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Hi, everyone!

I'm trying to explain the proof of Luroth theorem (every field $L$, s.t. $K\subset L\subset K(t)$, is isomorphic to $K(t)$) to the high-school audience. I'm not going to use such methods as algebraic extensions and complex analysis. Is there any way to prove this fact with only elementary methods?

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    $\begingroup$ There is an elementary proof in Winter: The structure of fields, Graduate Texts in Mathematics 16, Springer. It does use a little bit of field theory however but that could possible be whittled away. $\endgroup$ Jun 6, 2011 at 8:31
  • $\begingroup$ @Torsten: Also I do not want to use linear algebra $\endgroup$
    – zroslav
    Jun 6, 2011 at 9:22
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    $\begingroup$ I must say that I am very curious about a presentation of Luroth's Theorem for high schoolers which is not motivated by complex analysis or linear algebra. How did you hit upon this topic? What is your angle on it? $\endgroup$ Jun 6, 2011 at 11:58
  • $\begingroup$ What happens if you take an element of least degree (in the same sense of Charles Matthews's answer below) in the subfield $L$? Is there any chance of showing directly, by some sort of induction on the degree, that such an element generates $L$ as a field? $\endgroup$
    – Emerton
    Jun 6, 2011 at 12:59
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    $\begingroup$ @zroslav: The theorem just states that any subfield of the big field properly containing the ground field must be a purely transcendental extension (necessarily simple, by comparison of transcendence degrees). This requires some discussion of fields including the distinction between algebraic and transcendental extensions. $\endgroup$ Jun 6, 2011 at 17:38

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See here: http://math.berkeley.edu/~gbergman/grad.hndts/Luroth.ps

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I think this is something Gauss could have proved, and the point is to come up with his sort of proof. I'm not seeing that as too hard. To show a polynomial of degree at least 1 is transcendental over K is easy. The harder step is to do that for rational functions P/Q, where the degree should be defined as the maximum of degree P and degree Q. So by rationalising we need to look at F(P, Q) where F is a general binary form (well, you should start with a monic polynomial to get to F, is what I mean). In the case of unequal degrees there is an easy reason why this can't be zero, looking at the top power of t. So we should assume equal degrees. But then P/Q can be written as a constant plus a term with unequal degrees, by long division of polynomials. That looks like the inductive step in a proof, by induction on the degree.

Presumably you would want to express the idea that K(P/Q) and K(t) are then isomorphic in some other fashion, but I don't know quite what you have in mind.

Edit: In reply to the comment: we want to prove unirational implies rational for curves C, as geometers would put it. So far we have enough to prove C has a non-constant rational map to the projective line L. This clearly isn't enough yet, but the work involved can be reduced to some finiteness statement. It would be enough to show that there are only finitely many intermediate fields, i.e. "curves" C, for a given rational map L -> L. That statement is known to imply the primitive element theorem over infinite fields, by a box principle argument. Abstract field theory takes us to Steinitz, away from Gauss indeed (and Luroth, really). Without at least the ACC for subfields in this case, how are we going to prove that a general subfield is finitely generated? I ask because we don't really have a clear statement from the OP about how many limbs we have to tie behind our backs here.

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    $\begingroup$ Unless I am confused, you are not addressing the hard part of the theorem. See my comment to Emil. $\endgroup$ Jun 6, 2011 at 13:32
  • $\begingroup$ OK, I was trying to reconstruct an old question on an example sheet. You obviously need something like a primitive element theorem also, to show that a subfield generated by a couple of elements is also generated by a single element. $\endgroup$ Jun 6, 2011 at 14:25
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Okay, I want to finally close the question: I've read Schinzel's book and there I found Ostrovski's elementary and constructive proof. Actually, A_H's answer is very similar to Ostrovski's proof.

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  • $\begingroup$ I do not see why this needs to be closed. It is a natural question. Anyway, discussion of closure should be in the meta site, tea.mathoverflow.net , so you may want to open a thread there. $\endgroup$ Dec 28, 2012 at 15:48
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    $\begingroup$ Also, can you give a more precise reference to the proof you found? $\endgroup$
    – Ravi Vakil
    Dec 9, 2013 at 0:36

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