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Let $K$ be the rational function field $k (x)$ over a perfect field $k$ of characteristic $p > 0$. Let $F = k(u)$ for some $u$ in $K$, and write $u = \frac {f(x)}{g(x)}$ with $f$ and $g$ relatively prime. Show that $K$ is separable over $F$ if and only if $u$ is not in $K^p$.

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@ Biswa Suppose $u$ is in $K^p$ and then show that $K/F$ is not separable. For the other way suppose $u$ is not in $K^p$ and then show that $K/F$ is separable. –  Reader Nov 2 '12 at 8:47
What have you tried? Which direction can you do? What are you having trouble with? –  Alex B. Nov 2 '12 at 9:58
@Reader & @ Alex I got the solution that if u is in K^p then K/F is not separable. But I am not getting the other way. I have a rough idea. Suppose x is inseparable in F then min(x;F) is in F[y^p]. By an exercise in Lang's Algebra 3rd Edition(Chapter V, Problem 20, Page 254) we get min(x;F) is ug(y)-f(y). Now if we can claim (may be using u is not in K^p and (f,g)=1) that f(y),g(y) is in F[y^p] then actually we get f(y),g(y) is in k[y^p]. Since k is perfect we get u is in K^p(=k(x^p)). –  user41481 Nov 2 '12 at 13:52
Okay!! I have got it. Thanks, anyway! –  user41481 Nov 6 '12 at 11:27

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