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What is an example of a finite field extension which is not generated by a single element?

Background: A finite field extension E of F is generated by a primitive element if and only if there are a finite number of intermediate extensions. See, for example, [Lang's Algebra, chapter V, Theorem 4.6].

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Hello SE 2.0. RIP FTS. –  Vandermonde Jun 29 '13 at 0:01

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up vote 20 down vote accepted

Let F be a finite field with p elements. Let K=F(x,y) be the field of rational functions in two indeterminate variables over F. Consider the extension of K obtained by adjoining p-th roots of x and of y. More precisely, let k be an algebraic closure of K. In k we can solve the equation X^p=x in the variable X. Let a be a solution of this equation; so a is an element of k which satisfies a^p=x. Similarly find an element b which satisfies b^p=y.

Consider L=K(a,b). L is a finite extension of K, of order p^2 as you can check. However there is no element of degree p^2 in L, and a primitive element would have to have degree p^2.

This example is, in a sense, the simplest possible. Separable finite extensions are simple (contain a primitive element), so we must use a non-perfect base field. Also, extensions of degree p are also simple, so we must use p^2.

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Very nice. In case anybody by me was confused by the middle step, the reason there is no element of degree p^2 is that taking the p-th power of a rational function leaves all the coefficients alone (freshman's dream + x^p=x in F) and replaces a by a^p=x and b by b^p=y. So the p-th power of any element of L lies in K, so every element of L satisfies a degree p polynomial. –  Anton Geraschenko Sep 29 '09 at 14:30
    
Does there exist a finite extension of F_p(x) that is not primitive? –  Raju Dec 9 '09 at 16:33
    
RK, see my separate answer to Anton's question for an answer to your question. –  KConrad Mar 15 '10 at 0:31
    
Anton, since you cite Chapter V of Lang's Algebra as background for your question, note the field extension Alon gives is exercise 25 in Chapter V of Lang (revised 3rd edition). That exercise asks the reader to show there are infinitely many intermediate extensions. Alon points out that you can directly check there is no primitive element for the extension, but independently computing infinitely many intermediate fields is also instructive (see my answer below), just to see it can be done. –  KConrad Mar 15 '10 at 0:36

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: $$ \mathrm{Der}(K) = K\partial_u + K\partial_v. $$ 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 $K\delta$ in $\mathrm{Der}(K)$. Set $F$ to be the common kernel of the operators in $K\delta$, which is just the same thing as $\ker(\delta)$, so $F$ is a field between $K$ and $K^p$.
Clearly $$ K\delta \subset \mathrm{Der}_F(K) \subset \mathrm{Der}(K) $$ and the top space is 2-dimensional over $K$. Thus $\mathrm{Der}_F(K)$ is either $K\delta$ or $\mathrm{Der}(K)$. If $K\delta = \mathrm{Der}_F(K)$ then $K\delta$ is restricted since the space of derivations on $K$ vanishing on a subfield is restricted and then we're done. So assume instead that $K\delta \not= \mathrm{Der}_F(K)$, 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 $m\geq 0$ 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

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I think I've made a mistake, but I don't know what I've done wrong. I thought that $Der_R(S,M)$ is the set of $R$-linear derivations for $S$ an $R$-algebra, and $M$ an $S$-module. But $K$ is not an $F$-module for any $F$ a field extension of $K$. Did you mean to let $F\mapsto Der_F(K^p,K^p)$, or have I gotten something horribly wrong? It just seems strange because a derivation $K\to K$ doesn't seem to be a priori defined on a field extension. –  Harry Gindi Mar 14 '10 at 23:11
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Ah, I see, $K^p$ is a subfield of $K$, not the other way around. –  Harry Gindi Mar 14 '10 at 23:17
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By the way, when I write Der(K) without a subscript I mean just additive maps from K to K satisfying the product rule. Since we're in characteristic p that is the same as Der_{K^p}(K). More generally, the kernel of any set of derivations from a field to itself is a subfield, so there is always a field of "constants" of any derivation algebra of a field. –  KConrad Mar 14 '10 at 23:22
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Some background: when I took Galois theory, the class had as a homework problem the task of showing there are infinitely many intermediate fields between K = F_p(x,y) and K^p = F_p(x^p,y^p). The idea of considering a general derivation on a field was never introduced in the course (and the teacher was Dwork!). I guessed the fields K^p(x+y^{p^n+1}) should all be different as n runs through the nonnegative integers and to show that I used computations with bases. What a pain that was. –  KConrad Mar 15 '10 at 1:16
    
Hi Keith! It might also be worth pointing out that it's surely possible to approach purely inseparable Galois theory as a special case of faithfully flat descent of algebras, much like you can with usual Galois theory. I've never written down the details, though. –  JBorger Mar 15 '10 at 6:08

In case people are interested, here's a simple proof that there are infinitely many distinct sub-extensions. (This Lang-problem drove me out of my gourd for a few hours recently.)

Okay, so let $k=F_p$. We want to show that

$k(a,b)$ has infinitely many subextensions over $k(x,y)$. where $a=x^{\frac1p}$ and $y=b^\frac1p$

I claim that the extensions given by adjoining $a^m+b^m$, where $m \equiv 1 \text{mod} p$, are different for different $m$. It's enough to show that for any such $m$ and $n$, adjoining both $a^m+b^m$ and $a^n+b^n$ gives $k(a,b)$.

So suppose $m=n+pr$. Then $(a^n+b^n) * (a^p)^r = a^m + b^n*(a^{pr})$. Subtracting from the other one, we get: $(b^m - b^n) * a^{pr} = b^n * (b^{pr} - a^{pr})$

But $b^{pr} - a^{pr}$ is in the base field (each of $b^p$ and $a^p$ is), so we can invert it and get $b^n$ and thus $b$ ($n$ is congruent to 1 mod $p$)!

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Well, one needs to prove then that such an a^m+b^m is never a primitive element (unless one takes this for granted given the previous answers, but I don't know if that was the idea). –  David Sevilla Nov 10 '10 at 12:34

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