I want to prove or disprove the following:

Let $L_1$ and $L_2$ be two finite extensions of field $k$ inside of an extension $L/k$. Moreover, $L_1 \cap L_2 = k$. Then the degree of the composite field $L_1 L_2$ is $[L_1 L_2 : k] = [L_1 : k] [L_2 : k]$.

I want to solve this problem with basic field theory (I haven't studied Galois theory). Thanks in advance.

Below is what I've done so far.

Let $[L_1 : k] = m$, $[L_2 : k] = n$, and write $L_1$, $L_2$ as $L_1 = k(\alpha_1, \dots, \alpha_m)$, $L_2 = k(\beta_1, \dots, \beta_n)$, where $\alpha_1, \dots, \alpha_m$ is a $k$-basis for $L_1$, and $\beta_1, \dots, \beta_n$ is a $k$-basis for $L_2$. Then we can easily show that

$$[L_1 L_2 : k] \le [L_1 : k] [L_2 : k],$$

where the equality is achieved iff $\beta_1, \dots, \beta_n$ are linearly independent over $L_1$.

I tried to use $L_1 \cap L_2 = k$ to prove the linear independence but failed.


The answer turns out to be negative.

Consider $k = \mathbb{Q}$, $L_1 = \mathbb{Q}(\alpha)$, and $L_2 = \mathbb(\alpha \zeta_3)$, where $\alpha$ is the real root of $x^3 - 2$, and $\zeta_3$ is a primitive root of $x^3 - 1$. Then $[L_1 : k] = [L_2 : k] = 3$, and it is easy to show that $L_1 \cap L_2 = k$ (one way is to consider the imaginary parts). However, $L_1 L_2 = \mathbb{Q}(\alpha, \zeta_3)$ is the splitting field of $x^3 - 2$, which has degree 6 over $\mathbb{Q}$.

It turns out that if one of $L_1/k$ and $L_2/k$ is Galois, then the argument would be true.


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