# Field extension $F\subseteq L_1$ and $F\subseteq L_2$ and $[L_1L_2:F]<[L_1:F][L_2:F]$.

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 $[L_1 L_2:F]=[L_1:F][L_2:F].$

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

## 1 Answer

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 $[L_1L_2:F]$.