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Let $\mathbb F_p(t)$ the field of rational functions with coefficients in $\mathbb F_p$.

Is it true or not that every finite extension $K$ of $\mathbb F_p(t)$ is $K\cong\mathbb F_{p^m}(t)$ for some $m\ge 1$?

manifold
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  • I guess that $\Bbb F_{p^n}(t)$ is separable over $\Bbb F_p(t)$, while $\Bbb F_p(t)$ has non separable extensions. The notation $K \cong \Bbb F_{p^m}(t)$ is a bit ambiguous : it can be an isomorphism over $\Bbb F_p$ (just a field isomorphism), or an isomorphism over $\Bbb F_p(t)$ (i.e. as $\Bbb F_p(t)$-algebras). – Watson Oct 07 '17 at 10:07
  • Possibly related: https://math.stackexchange.com/questions/101406 (see "So given a function field over a perfect field K, the primitive element theorem holds for K"). – Watson Oct 07 '17 at 10:16

2 Answers2

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No. Other such extensions are fields of functions on some curve. For that to make sense you need to know a few basics from algebraic geometry. The idea is that the field of rational functions corresponds to the line, but more complicated curves have more complicated function fields.

For example, if $p>3$, then the equation $$ u^2=t^3+at+b,\qquad(*) $$ with $a,b$ some constants from $\Bbb{F}_p$, chosen in such a way that the cubic has no multiple roots, defines an elliptic curve $E$. The related function field is simply the algebraic extension $\Bbb{F}_p(E)=\Bbb{F}_p(t)[u]$, where $(*)$ gives the minimal polynomial of $u$ over $\Bbb{F}_p(t)$.

Jyrki Lahtonen
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    The proof that the field $\Bbb{F}_p(E)$ is not isomorphic to $\Bbb{F}_p(t)$ (or any of its constant field extensions) is a bit subtle. Stichtenoth's book has a relatively accessible one. May be there is a simpler case/way? Anyway, it is easy to show that $t^3+at+b$ has no square roots in $\Bbb{F}_p(t)$ (it's an odd degree polynomial). Proving that there is no element $z\in \Bbb{F}_p[E]$ such that $\Bbb{F}_p[E]\cong \Bbb{F}_p(z)$ is not trivial. – Jyrki Lahtonen Jan 21 '16 at 11:51
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    Right. I wonder whether the quickest way to exhibit two different fields, in this situation, might be to show that their automorphism groups are different. Over $k=\Bbb F_2$, for instance, $k(t)$ has exactly $6$ automs, but the fraction field of $k[x,y]/(y^2+y+x^3 +x)$ has many more, since it has five $k$-rational points, and you have the translations by these, thus a subgroup of order $5$. – Lubin Jan 23 '16 at 21:44
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The answer is no. For example, you can consider the fraction field of $\Bbb{F}_p[x,t]/(x^2-t)$: this is a quadratic extension of $\Bbb{F}_p(t)$, where the element $x= \sqrt{t}$ is added.

Crostul
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    A good idea! But my inner nitpicker wants to comment that your field $\Bbb{F}_p(t)[x]$ actually is $\Bbb{F}_p(x)$. And hence isomorphic to $\Bbb{F}_p(t)$, an isomorphism mapping $t$ to $x$. That is why I went with an elliptic curve. Genus $1$ makes sure that I'm out of the area of rational function fields. Anyway, your point of adjoining zeros of polynomials with coefficients in $\Bbb{F}_p(t)$ is correct, and it may be pedagogically better to start out with a simpler quadratic extension. IOW +1 – Jyrki Lahtonen Jan 21 '16 at 11:47