Function field over a perfect field can be generated by two elements

Question

I have two questions about the following theorem:

Theorem: Let $K$ be a perfect field, $F$ a function field in one variable over $K$ (i.e., a finite algebraic extension of $K(t)$). Then there is $x \in F$ such that $F$ is finite and separable over $K(x)$.

A corollary of the above theorem and the primitive element theorem is:

Corollary: Any function field in one variable over a perfect field can be generated by two elements.

$$\underline{\qquad \qquad \qquad \qquad \qquad}$$

Below, I'll write down a proof of the above theorem, which is also found here for $K = \mathbb F_p$.

Questions:

1. Is the hypothesis that $K$ is perfect really needed for the Theorem or the corollary? (I'll indicate where it is used in the proof below.) I've seen several sources (e.g. Stichenoth's book on function fields, after Remark 1.1.3) claim the corollary holds for every ground field (not only perfect ones), but I've seen no reference or proof.
2. There is a detail I don't understand in the proof, namely, why the polynomial $h(X)$ defined near the end is separable.

$$\underline{\qquad \qquad \qquad \qquad \qquad}$$

Proof of the Theorem: The statement is clear in characteristic $0$, so let's assume $\mathrm{char}(K) = p > 0$. Let $F' \subset F$ be a subfield satisfying the following two properties:

1. there is $x \in F$ such that $F'/K(x)$ is algebraic and separable,
2. the degree $[F:F']$ is minimal.

Obviously, the goal is to show that $F = F'$. Let us assume that $F \neq F'$. Observe that every element of $F \setminus F'$ is inseparable over $F'$, by property 2. In other words: $F/F'$ is purely inseparable. We thus find $y \in F \setminus F'$ such that $y^p \in F'$. Our main claim is that $F'(y)/K(y)$ is separable, contradicting property 2, since $[F:F'(y)] < [F:F']$.

Let $$f(T) := T^n + a_{n-1}T^{n-1} + \ldots + a_0 \in K(x)[T]$$ be the separable (!) minimal polynomial of $y^p$ over $K(x)$. By choosing $y$ appropriately, we may assume that $a_{n-1}, \ldots, a_0 \in K[x]$.

We claim that there is an index $i$ such that $a_i \notin K[x^p]$. Assume for a contradiction that $a_{n-1}, \ldots, a_0 \in K[x^p]$. Since $K$ is perfect, there are $b_i \in K[x]$ such that $b_i^p = a_i$ (since $K$ being perfect is equivalent to the Frobenius $K \to K$ being surjective). Now, let $$g(T) := T^n + b_{n-1}T^{n-1} + \ldots + b_0 \in K(x)[T].$$ Then $g(T^{1/p})^p = f(T)$. This formula shows that $g(T)$ is irreducible, since $f$ is. Moreover, $f(y^p) = g(y) = 0$. But $f$ is separable, hence $f' \neq 0$. We obtain $g' \neq 0$, a contradiction, since then $y$ would be the root of an irreducible separable polynomial.

Thus there is at least one $i$ such that $a_i = a_i(x) \notin K[x^p]$. Consider now the polynomial $$h(X) := y^{pn} + y^{p(n-1)}a_{n-1}(X) + \ldots + a_0(X) \in K(y)[X].$$ We have $h(x) = 0$, and since there is at least one $a_i$ which is not a polynomial in $X^p$, we obtain that $h'(X) \neq 0$.

If we could show now that $h(X)$ is irreducible, we would be done: then $x$ is separable over $K(y)$, so that we have the two separable extensions $K(x,y)/K(y)$ and $F'(y)/K(x,y)$. Then $F'(y)/K(y)$ is also separable, which is a contradiction, as explained above. $\Box$

So how do we show that $h(X)$ is irreducible? Irreducibility over $K(y^p)$ is easy, but not sufficient. Note that the implication $$h' \neq 0 \implies h \text{ has no multiple roots}$$ is only true if $h$ is irreducible, so I think this is what we need to show. I might be missing something simpler, though.

Answer 1

First, let us answer question (2) by using Gauss lemma to show $h(X)$ is irreducible over $K(y)$ as follows. Since $[K(x,y):K(x,y^p)]=p$ and $[K(x,y^p):K(x)]=n$, we know that $f(T^p)$ is the minimal polynomial of $y$ over $K(x)$. Since $f(T^p)$ is a monic polynomial with coefficients in K[x], by Gauss lemma $f(T^p)$ is irreducible over $K(x)$ if and only if $f(T^p)$ is irreducible over $K[x]$. Since $f(T^p)$ is the minimal polynomial of $y$ over $K(x)$, it follows that $f(T^p)$ is irreducible over $K[x]$, i.e. $f(T^p)$ is irreducible in the polynomial ring $K[x,T]$ of two variables, or equivalently $f(T^p)$ is irreducible in $K[T][x]$. Since $y$ is transcendental over $K$ and $f(y^p)=h(x)$, we know $h(X)$ is an irreducible polynomial in $K[y][X]$, which implies $h(X)$ is irreducible in $K(y)[X]$. Since $h'(X)\ne 0$ we know that $K(x,y)$ is separable over $K(x)$. This concludes your proof of Theorem.

Next, let us answer question (1). I think the assumption $K$ is perfect is essential for both Theorem and Corollary. I don't know the subject well, in particular I don't know if this question has been mentioned in any reference. But I took a look at the remark after Example 1.1.3 in Stichenoth's book on function fields, which wrote "A function field $F/K$ is often represented as a simple algebraic field extension of a rational function field $K(x)$." Since I am not a native English speaker, I am not sure if this sentence means implicitly any function field $F/K$ can be represented in the way as above. But never mind about it, since we can give counterexamples as follows to both Theorem and Corollary if we don't assume $K$ is perfect.

Here is a counterexample to Theorem if we don't assume $K$ is perfect. Take $F:=\mathbb F_p(t,x)$ and $K:=\mathbb F_p(t^p)$, where $\mathbb F_p$ is the finite field with $p$ elements for some prime $p$, and $t$ and $x$ are algebraically independent over $\mathbb F_p$. Thus $F$ is a function field in one variable over $K$. It remains to show that for any transcendental $y$ over $K$, we have $F$ is not separable over $K(y)$. Suppose $y$ is transcendental over $K$, since the transcendental degree of $F$ over $\mathbb F_p$ is $2$, it follows that $t^p$ and $y$ are algebraically independent over $\mathbb F_p$, which implies that $t$ is not in $K(y)=\mathbb F_p(t^p,y)$. Since $t^p\in K(y)$ we know $t$ is purely inseparable over $K(y)$ of degree $p$, so that $F$ is not separable over $K(y)$.

One may argue that the above example is not a counterexample to Corollary, since $F:=\mathbb F_p(t,x)$ is indeed generated over $K:=\mathbb F_p(t^p)$ with two elements $t$ and $x$. But we can give a counterexample to Corollary by adding one more transcendental element. More precisely, here is a counterexample to both Theorem and Corollary. Take $F:=\mathbb F_p(t,s,x)$ and $K:=\mathbb F_p(t^p,s^p)$, where $\mathbb F_p$ is the finite field with $p$ elements for some prime $p$, and $t,s,x$ are algebraically independent over $\mathbb F_p$. As above we can show that for any transcendental $y$ over $K$ the elements $t^p,s^p,y$ are algebraically independent over $\mathbb F_p$, so that $t,s\notin K$ with $[F:K]=p^2$. Since there are infinitely many intermediate fields between $F$ and $K(y)$ of the form K(y)(t+as) with $a\in K(y)^*$, it follows that $F$ cannot be generated by one single element over $K(y)$. Thus it is a counterexample to both Theorem and Corollary.