I have just begun to read Shafarevich's Basic Algebraic Geometry. In the first section of the first chapter, he quotes Luroth's theorem, which states that any subfield of $k(x)$ that is not just $k$ is isomorphic to $k(x)$, i.e. is generated as a field over $k$ by a single rational function of $x$. I have been trying to find a proof. I am stuck, and would appreciate any hints to fill in the argument. (I have consulted Wikipedia, Wolfram Mathworld, and this MathOverflow question, but so far haven't been able to satisfy myself.)
I have thought about two approaches so far. My question would be answered by a suggestion about how to complete either one of these ideas. Here they are:
Let $k\subset L \subset k(x)$ be an intermediate field not equal to $k$.
Approach #1: Any element of $k(x)$ not in $k$ is transcendental over $k$; meanwhile, $k(x)$ has transcendence degree 1 over $k$; it follows that $L$ has transcendence degree 1 over $k$. Thus $k(x)$ is algebraic over $L$.
Let $p(t)$ be the minimal polynomial of $x$ over $L$.
where $l_1,\dots,l_n\in L$ (and are thus rational functions of $x$). Now if the theorem is really true, $L=k(f)$ for some $f\in k(x)$; and $f=r/s$, with $r,s\in k[x]$. Then $p(t)=r(t)-fs(t)$. This is degree $n=\max(\deg r,\deg s)$ in $t$. Any coefficient of any power of $t$ in $p(t)$ is actually either in $k$ (if this power of $t$ does not appear in $s$), or else it is a linear function of $f$ and thus a field generator for $L$ and degree $n$ as a rational function of $x$. Thus I expect to be able to prove that, with $p(t)$ defined as above, actually any of the coefficients $l_1,\dots,l_n$ not contained in $k$, i.e. any of them (say $l_i$) that is a nonconstant function of $x$, is degree $n$ as a function of $x$ and is thus a field generator for $L$. (It would be sufficient to prove that it is degree $n$ as a function of $x$, because then $k(x)\supset L \supset k(l_i)$, but $[k(x):L]=[k(x):k(l_i)]=n$.) One internet source I found suggested that this is the right approach, but I can't seem to fill it in. Here's what I've got:
$p(t)$ is divisible by $t-x$ over $k(x)$ (since $x$ is a root), and over $k(l_1,\dots,l_n)$ it is irreducible (since this field is contained in $L$). I can't see that there is anything else I know about it for sure. It must be that irreducibility over $k(l_1,\dots,l_n)$ implies that $l_1,\dots,l_n$ are all either degree $n$ or else in $k$; but I haven't figured out how. From examples I have worked out (in which I chose $l_1,\dots,l_n$ semi-arbitrarily to fulfill $(t-x)\mid p(t)$), this seems to be true; if I make any of them different in degree from $0$ or $n$, then usually I can also get $x$ as a rational function of them, thus in these examples $k(l_1,\dots,l_n)=k(x)$ and $p(t)$ is divisible by $t-x$ over $k(l_1,\dots,l_n)$. Of course I assume it can also happen that I choose $l_1,\dots,l_n$ so that $k(l_1,\dots,l_n)\neq k(x)$, but $p(x)$ will still factor over $k(l_1,\dots,l_n)$ as long as any of the $l_i$ not in $k$ differ in degree from $n$. In any case all the calculations have felt ad-hoc and I haven't so far seen a reason for what is happening. So any hints here would be appreciated.
Approach #2: Because the theorem reminds me of the result that $k[x]$ is a p.i.d., I have also been unable to escape the following thought: let $f\in L$ be an element of $L$ of minimal degree as a function of $x$, and suppose that there is some other element $g\in L$ not in $k(f)$. Can I construct some element of $L$ using $f$ and $g$ (i.e. an element of $k(f,g)$) that contradicts $f$'s minimality in degree? I have not given this approach as much thought as the above, but again, so far I have not seen how to carry out the construction. The Euclidean-algorithm trick that proves $k[x]$ is a p.i.d. is unavailable here because I can't multiply $f$ or $g$ by anything that is not a rational function of one or the other of them. (In particular I can't see how to pass to a polynomial ring in $x$ but make sure I've stayed inside $k(f,g)$.) $g$ does have a minimal polynomial over $k(f)$, and if $g\notin k(f)$ then its degree is $>1$, so this could be a starting point for trying to construct the lower-degree element of $k(f,g)$, but again I haven't seen how to make this work. So here again, I would appreciate any thought that could be used to complete the argument.
Thanks in advance!