On a proof of Lüroth's theorem While looking for a generalization of the result mentioned in my recent answer I came across this theorem in Galois Theory by David Cox (page 176):

Theorem 1: Let $F$ be a field and $F(t) $ denote the function field of rational functions in indeterminate $t$ with coefficients in $F$. Assume $\alpha \in F(t) \setminus F$ and write $\alpha =a(t) /b(t) $ where $a(t), b(t) \in F[t] $ are relatively prime. Then

*

*$\alpha$ is transcendental over $F$.

*The polynomial $a(x) - \alpha b(x) \in F(\alpha) [x] $ is irreducible over $F(\alpha) $.

*$F(\alpha) \subseteq F(t)$ is a finite extension with degree $[F(t) : F(\alpha)] =\operatorname{max} (\operatorname {deg} a(t), \operatorname {deg} b(t)) $.


A few pages later Cox mentions another theorem famous by name of Lüroth:

Lüroth's theorem: Let $F$ be a field and $t$ be an indeterminate and assume $K$ is a field with $F\subset K\subseteq F(t) $. Then there is a rational function $\alpha=f(t) \in F(t) $ such that $K=F(\alpha) $.

While Cox does not prove the above theorem in his book, I think it should be an easy corollary of the theorem 1 mentioned in the beginning of this post.
Using theorem 1 we can define a function $D:F(t) \setminus F\to\mathbb {N} $ by $$D(\alpha) =\operatorname {max} (\operatorname {deg} a(t), \operatorname {deg} b(t)) $$ where $\alpha =a(t) /b(t) $ with $a(t), b(t) \in F[t] $ being relatively prime.
Next if $\alpha \in K\setminus F$ then we have $F(\alpha) \subseteq K\subseteq F(t) $ so that $K\subseteq F(t) $ is also a finite extension whose degree does not exceed $D(\alpha) $.
Since $D(\alpha) $ is a positive integer we can find an $\alpha \in K\setminus F$ which minimizes $D(\alpha) $. For such a choice of $\alpha$ we must have $K=F(\alpha) $.
For if it were not so we would have $F(\alpha) \subset K$ and we can find a $\beta\in K\setminus F(\alpha) $ and then $$F(\alpha) \subset F(\beta) \subseteq K\subseteq F(t) $$ so that $[F(t):F(\alpha)]>[F(t):F(\beta)] $ ie $D(\alpha) >D(\beta) $. And this contradicts the fact that $\alpha$ is chosen so that $D(\alpha) $ is minimal.
The above does not work as we don't know if $F(\alpha) $ is contained in $F(\beta) $.
Another try: Let $D(\alpha) =n, \alpha=a(t)/b(t) $ with $a(t), b(t) \in F[t] $ being relatively prime. Also$[F(t) :K]$ is finite and $F(t) =K(t) $ it follows that $t$ is algebraic over $K$. Let  $g(x) \in K[x] $ be its minimal polynomial over $K$. Then since $t$ is also a root of $a(x) - \alpha b(x) \in K[x] $ we have $$a(x) - \alpha b(x) = g(x) h(x) \tag{1}$$ with $h(x) \in K[x] $.
Since $g(x) $ is monic we can write it as $g(x) =g(x, t) /g_1(t)$ where $g(x, t)\in F[t] [x] $ is primitive and $g_1(t)\in F[t]$ is the leading coefficient of $g(x, t) $ (viewed as a polynomial in $x$ ie as a member of $F[t] [x] $).
Further since $t$ is transcendental over $F$ at least one of the coefficients of $g(x) $, say $a_i$, of $g(x) $ is in $K\setminus F$ and hence $D(a_i) \geq n$. This means that $g(x, t) $ when viewed as a polynomial in $t$ ie as member of $F[x] [t] $ has degree greater than or equal to $n$.
We can also rewrite $h(x) \in K[x] $ as $h_1(t)h(x,t)/h_2(t)$ where $h(x, t) \in F[t] [x] $ is primitive and $h_1(t),h_2(t)\in F[t] $. Multiplying $(1)$ by $b(t) $ we get $$b(t) a(x) - a(t) b(x) =\left(\frac{b(t) h_1(t)}{g_1(t)h_2(t)}\right) g(x,t)h(x,t)\tag{2}$$ Treating both sides of above equation as members in $F[t] [x] $ we see that left side is primitive and $g(x, t) h(x, t) $ is also primitive. Hence the expression in large parenthesis on right side is just a member, say $c$, of $F$. Further the degree of left side viewed as a member of $F[x] [t] $ is $n$ and degree of $g(x, t) $ as member of $F[x] [t] $ is not less than $n$. Thus degree of $h(x, t)$ as a polynomial in $t$ is $0$ so that $h(x, t)=h_3(x) \in F[x]$ and we get $$b(t) a(x) - a(t) b(x) =ch_3(x)g(x,t)\tag{3}$$ The left hand side when viewed as a member of $F[x] [t] $ is primitive and hence we must have $h_3(x)\in F$ and thus $$b(t) a(x) - a(t) b(x) =dg(x, t) \tag{4}$$ for some $d\in F$. Then the degree of $g(x, t) $ viewed as a polynomial in $x$ is $n$. It follows that the minimal polynomial $g(x) \in K[x] $ of $t$ over $K$ is of degree $n$ and hence $K(t) =F(t) $ is of degree $n$ over $K$. But we have $F(\alpha) \subseteq K\subseteq F(t) $ and $F(t) $ is of degree $n$ over $F(\alpha) $. Hence $K=F(\alpha) $.

It took me quite sometime to establish that $a(x) - \alpha b(x)\in F(\alpha) [x] $ is not only irreducible over $F(\alpha) $ but also over $K$ and thus Lüroth theorem is not really a simple corollary of theorem 1, but rather a somewhat difficult result.
I may have missed some subtleties in the above proof but I think it is in much better shape now. Please let me know if there are any issues which need to be fixed.
The argument using primitive polynomials is based on Gauss Lemma and I hope I have used it correctly.
 A: This is a CW self-answer designed to eventually put the question out of unanswered queue.
The answer by Arturo Magidin which was also referenced in comments to the question gives an outline of the proof given by Bergman. The post does not include all the details (they form a part of postscript file linked there), however I could gather that the idea used is essentially the same as in my question.
The function $D$ of my post is called height in Arturo's answer and the proof establishes that the polynomial $a(x) - \alpha b(x) $ is irreducible over $K$ as well as over $K(\alpha) $. Irreducibilty over $K(\alpha) $ is a part of theorem 1 in my question and is available in Cox's book (and one of the approaches uses Gauss Lemma, another uses the fact that it is irreducible in $F[x, \alpha ] $).
The irreducibilty over $K$ as given in my question uses the fact that $b(t) a(x) - a(t) b(x) $ is primitive as a polynomial in $x$ and primitive as a polynomial in $t$ also and is of same degree in $x$ as well as in $t$.
