3
$\begingroup$

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.

  • 0
    I think though, that is the only possible value for E.2011-06-08

1 Answers 1

4

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.

  • 0
    Ah, never mind it is easy to prove it for all K.2011-06-08