# Let $k^p = k$, $K = k(x)$ and $F = k(u)$ for some $u \in K$. Show that $K/F$ is separable $\iff u \not\in K^p$.

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 $u = f(x)/g(x)$ with $f$ and $g$ relatively prime. Show that $K/F$ is a separable extension if and only if $u\notin K^p$.

There is already a post on this but it has not yet been answered : Patrick Morandi- Field and Galois Theory- Section 4- Exercise 11

The truth is that I am very confused with this problem, I would like someone please help me to understand what happens with this exercise and explain to me how it could be done by giving me a help or something. I already have an approach for the left-to-right implication, but I do not know very well how good it is. Could someone help me with the other implication please? Thank you very much.

• If $u = v^p$ with $v \in k(x) \setminus k$ then you have the tower $k(x)/k(v)/k(v^p)$. Possibly $k(x)/k(v)$ is separable. What about $k(v)/k(v^p)$ ? – reuns Oct 18 '17 at 22:20

## 1 Answer

Try this approach:

Consider the minimal $F$-polynomial for $x$, call it $H(T)$. It’s $F$-irreducible, and so of form $h(T^q)$ with $q=p^s$ for an $s\ge0$ and $h(T)$ an $F$-irreducible polynomial that’s separable.

If $s>0$, then $x$ is not separable over $F$ (and so $K$ is not separable over $F$), but $x^q$ is separable over $F$, as a root of the separable polynomial $h(T)$. We have $F\subset F(x^q)=K^q\subset F(x^p)=K^p\subset F(x)=K$. The last (rightmost) inclusion is strict, since $s>0$, and so $u\in K^p$.

On the other hand, if $s=0$, then $x$ is separable over $F$, the extension $K\supset F$ is separable, and $F\not\subset K^p=k(x^p)$: if we had had $F\subset k(x^p)$, then the minimal $F$-polynomial for $x$ as above would have $s>0$. Since $F\not\subset K^p$, it follows that $u\notin K^p$, since $u\in K^p\Rightarrow k(u)=F\subset K^p$.

So there are two mutually exclusive cases: $s>0$ with $K$ not separable over $F$ and $u\in K^p$; and $s=0$ with $K$ separable over $F$ and $u\notin K^p$.

This argument is not as neat and clear as I’d have liked. Maybe someone can do better.

• I want to ask where do we need k is a perfect field? – Peter Liu Dec 23 '17 at 0:38
• (at least) one place where perfectness is used is in the fourth paragraph (the one starting “On the other hand”) where I say that $K^p=k(x^p)$. Uses the fact that $k^p=k$. – Lubin Dec 23 '17 at 6:19