Lüroth's Theorem

I have just begun to read Shafarevich's Basic Algebraic Geometry. In the first section of the first chapter, he quotes Lüroth'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 $L$.

$$p(t)=t^n+l_1t^{n-1}+\dots+l_n$$

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!

Solution 1:

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 $L=k(u)$.

Solution 2:

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 $k$ .