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F is a field and F(x) is the field of rational functions in one variable x over F. Are there some clever ways to prove that some field extension K/F is (or is not) isomorphic to F(x)? One way is to try to see if I can embed K into some field which is isomorphic to F(x). Then K would also be isomorphic to F(x), by Luroth's theorem. But are there other ways to do that? It's ok to suupose that the transcendence degree of K over F is 1.

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You could look at unramified cohomology? See… – Mikhail Bondarko Aug 24 '10 at 10:07
Thanks, I will. – iravan Aug 24 '10 at 12:20

I think that your question is formulated to broad to give a precise answer.

I assume that K is given in the form F(x1,...,xk)/(f1,..,fm). If your equations f_1...f_m are nice enough (low degre etc.) then several computer algebra packages can tell you transcendence degree of K/F using e.g. Groebner Bases. If this tr. degree is different from one then you are done, otherwise K is the function field of a smooth projective curve C/F.

Now the function field of C/F is isomorphic to F(x) if and only if C is isomorphic to ℙ1 .

If F is algebraically closed then it suffices to show that C has genus 0. If char(K)=0 this is can be done relatively easy: following the proof of the lemma of the primitive element you can write K=F(x,y)/f. One can consider y as a function on C and this yields a morphism g:C→ ℙ1. The Riemann-Hurwitz formula (RH) gives you an easy recipe to calculate the genus of C. (Most of the details are explained in Fulton's book on algebraic curves, if only care about RH you might also read the first two chapters of Silverman's book on Arithmetic of elliptic curves.) If F is alg. closed, but char(K)>0 this recipe also works, except that the calculation of the entries in the RH formula is harder. So if F is alg closed then a computer can do the job.

If F is not alg. closed you have a harder problem. You need to use the following criterion a smooth projective curve C is isomorphic to ℙ1 if and only if the genus of C is zero and C has a point with coordinates in F.

So besides the genus calculation you need to show that the curve C has at least one point with coordinates in F. This is in general a hard problem, if F is finite this is known and if F is a number field there is a good criterion to check whether C has a point with coordinates or not, for arbitrary F this seems hard.

Examples ℂ[x,y]/(x^2+y^2+1) is isomorphic to ℂ[z], but ℝ[x,y]/(x^2+y^2+1) is not isomorphic to ℝ[z], since the conic x^2+y^2+1 has non ℝ-points.

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I don't have f_1, ... , f_m but still I like to try to understand your answer better. Thank you. – iravan Aug 24 '10 at 12:24
And yes, it's ok to take F to be algebraically closed of characteristic 0. I think I can even assume that F is the field of complex numbers. – iravan Aug 24 '10 at 12:46
Maybe you could give a hint what you know about K, (since you claim you do not have a presentation in terms of generators and relations). – Remke Kloosterman Aug 24 '10 at 13:02
Nevermind, sadly I just found out that K cannot be isomorphic to F(x), which is a big disappointment. Thank you. – iravan Aug 24 '10 at 20:59

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