L Roth S Theorem

\href{https://ajwilson.co.uk/files/maths/luroth-problem.pdf{Theorem 1.}} (Lüroth) Suppose we have an inclusion of fields $k \subset L \subseteq k(t)$, where $k \neq L$ and $k(t)$ is the field of rational functions in one variable $t$. Then $L$ is isomorphic to $k(t)$. We now sketch an elementary algebraic proof of Lüroth's theorem. For details see [Waerden] Section 63. Considering any element $\lambda \in L \backslash k$, we observe that $t$ is an algebraic element of $k(\lambda)$ and so an algebraic member of $L$. Next, examine the polynomial \[ f(z)=z^n+a_1 z^{n-1}+\cdots+a_n \in L[z] \] where the $a_i$ are rational functions in $x$. Multiplying through by the lowest common denominator yields polynomials and we may write \[ f(x, z)=b_0(x) z^n+b_1(x) z^{n-1}+\cdots+b_n(x) \] Let the degree of $f$ with respect to $x$ be $m$. Note that not all the coefficients $a_i=\frac{b_i}{b_0}$ in $f(x)$ can be independent of $x$, since that would imply that $x$ is algebraic with respect to $k$. Thus at least one of the terms $a_i=\theta$ must be dependent on $x$. To complete the proof, one uses elementary field extension properties to show that (i) $m=n$, and (ii) $\theta$, as a function of $x$, is of degree $m$. It follows that \[ [k(x): k(\theta)]=m=[k(x): L] \] and as $L \supseteq k(\theta)$, we have $[L: k(\theta)]=1$. Therefore \[ L=k(\theta) \cong k(t) \] by a change of variables.