I'm searching for an example of field extensions $L1$, $L2$ of $F$ for which $[L_1L_2:F]<[L_1:F][L_2:F]$.
Infact I'm trying prove the problem below. So any hint can be helpful.

Let $K$ be a finite extension of $F$. If $L_1$ and $L_2$ are subfields of $K$ containing $F$, show that $[L_1L_2:F]\leq [L_1:F][L_2:F]$. If $\gcd([L_1:F],[L_2:F])=1$, prove that $[L_1 L_2:F]=[L_1:F][L_2:F].$

  • $\begingroup$ $L_1 = \mathbb{Q}(\sqrt{2},\sqrt{3})$, $L_2 = \mathbb{Q}(\sqrt{2},\sqrt{5})$. Consider some intermediate fields of $L_1L_2 \supset F$. $\endgroup$ – Daniel Fischer Nov 20 '13 at 11:41
  • 1
    $\begingroup$ If $L_1=L_2\ne F$ you'll have an example of the inequality. $\endgroup$ – Gerry Myerson Nov 20 '13 at 11:43

The easiest example is when $L_1$ is contained in $L_2$ and both are non-trivial extensions of $F$.

To prove the first part of the exercise take a basis $a_1,\ldots,a_n$ of $L_1$ and a basis $b_1,\ldots, b_m$ of $L_2$ and prove that $a_ib_j$ generates $L_1L_2$ (everything here as $F$ vector spaces).

For the second part just note that both $[L_1:F]$ and $[L_2:F]$ must divide $[L_1L_2:F]$, so the least common multiple of those divides $[L_1L_2:F]$.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.