Can a field extension still have "non-separability" above its maximal purely inseparable subextension?

Question

Question 1

Let $E/F$ be an algebraic field extension. Let $K$ be the set of all elements of $E$ that are purely inseparable over $F$. Then, $E/K/F$ is a tower of fields, and $K/F$ is purely inseparable.

In this case, is $E/K$ always separable? If not, what is a counterexample?

Moreover, let $K^s$ and $F^s$ be the separable closures in $E$. Then, $E/K^s$ is purely inseparable. How different are $K^s$ and $F^s$ as fields? That is, how much nicer is the extension $K^s/K$ compared to $F^s/F$?

Question 2

Let $E/F$ be a normal field extension. Let $E^G$ denote the fixed field of $\operatorname{Aut}(E/F)$. Then, $E/E^G$ is separable and $E^G/F$ is purely inseperable.

Is $E^G$ the unique such subextension? That is, is there a subextension $K$ of $F$ such that $E/K$ is separable and $K/F$ is purely inseparable, but $K\neq E^G$?

Answer 1

The extension $E/K$ need not be separable. Here is the example I learned from a note by J. Lipman.

Consider the rational function field $F=\mathbb{F}_2(y,z)$ and the extension $E=F(x)$, where $x$ is a root of $$f(t)=t^4+yt^2+z\in F[t].$$ If $E/K$ was separable, we would have $f=g^2$, for $g\in K[t]$. We have $g=t^2+\sqrt{y}t+\sqrt{z}$, which means that $\sqrt{x},\sqrt{y}\in K$. The latter condition says that $K/F$ is a degree-four extension, and so it is not purely inseparable.

Answer 2

Answer to Question 1.

Here is another example of an extension $K/F$ such that $K/I$ is not separable, where $I$ is the purely inseparable closure of $F$ in $K$. This is Example 4.24 from Patrick Morandi's Field and Galois Theory (pages 48-49), and it is similar in spirit to the one mentioned in @user2097's answer.

Let $k$ be a field of characteristic $2$. Let $F = k(x,y)$ where $x$ and $y$ are indeterminates. Let $S = F(u)$, where $u$ is a root of the polynomial $$ t^2 + t + x \in F[t]. $$ Let $K = S(\sqrt{uy})$. Then $K/S$ is purely inseparable and $S/F$ is separable, so $S$ is the separable closure of $F$ in $K$. But the purely inseparable closure of $F$ in $K$ equals $F$ itself, and so $K$ is not separable over $I$.

I cannot provide a strong answer for how much nicer $I^s/I$ is compared to $F^s/F$. Perhaps you would receive better answers if you asked it separately, either here or on MathOverflow.

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Answer to Question 2.

Note that for any algebraic extension $E/F$, we have that $E/E^G$ is separable and that $E^G \supset I$, and if $E/F$ is normal, then $E^G = I$.

Now, let $E/F$ be a normal extension, and let $K$ be an intermediate extension such that $E/K$ is separable and $K/F$ is purely inseparable. Then, we have the chain of extensions $$ F \subset K \subset I \subset E. $$ Since $E/K$ is separable and $E/I = E/E^G$ is separable, $I/K$ is also separable. But $I/F$ is purely inseparable, so $I/K$ is also purely inseparable. Hence, $I = K$. Thus, if $E/F$ is normal, then there is a unique intermediate extension such that $E/F$ splits into a purely inseparable extension followed by a separable extension, namely $I = E^G$.

If $E/F$ is not necessarily normal, then there may not exist such an intermediate extension at all. If it does exist, then that intermediate extension is unique and equals $I$, but $E^G$ need not equal $I$ in this case! This follows from the results in the note of J. Lipman that @user2097 mentioned in their answer.

Answer 3

This is to complement user2097's answer.

Lemma: Let $F=\mathbb{F}_2(y,z)$ and $g(t)=t^2+yt+z\in F[t]$. Then

1. $g(t)$ is an irreducible separable polynomial over $F$, let $\alpha$ be a root of $g$ in an algebraic closure of $F$, so $F(\alpha)/F$ is separable of degree 2.
2. $f(t):=g(t^2)=t^4+yt^2+z$ is irreducible over $F$ of degree 4 with a root $x=\alpha^{1/2}$, so $F(\alpha^{1/2})/F$ is of degree 4 and $F(\alpha^{1/2})/F(\alpha)$ is purely inseparable of degree 2. Let $E=F(\alpha^{1/2})$, then the separable closure $E^s$ of $F$ in $E$ is $E^s=F(\alpha)$.
3. Let $K$ be the set of all elements of $E$ that are purely inseparable over $F$, then $K=F$ and thus $E/K$ is not purely inseparable.

Proof. (1) As $\deg g>0$, the irreducibility of $g$ can be checked over $\mathbb{F}_2[y,z]$. Quotient out the prime ideal $(y)$, we have $\overline{g}(t)=t^2+z\in \mathbb{F}_2[z][t]$ which is irreducible by Eisenstein's Criterion with the prime ideal chosen to be $(z)$. So $g$ is irreducible over $F$. An irreducible polynomial over a field is separable iff its formal derivative is non-zero. Plus that $g^\prime(t)=y\neq 0$, so $g$ is separable.

It follows that $F(\alpha)/F$ is separable of degree 2.

(2) Here we use a lemma [Let $F$ be a field of characteristic $p>0$ and $f(x)$ be an irreducible polynomial over $F$, if one of the coefficients of $f$ is not in $F^p$, then $f(x^q)$ is irreducible for all $p$-power $q=p^r$ with $r\geq 0$. See this answer ].

Clearly $y\notin \mathbb{F}_2[x,y]^2$ so $y\notin \mathbb{F}_2(x,y)^2$ (We use this result: Let $R$ be an UFD with fraction field $K$ and $a\in R,n\geq1$, then $a\in R^n$ iff $a\in K^n$. It can be proved easily with the unique factorization.) Hence $f(t)=g(t^2)$ is irreducible of degree 4. It follows that $F(\alpha^{1/2})/F$ is of degree 4 and thus $F(\alpha^{1/2})/F(\alpha)$ is purely inseparable of degree 2.

Let $E=F(\alpha^{1/2})$. Since $E/F(\alpha)$ is purely inseparable and $F(\alpha)/F$ is separable, we must have $F(\alpha)=E^s$ by the uniqueness of separable closure.

(3) Let $K$ be the set of all elements of $E$ that are purely inseparable over $F$. Then $K$ is a sub-extension of $E/F$. Then $[E:K]$ divides $[E:F]=4$, i.e. $[E:K]=1,2$ or $4$.

Case $[E:K]=1$: we have $E=K$ then $F(\alpha)/F$ is both separable and purely inseparable which can only be true if $F(\alpha)=F$, contradiciton.

Case $[E:K]=2$: we have $[K:F]=2$ which must be of the form $F(\sqrt{c})/F$ for some $c\notin F^2$ by the characterization of finite purely inseparable extensions. In particular $K^2\subset F$. Clearly $E=K(\alpha^{1/2})$ as well, let $m(t)$ be the minimal polynomial of $\alpha^{1/2}$ over $K$, then $\deg m=2$ and $m|f$. As $m^2\in F[t]$ (since $K^2\subset F$) is monic and $\deg m^2=4$, we have $m^2=f$, in particular, $m(t)=t^2+y^{1/2}t+z^{1/2}$ so $y^{1/2},z^{1/2}\in K$.

We have $y\notin \mathbb{F}_2[y,z]^2\Rightarrow y\notin \mathbb{F}_2(y,z)^2\Rightarrow [F(y^{1/2}):F]=2$.

We also have $z\notin \mathbb{F}_2[y^{1/2},z]^2\Rightarrow z\notin \mathbb{F}_2(y^{1/2},z)^2 \Rightarrow [F(y^{1/2},z^{1/2}):F(y^{1/2})]=2$.

So $2=[K:F]\geq [F(y^{1/2},z^{1/2}):F]=4$, contradiction.

Hence we must have $[E:K]=4$ and thus $K=F$, so $E/K$ is not separable. $\square$