Lüroth's Theorem

I think that Bergman's suggested approach (from his handout in postscript) follows along the lines of your first approach, though perhaps organized a bit differently.

(For the benefit of those lacking a Postscript reader)

Preliminaries:

Every element of $k(x)[t]$ can be written as $$\frac{P(x,t)}{Q(x)}$$ where $P(x,t)$ and $Q(x)$ are relatively prime in the UFD $k(x)[t]$, and $Q$ is monic in $x$.

Given such an expression for an element of $k(x)[t]$, define its height to be the maximum of the degree of $P$ in $x$ and the degree of $Q$ in $x$. This also applies to elements of $k(x)$.

• If $u=P(x,t)/Q(x)$ is monic in $t$ (viewed as an element of $k(x)[t]$, then the height of $u$ equals the degree of $P$, and $P$ is not divisible by any nonunit element of $k[x]$.

• If $f,g\in k(x)[t]$ are both monic as polynomials in $t$, then $\mathrm{height}(fg) = \mathrm{height}(f)+\mathrm{height}(g)$.

• If $u\in k(x)$, $u\notin k$, then there exists $u'\in k(x)$, such that $\mathrm{height}(u')=\mathrm{height}(u)$, with $k(u)=k(u')$, and such that when we write $u'=P'(t)/Q'(t)$, $P'$ and $Q'$ coprime, we will have $\deg(P')\gt \deg(Q')$, and both are monic. In fact, $u'$ can be taken of the form $\alpha u$ or $\alpha/(u-\beta)$, $\alpha,\beta\in k$.

• If $u\in k(x)-k$, $u=P(x)/Q(x)$, then $x$ is a root of $P(t)-uQ(t)\in k(u)[t]$. If $\deg_x(P)\gt \deg_x(Q)$ and $P$ is monic, then the polynomial $P(t)-uQ(t)$ is monic.

Argument.

Let $L$, $k\subseteq L\subseteq k(x)$, and pick $u\in L-k$, $u=P(x)/Q(x)$, that minimizes the height; let $\mathrm{height}(u)=n$.

• Show $P(t)-uQ(t)$ is either irreducible over $L$, or divisible by a nonunit element of $k[t]$ in $L[t]$.

• Show that if $P(t)-uQ(t)$ is divisible by a nonunit element of $k[t]$ in $L[t]$, then the element divides both $P(t)$ and $Q(t)$.

• Conclude that $P(t)-uQ(t)$ is the minimal polynomial of $u$ over $L$.

• Show that $P(t)-uQ(t)$ is the minimal polynomial of $x$ over $k(u)\subseteq L$, and conclude that $L=k(u)$.

Since this is clearly a tough technical result, it might be of some interest to know why we should care about it.

The geometric interpretation is that if $f: \mathbb P^1_k\to X$ is any non constant morphism from the projective line to any complete nonsingular algebraic curve over $k$, then $X$ is actually another copy of the projective line, $X=\mathbb P^1_k$, and $f$ is a rational function.

Over $\mathbb C$ the analogous result for Riemann surfaces is true and can be proved as follows:
We can lift $f$ to the universal cover of $X$ (because $\mathbb P^1(\mathbb C)$ is simply connnected) and obtain a holomorphic map $\tilde f:\mathbb P^1(\mathbb C) \to \tilde X$.
But if $X$ had genus $g\gt 0$, its univeral cover $\tilde X$ would be a disc or $\mathbb C$ (according to the difficult Riemann uniformization theorem) and since $P^1(\mathbb C)$ is compact, $\tilde f$ would be constant and $f$ would be constant too: contradiction.

Finally, let me make three little comments:
1) The field $k$ in Lüroth's theorem is completely arbitrary and needn't be algebraically closed.
2) There are purely geometric proofs of Lüroth for arbitrary fields (not only for $\mathbb C$) but they assume some algebraic geometry, Riemann-Roch for example.
3) The analogue of Lüroth is in general false for the rational function fields $k(x_1,...,x_n) \; (n \gt 1)$ : its subfields are not all purely transcendental extensions of $k$ .