Patrick Morandi- Field and Galois Theory- Section 4- Exercise 11 Let $K=k(x)$ be the rational function field 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 $K^p$.
 A: It is necessary to assume that $u \in K \setminus k$. If $u \in k$, then $K/F$ is not an algebraic extension, so it does not make sense to ask whether $K$ is separable over $F$.

Note that $F(x) = K$, so by Corollary 4.10(2) it suffices to show that $x$ is separable over $F$ if and only if $u \not\in K^p$. In Example 1.17, it was discussed that $p(t) = ug(t) - f(t) \in F[t]$ is an irreducible polynomial satisfied by $x$, so it is a scalar multiple of $\min(F,x)$.
Suppose $u \in K^p$. Then, $u = \left( \frac{f_1(x)}{g_1(x)} \right)^p$ for some $f_1,g_1 \in k[x]$ that are relatively prime. Since $k[x]$ is a UFD, $f(x) = f_1(x)^p$ and $g(x) = g_1(x)^p$. Let $u_1 = \frac{f_1(x)}{g_1(x)} \in K$. Then,
$$
ug(t) - f(t) = (u_1 g_1(t) - f_1(t))^p.
$$
This shows that $x$ is a repeated root of its minimal polynomial over $F$, and so $x$ is not separable over $F$.
Conversely, suppose that $x$ is not separable over $F$. Let
\begin{align}
f(x) &= a_0 + a_1 x + \dots + a_n x^n, \quad \text{and}\\
g(x) &= b_0 + b_1 x + \dots + b_n x^n.
\end{align}
We do not assume that $a_n$ and $b_n$ are nonzero. By Proposition 4.6(2), there is an irreducible separable polynomial $h(t) \in F[t]$ such that $h(t^{p^m}) = p(t)$ for some $m \geq 0$.
Since $x$ is assumed to not be separable over $F$, $m$ is in fact strictly greater than $0$.
This implies that $f(t),g(t) \in F[t^p]$. For, if $0 \leq l \leq n$ and $p \nmid l$, then the coefficient of $t^l$ in $h(t^{p^m})$ is zero. So, the coefficient of $t^l$ in $p(t)$ is zero. This coefficient is $$ub_l - a_l,$$ and $ub_l - a_l = 0 \iff u \in k$, whereas $u \not\in k$. So, $f(t),g(t) \in F[t^p]$.
Finally, since $k$ is perfect, there are polynomials
$f_1(t),g_1(t) \in k[t]$ such that $f(t) = f_1(t)^p$ and $g(t) = g_1(t)^p$. Let $u_1 = \frac{f_1(x)}{g_1(x)} \in K$. Then,
$$
u = u_1^p \in K^p.
$$


Proposition 4.6(2). Let $f(x) \in F[x]$ be a nonconstant irreducible polynomial. If $\mathrm{char}(F) = p > 0$, then $f(x) = g(x^{p^m})$ for some integer $m \geq 0$ and some $g(x) \in F[x]$ that is irreducible and separable over $F$.
Corollary 4.10(2). Let $L$ be a finite extension of $F$. If $L = F(\alpha_1,\dots,\alpha_n)$ with each $\alpha_i$ separable over $F$, then $L$ is separable over $F$.

