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Suppose $K$ is finitely generated extension of transcendence degree 1 of the finite field $F$. Show that there exists an element $u \in K$ such that $K$ is a finite separable extension of the field $F(u). $

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Having stumbled across a question that I believe I can solve, I've decided to give an answer a go.

If $K = F$ or $K = F(u)$, we're done; so assume that $K$ has, at minimum, two generators over $F$.

Firstly, we claim that we can always write $K = F(u, \beta)$ for some pair of elements $u,\beta\in K$ that are both transcendental over $F$, and where moreover, $\beta$ is separable over $F(u)$. We'll show this fact in a moment, but for now, we'll prove $\beta$ is separable over $F(u)$ in the case of two generators.

Note that we must have a polynomial $f\in F[X,Y]$ for which $f(u,\beta) = 0$, due to the transcendence degree being $1$. WLOG assume $f$ is irreducible.

Finally, note that since $F[X^p,Y^p] = \left(F[X,Y]\right)^p$ where $p$ is the characteristic of $F$, it must be the case that $f\not\in F[X^p,Y^p]$, as it is irreducible. Hence, one of $f(u, Y)$ or $f(X, \beta)$ is a separable polynomial. This follows from the fact that inseparable polynomials must be of the form $$\sum_{i=0}^n a_ix^{p^i},$$

and up to relabeling, the conclusion is proved in the case of $K = F(u, \beta).$

We now reduce to this case.

We proceed by induction on $n$. The base case of $n=2$ has been proved above.

Assume $K = F(x_1,\ldots, x_n)$ for some $n \geq 3$. Then, there exist $y,\gamma\in K$ such that $F(x_1,\ldots, x_{n-1}) = F(y, \gamma)$ with $\gamma$ separable over $F(y)$ by the inductive hypothesis. Then, certainly $K=(y,x_n,\gamma)$ is a separable extension over $F(y,x_n)$, and again, we know that $F(y,x_n) = F(u, \beta')$ with $\beta'$ separable over $F(u)$ by induction. Hence, $K = F(u)(\beta',\gamma)$, and one may then use the primitive element theorem to find an element $\beta$ such that $K = F(u)(\beta) = F(u, \beta)$, proving the claim.

(Sorry for the strange notation... Things might have gotten a bit confusing at the end there...)

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