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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 $L$.

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

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!

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George Bergman has a 2 page handout which develops Luroth's Theorem as a sequence of exercises; it is in Postscript, and can be found here. –  Arturo Magidin Feb 13 '12 at 4:24

2 Answers 2

up vote 5 down vote accepted

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 $L=k(u)$.

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Let me process this: I pick $u=P/Q\in L$ such that $u$ is of minimal height (isn't this also called degree?) and $P,Q$ are monic, $\deg P>\deg Q$, and of course $(P,Q)=1$. Then $f(t)=P(t)-uQ(t)$ is going to be irreducible in $L[t]$ because (since it is monic in $t$), if it factored nontrivially then either (a) a factor would be in $k[t]$ but then this factor would also divide both $P,Q$, impossible, or (b) both factors would contain $x$, but then their heights would be lower, implying coefficients $\in k(x)$ of lower height, contradicting $u$'s minimality in $L$. –  Ben Blum-Smith Feb 15 '12 at 22:36
    
But since $f$ clearly has $x$ as a root, irreducibility in $L[t]$ means it is $x$'s minimal polynomial over $L$, and since it is clearly also irreducible over $k(u)$ then it is $x$'s min. poly. over this too, so $k(x)$ has the same degree over $L$ as over $k(u)\subset L$, and we conclude $L=k(u)$. That's it, right? The proof strategy seems to be the union of both my approaches (which I was not thinking of trying to unify) together with the idea of normalizing $u$ in the specified ways. So cool how it all comes together. Thanks! –  Ben Blum-Smith Feb 15 '12 at 22:40
    
@BenBlum-Smith: Height is not the degree, because it is the maximum of the degrees of $P$ and $Q$. So height does not satisfy the same rules as degrees; e.g., $\frac{1}{t^3}$ has height $3$, so does $\frac{t^3}{1}$, and so does their product. –  Arturo Magidin Feb 16 '12 at 3:58
    
No it's true, I checked! At least in complex analysis, if $f=p(x)/q(x)$ is a rational function written in lowest terms i.e. $(p,q)=1$, then $\max(\deg p,\deg q)$ is called the degree of $f$. See en.wikipedia.org/wiki/…. It is the degree in the topological sense of $f$ as a map from the Riemann sphere to itself. It seems to me that it is also (from an algebraic point of view) the degree of the field extension $k(x)/k(f)$, do you agree? –  Ben Blum-Smith Feb 16 '12 at 20:16
    
@BenBlum-Smith: Oh, you meant "degree" in a different context... well, okay. But it's confusing to use "degree" to talk about what Bergman calls the "height", when we are also going to talk about "degree" of extensions and "degree" of polynomials. "Height" is a good name, because it corresponds to the notion of height of the rationals, where $\mathrm{height}(p/q) = \max\{|p|,|q|\}$; so it generalizes naturally to the field of quotients of any Euclidean domain, which is what we are doing here. –  Arturo Magidin Feb 16 '12 at 20:21

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 $k$ .

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+1 Context always appreciated! –  Ben Blum-Smith Feb 15 '12 at 22:41
    
@Arturo Magidin I have a question with this proposition . "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)". It's difficult to prove it? Because I can't :/! –  Daniel Oct 1 '12 at 0:34

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