Let $L, K, F = L \cap K$ be fields such that $[L:F] , [K:F] < \infty$. $LK$ is defined as the compositum of the two fields(shortest field containing $K$ and $L$ in some algebraic closure).

Also, assume hat $L$ is Galois. Then, I can prove that $[L:F][K:F] = [LK:F]$ assuming that $L = F[\theta]$ using something like the primitive element theorem.

Does the result hold in more generality($L \neq F[\theta])$ and how does one prove it?

  • $\begingroup$ Not clear what your “more general” statement would look like. If neither field is Galois over $F$, you certainly don’t have that relationship. $\endgroup$ – Lubin May 12 '15 at 3:22
  • $\begingroup$ I meant without the conditions for the primitive element theorem holding, sorry! $\endgroup$ – Asvin May 12 '15 at 5:57
  • $\begingroup$ Chiming in with Lubin. The standard counterexample is $L=\Bbb{Q}(\root3\of2)$, $K=\Bbb{Q}(\omega\root3\of2)$ where $\omega$ is a primitive third root of unity. Because $p(x)=x^3-2$ is irreducible $[L:F]=3=[K:F]$, but $LK$ is the splitting field of $p(x)$, so $[LK:F]=6$. $\endgroup$ – Jyrki Lahtonen May 12 '15 at 5:57
  • $\begingroup$ I have clarified, sorry for the confusion. $\endgroup$ – Asvin May 12 '15 at 6:00
  • $\begingroup$ The concept you are looking for is linearly disjoint extensions. If you can locate for example Pete L. Clark's lecture notes (on University of Georgia webpages, sorry I don't have the link), there is more extensive information there. $\endgroup$ – Jyrki Lahtonen May 12 '15 at 6:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.