# Field extension which is not separable but still simple

Let $E$ be a finite extension of a field $k$ of characteristic $p > 0$, and let $p^r = [E:k]_{i}$. Assume that there is no exponent $s <r$ such that $E^{p^s}k$ is separable over $k$ (i.e., such that $a^{p^s}$ is separable over $k$ for each $a$ in $k$). Show that $E$ can be generated by one element over $k$. (Here $[E:k]_i=\dfrac{[E:k]}{[E:k]_{s}}$, adapting Lang's notation.)

My try: Take $a\in E$. Now from the fact that there is no exponent $p^s$ with $s <r$ such that $E^{p^s}k$ is separable over $k$ we can say $[k(a):k]_i \geq p^r$. But from the fact $p^r=[E:k]_{i}$ we must have $[E:k(a)]_i=1$. So, $E$ is finite separable extension of $k(a)$. So, we can write $E=k(a,b)$. Now I'm stuck. Can anyone please help me out ?

This is proof is kind of tricky, I hope someone comes up with a more convenient one:

Let $b$ be a primitive element of $E^{sep}/k$. Since $b$ is separable over $k$ it is well known that we have $k(b)=k(b^{p^n})$ for all $n > 0$. In particular $b^{p^r}$ is also a primitive element of $E^{sep}/k$.

Now take $a \in E$ such that $a^{p^{r-1}}$ is not separable $/k$ (exists by assumption). We have $a^{p^r} \in E^{sep}$ and clearly we have $$E^{sep} = k(b^{p^r}-a^{p^r},a^{p^r})$$

By the well known proof of the "primitive element theorem", we obtain that for all but finitely many $c \in k$ the element $b^{p^r}-a^{p^r}+ca^{p^r} = b^{p^r}+(c-1)a^{p^r}$ is a primitive element of $E^{sep}/k$.

Of course we can assume $k$ to be infinite (otherweise $k$ is perfect and we have nothing to show), in particular $k^{p^r}$ is infinite. Hence we can choose $c-1$ to be a $p^r$-th power. We end up with $E^{sep} = k(b^{p^r}+d^{p^r}a^{p^r})$ with some $d \in k$.

Consider the element $x = b+da$. I claim $k(x)=E$.

$b^{p^{r-1}}$ and $d^{p^{r-1}}$ are separable over $k$, $a^{p^{r-1}}$ is not. Thus $x^{p^{r-1}}$ is not separable over $k$.

In particular we obtain for the minimal polynomial $f_x$ of $x$, that $f_x(T)=g(T^{p^r})$ for some $g \in k[T]$. We have $g(x^{p^r})=f_x(x)=0$. But $x^{p^r}$ is a primtive element of $E^{sep}/k$, hence the degree of $g$ is at least $[E:k]_s$, which finally leads us to the desired $$\operatorname{deg}(f_x) = p^r\operatorname{deg}(g) \geq p^r [E:k]_s = [E:k].$$

• Meanwhile I came up with a shorter and more conceptual proof. I will write it down tomorrow. – MooS Nov 10 '15 at 22:48
• So, you get $[k(x):k] \geq [E:k]$. How does it imply $E=k(x)$ ? If we know $x\in E$ then we can say this, but $x\in E \implies b \in E$ – dragoboy Nov 11 '15 at 8:19
• $x$ is in $E$ since $a,d,b \in E$. $b \in E$ by definition since $b \in E^{sep} = \{y \in E | y \text{ separable over } k \}$ – MooS Nov 11 '15 at 8:45
• Ohh, okay, I thought $E^{sep}$ something else. – dragoboy Nov 11 '15 at 11:37

Ok here is the shorter proof, which only uses the result of the primitive element theorem, but not its proof.

Recall the following: If $k$ is a field of characteristic $p$, we have that $x \in \overline k$ is separable over $k$ if and only if $k(x)=k(x^p)$. If $x$ is not separable, we have that $k(x)/k(x^p)$ is an extension of degree $p$.

Now to the proof:

Again, take $a \in E$, such that $a^{p^{r-1}}$ is not separable over $k$. Clearly it is not separable over $E^{sep}$ either, because otherwise we have two separable extensions $E^{sep}(a^{p^{r-1}}) \supset E^{sep} \supset k$, which would give us separability of $a^{p^{r-1}}$ over $k$.

Consider

$E^{sep}(a) \supset E^{sep}(a^p) \supset \dotsc \supset E^{sep}(a^{p^{r-1}}) \supset E^{sep}(a^{p^r})=E^{sep}$

By the above fact, all those extensions have degree $p$, in particular we get $$[E^{sep}(a):E^{sep}]=p^r=[E:E^{sep}],$$

hence $E^{sep}(a)=E$. By the primitive element theorem we have $E^{sep}=k(b)$ with $b$ separable over $k$. We obtain $E=k(a,b)$ with $b$ separable over $k$. By the primitive element theorem (stronger version!), we obtain that $E/k$ is simple.

• What do you mean by stronger version of primitive element theorem ? – dragoboy Nov 11 '15 at 11:46
• And how do you get $p^r=[E:E^{sep}]$ ? which is essentially equivalent saying $[E^{sep}:k]=[E:k]_{s}$ – dragoboy Nov 11 '15 at 12:10
• We have $[E:k]_s=[E:E^{sep}]_s \cdot [E^{sep}:k]_s$. The first factor is $1$, since the extension is purely inseparable. We deduce $[E:k]_s= [E^{sep}:k]_s=[E^{sep}:k]$, where the latter equality holds, because the extension is separable. – MooS Nov 11 '15 at 12:28
• The primitive element theorem states that any finite extension $k(a,b_1, \dotsc, b_n)/k$ is simple, whenever the $b_i$ are separable. $a$ can actually be inseparable. The corollary that any finite separable extension is simple, is what many people call the primitive element theorem. But one should keep in mind that the original theorem is actually a little bit stronger. Here is a source, that states this stronger version: math.cornell.edu/~kbrown/6310/primitive.pdf – MooS Nov 11 '15 at 12:31