ac.commutative algebra - Finite extension of fields with no primitive element

The example given by Alon is not only a finite extension without a primitive element, but it's the simplest example of a substitute for Galois theory when the field extensions are purely inseparable. In the example, the top field is $L = {mathbf F}_p(a,b)$ and the base field is $L^p = {mathbf F}_p(a^p,b^p)$, where $a$ and $b$ are algebraically independent. Since the extension $L/L^p$ is purely inseparable, we can't describe the intermediate fields using Galois theory. Jacobson found a way to describe them using Lie algebras of differential operators: fixed points of field automorphisms are replaced by kernels of differential operators.

Theorem (Jacobson). Let $K$ be a field of characteristic $p$ and $mathrm{Der}(K)$ be the space of derivations $K rightarrow K$, which is a Lie algebra under the bracket operation on derivations. To each field $F$ lying between $K$ and $K^p$, let $mathrm{Der}_F(K)$ be the subspace of derivations on $K$ which vanish on $F$ (the $F$-linear derivations on $K$). Then $mathrm{Der}_F(K)$ is a restricted Lie subalgebra of $mathrm{Der}(K)$ (the label "restricted" means the $p$th power of each element in the subalgebra is also in there).
If $[K:F]$ is finite then $mathrm{Der}_F(K)$ is finite-dimensional as a (left) $F$-vector space. Sending $F$ to $mathrm{Der}_F(K)$ is an inclusion-reversing bijection from the fields between $K$ and $K^p$ over which $K$ is finite-dimensional and the restricted Lie subalgebras of $mathrm{Der}_F(K)$ which are finite-dimensional over $K$, with $[K:F] = p^{dim_K(mathrm{Der}_F(K))}$. The inverse map associates to any restricted Lie subalgebra of $mathrm{Der}(K)$ which is finite-dimensional over $K$ the elements of $K$ on which the entire subalgebra vanishes.

Let's see how this theorem manifests itself in the example $K = mathbf{F}_p(u,v)$, where $u$ and $v$ are algebraically independent. Here $K^p = mathbf{F}_p(u^p,v^p)$ and $[K:K^p] = p^2$. (In other examples, $[K:K^p]$ could be infinite.) The partial derivative operators $partial_u = partial/partial u$ and $partial_v = partial/partial v$ are a basis for $mathrm{Der}(K)$ as a left $K$-vector space:
$$
mathrm{Der}(K) = Kpartial_u + Kpartial_v.
$$
An example of a field between $K$ and $K^p$ is $K^p(u^m + v^n)$ for positive integers $m$ and $n$. If $m$ and $n$ are both multiples of $p$ then $K^p(u^m + v^n) = K^p$, which is dull and all derivations are 0 on it: it corresponds to $mathrm{Der}(K) = mathrm{Der}_{K^p}(K)$. If at least one of $m$ or $n$ is not a multiple of $p$ then the set of derivations that vanish on $K^p(u^m + v^n)$ is $K(nv^{n-1}partial_u - mu^{m-1}partial_v)$. If we insist now that $m$ and $n$ are both not multiples of $p$, then changing $m$ or $n$ changes that line of derivations, so the corresponding fields $K^p(u^m+v^n)$ are not the same (you don't need Jacobson's bijection to see that: if one field is killed by a certain differential operator and another isn't, they are not the same field).

We haven't seen the restricted Lie algebra aspect playing an essential role here. In fact all one-dimensional $K$-subspaces of this two-dimensional space $mathrm{Der}(K)$ are restricted. To see this, pick a one-dimensional $K$-subspace $Kdelta$ in $mathrm{Der}(K)$. Set $F$ to be the common kernel of the operators in $Kdelta$, which is just the same thing as $ker(delta)$, so $F$ is a field between $K$ and $K^p$.
Clearly
$$
Kdelta subset mathrm{Der}_F(K) subset mathrm{Der}(K)
$$
and the top space is 2-dimensional over $K$. Thus $mathrm{Der}_F(K)$ is either $Kdelta$ or $mathrm{Der}(K)$. If $Kdelta = mathrm{Der}_F(K)$ then $Kdelta$ is restricted since the space of derivations on $K$ vanishing on a subfield is restricted and then we're done. So assume instead that $Kdelta not= mathrm{Der}_F(K)$, and then for dimensional reasons we must have $mathrm{Der}_F(K) = mathrm{Der}(K)$, which by Jacobson's correspondence implies $F = K^p$, so $ker(delta) = K^p$. But this is impossible: any nonzero $delta$ in $mathrm{Der}(K)$ is nonvanishing at either $u$ or $v$ (if $delta(u) = 0$ and $delta(v) = 0$ then $delta$ vanishes on $K^p(u,v) = K$), so $ker(delta)$ is not $K^p$ and thus we have a contradiction, which wraps up this little argument.

That all 1-dimensional $K$-subspaces of $mathrm{Der}(K)$ in this particular example are restricted Lie subalgebras sounds analogous to the fact that all one-dimensional subspaces of a Lie algebra are Lie subalgebras, although note $mathrm{Der}(K)$ is not a Lie algebra over $K$, only over $K^p$. Still, if anyone who knows about restricted Lie algebras can indicate in a comment if there is some general theorem about certain subspaces automatically being restricted subspaces, please speak up.

It is a theorem of Gerstenhaber (1964) that the Lie algebra aspect in Jacobson's theorem is always automatic from the restricted aspect: for any field $K$ of characteristic $p$, any $K$-subspace of $mathrm{Der}(K)$ which is closed under $p$th powers is closed under the Lie bracket on derivations.

This approach to a Galois theory for inseparable extensions has been extended to the case of inseparable $K/F$ where $K^{p^r} subset F$ some some $r > 1$, using higher derivations.
See "Higher derivation Galois theory of inseparable extensions" pp. 187--220 of Handbook [sic] of Algebra, Volume 1. (An account just of Jacobson's case $r = 1$ is in his Basic Algebra II, and in somewhat more detail -- including Gerstenhaber's result -- in Chapter 4 of Karpilovsky's Topics in Field Theory.)

New topic! RK asked, as a comment to Alon's answer, if there is a finite extension of $mathbf F_p(x)$ which is not primitive. No.

Theorem: If $K$ is a field of characteristic $p$ such that $[K:K^p] = p$, then every finite extension of $K$ has a primitive element.

We can apply this theorem to any rational function field over a perfect field of characteristic $p$. The conclusion of the theorem is stable under passage to finite extensions, so it applies to any function field over a perfect field of characteristic $p$.

Proof: Let $L/K$ be a finite extension. From the hypothesis that $[K:K^p] = p$, $[K:K^{p^m}] = p^m$ for all $m geq 0$. Then $[K^{1/p^m}:K] = p^m$, so if $K^{1/p^m} subset L$ we get $p^m|[L:K]$, which means $m$ is bounded. Taking $mgeq 0$ maximal such that $K^{1/p^m} subset L$, it can be shown that $L/K^{1/p^m}$ is separable.

Let $F$ be any field such that $K subset F subset L$. Letting $n geq 0$ be maximal such that $K^{1/p^n} subset F$, we have $n leq m$ and $F/K^{1/p^n}$ is separable.
Thus every field between $K$ and $L$ is a separable extension of one of the finitely many fields $K^{1/p^n}$ for $n = 0,1,dots,m$. Note the tower $F supset K^{1/p^n} supset K$, which is a separable extension on top of a purely inseparable extension, is backwards compared to what general field theory tells us: any finite extension can be expressed as a purely inseparable extension on top of a separable extension. We will now bring in that general field theory point of view, in a more precise form.

Any finite extension of fields admits a maximal separable subextension (containing all other separable extensions of the base field in the top field). Moreover, a separable extension of fields has only finitely many intermediate fields (say, by Galois theory). Therefore each $K^{1/p^n}$, for $n = 0,1,dots,m$ has only finitely many separable extensions inside $L$. Since every field between $K$ and $L$ is a separable extension of $K^{1/p^n}$ for some $n$ from 0 to $m$, there are only finitely many fields between $K$ and $L$. Therefore by Steinitz's theorem on primitive elements, $L$ is a primitive extension of $K$.

QED