# Prime degree ⇒ linearly disjoint?

If F is a field of characteristic 0 with subfields K, L such that F is the compositum of K and L and [ L : LK ] is prime, must be K and L be linearly disjoint over LK?

In other words, must [ KL : K ] = [ L : LK ] if [ L : LK ] is prime?

Here K, L are fields of characteristic 0 (definitely not p), but I don't assume [ K : KL ] is finite (or normal or anything). I think it doesn't have to be true if that index [ L : LK ] is not prime and the field extension KLK is not normal. I am hoping it is true when the index [ L : LK ] is prime.

The motivation is to show either every non-identity p-subgroup of PGL(2, K) is reducible or every non-identity p-subgroup of PGL(2, K) is irreducible.

-
A pair of fields $L,K$ are said to be linearly disjoint over $E$ if $L\otimes_{E}K$ is a field. Is this what you are interested in? If so, what is E? – jspecter Jun 8 '11 at 16:51
E = L ∩ K, sorry. – Jack Schmidt Jun 8 '11 at 16:53
I think though, that is the only possible value for E. – Jack Schmidt Jun 8 '11 at 16:54

## 1 Answer

The answer is no. Let $L = \mathbb{Q}(\sqrt[3]{5})$ and $K = \mathbb{Q}(\zeta_3\sqrt[3]{5}).$ Then $[LK:L] = 2 \neq 3 = [L: K\cap L]$ and $L\otimes_{\mathbb{Q}} K$ is not a field.

Edit in responce to the comment: For any fields $L$ and $K,$ if $L/L\cap K$ is finite, Galois, then so too is $KL/K$ and $[LK : K]$ divides $[L:L\cap K].$

Consequently, if $[L:L\cap K]$ is prime then $L\neq K$ and $[LK : K] = [L:L\cap K].$

Linear disjointness follows, assume momentarily that $K/L\cap K$ is finite and set $E = L\cap K.$ Then $$\mathrm{dim}_{E}(L\otimes_{\mathbb{E}} K) = [L:E][K:E] = [LK : K][K:E] = dim_{E}(LK).$$ Hence, the natural ring epimorphism from $L\otimes_{\mathbb{E}} K$ to $LK$ is an isomorphism of $E$-vector spaces and hence a ring isomorphism. It follows $L\otimes_{\mathbb{E}} K$ is a field.

For the case where $K/E$ is infinite,

Let $\mathcal{K} :=\{K'\}$ be the collection of fields which are finite extensions of $E$ and contained in $K.$ Then

$L\otimes_{\mathbb{E}} K \cong L\otimes_{\mathbb{E}} (\displaystyle\lim_{\rightarrow}K') \cong \displaystyle\lim_{\rightarrow}(L\otimes_{\mathbb{E}} K') \cong \displaystyle\lim_{\rightarrow} LK' \cong LK.$

where colimits are taken over $\mathcal{K}.$ It follows $L\otimes_{\mathbb{E}} K$ is a field and $K$ and $L$ are linearly disjoint.

-
Crud. This doesn't answer my hidden question though. I do have that L∩K ≤ L is Galois. Probably should have mentioned that. If it is easy to adjust to this, could you please? Otherwise, I'll ask as another question. – Jack Schmidt Jun 8 '11 at 17:02
If both L and K are Galois over the intersection, then it is not difficult to show that they are linearly disjoint (irrespective of whether the extension degree is a prime or not). I need to think about it more, if only one is Galois. – Jyrki Lahtonen Jun 8 '11 at 17:31
@Jyrki: thanks! If K is finite dimensional over K∩L, then they are linearly disjoint, I believe. I just have no references for the infinite dimensional case. – Jack Schmidt Jun 8 '11 at 17:34
@Jack: Sorry, I meant to write: if both $K$ and $L$ are finite and Galois over $K\cap L$, then they are linearly disjoint. Don't know if the proof works for infinite extensions :-( – Jyrki Lahtonen Jun 8 '11 at 17:48
@jspecter: Thanks! Do you have a reference for the first sentence of your edit? (KL/L is Galois). I only know it for K/E finite. – Jack Schmidt Jun 8 '11 at 18:18