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If $F$ is any algebraically closed field, and $L \supset F(X)$ is a finite extension of the purely transcendental extension of $F$ of transcendence degree $1$, then can $L$ necessarily be embedded into $F(Y)$ (so that $X$ becomes a rational function in $Y$)?

It seems like this ought to be an easy question to google, but I haven't found anything. I did find this wikipedia article which indicates a connection to algebraic geometry, so if my grasp of algebraic geometry were strong enough, I would probably be able to answer this question on my own. Alas, I am not there yet. :)

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A field extension in this situation corresponds to a ("ramified") cover of an algebraic curve by another. $F(X)$ is the function field of the projective line. Genus is a useful invariant of these things: it can only increase (or not decrease, to be precise) in an extension, by the Riemann-Hurwitz relation. So, yes, there are some extension that do not increase genus (like $Y^2=X$), but in general the genus goes up. So the extension cannot be written as $F(Y)$ for any $Y$... or else it'd be the projective line, of genus $0$.

But, also, this is not quite to say that there's no $Y$ in the extension so that $X$ is a rational function of it. But maybe that's not what you intended to ask.

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  • $\begingroup$ To clarify, you're saying that since the Genus may only stay the same or go up in a field extension, as soon as the genus strictly increases from $F(X)$ to $L$, there cannot be an embedding $L \hookrightarrow F(Y)$ (fixing $F$, of course)? The embedding in that direction was my intended question, not $F(X) \hookrightarrow F(Y) \hookrightarrow L$. $\endgroup$ – Dustan Levenstein May 7 '14 at 23:06
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    $\begingroup$ Yes, exactly: an imbedding $L\to F(Y)$ would say that the projective line parametrized by $Y$ (or whatever one wants to say) lies over the curve corresponding to $L$. So if that curve has genus $>0$, this is impossible. $\endgroup$ – paul garrett May 7 '14 at 23:41
  • $\begingroup$ @DustanLevenstein: Dear Dustan, This result is known as Luroth's Theorem. Regards, $\endgroup$ – Matt E May 9 '14 at 1:52

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