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

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 $$K\neq E^G$$?

• +1, very interesting question - I made some minor changes, reworded the title a bit, but if you don't like them feel free to undo. – Zev Chonoles May 23 '15 at 16:47
• – Brahadeesh Jun 1 at 16:27

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.

• Isn’t $K=F$ in this example? – Lubin May 24 '15 at 22:43
• @Lubin Yes. In other words, no element of $E$ is purely inseparable over $F$. – user2097 May 25 '15 at 19:43

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.

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.