(t) = \max\{\deg f(x), \deg g(x)\}$. The claim is that $[K:L] = \max\{\deg f(x), \deg g(x)\}$. This will follow by showing that $q$ is irreducible over $L$. Since $[K:L]$ is finite and $[K:F]$ is infinite, then $[L:F]$ is infinite, so $r$ is transcendental over $F$. Hence, $F$-polynomials in variable $t$ can be identified with $F$-polynomials in $r$.

As a polynomial in $r$, $q$ is a degree one polynomial with coefficients in $F(t)$. In particular, $q$ is an element of $F(t)[r]$. So $q$ is irreducible over $F(t)$. In fact, $q$ is irreducible over $F[t]$ since it is primitive in $F[t][r]$ (as $\gcd(f(x),g(x)) = 1$) and irreducible over $F(t)$. Now as $F[r][t] = F[t][r]$, $q$ is irreducible in $F[r][t]$, and thus $q$ is irreducible over $F(r)$ , which is $L$.