$[L:K]\leq [K_1:K][K_2:K]$

Let $K_1,K_2$ be intermediate fields of field extension $K\subseteq L$ such that $L=K(K_1,K_2)$ ($L$ is the smallest field containing $K_1$ and $K_2$). I would like to prove that $[L:K]\leq [K_1:K][K_2:K]$. My idea is by starting with a basis $\{x_1,\ldots, x_m\}$ of $K_1/K$ and $\{y_1,\ldots, y_n\}$ of $K_2/K$ and show that $\{x_iy_j\}_{i,j}$ generates $L/K$. The problem with this approach is that I don't know the explicit description of elements in $L$ in terms of $K_1$ and $K_2$.

Any hint to solve this problem is appreciated. Thanks.

• the idea is right... – user175968 Jun 2 '16 at 4:58
• Really, the only interesting obstruction is that $K(K_1) \cap K(K_2)$ might not be trivial. (Thinking about these as vector spaces is the right way to go.) For instance, what if $K(K_1,K_2) = K(K_2) \supset K(K_1)$? A concrete example is $\Bbb{Q}(\mathrm{i}, \mathrm{\sqrt{i}})$. – Eric Towers Jun 2 '16 at 5:06
• there is a general notion for two vector spaces of largest common subspace, and smallest vector space containing both – reuns Jun 2 '16 at 5:13

Let's assume for a second that we have $[L:K_2]\leq [K_1:K]$. We start from $[L:K]=[L:K_2][K_2:K]$ to get immediately that $[L:K]\leq [K_1:K]\cdot [K_2:K]$
Now let's prove the first assertion in the most general case. Consider $R$ the subset of finite sums of products $k_1\cdot k_2$ with $k_1\in K_1$ and $k_2\in K_2$. Obviously $R$ is a subring of $L$. Any basis of $K_1/K$ generates $R$ as a $K_2$ vector space. So in particular $[R:K_2]\leq [K_1:K]$. Now if $[K_1:K]$ is finite then $R$ is a field containing $K_1$ and $K_2$ Therefore $R=L$ and we're done.
• Thanks. I learn a lot from your post. What happen in both $[K_1:K]$ and $[K_2:K]$ are infinite? Surely $[L:K]$ will be infinite too. But why the type of infinity of $[L:K]$ is smaller than that of $[K_1:K][K_2:K]$ ? Why we can't have say that $[L:K]$ is uncountably infinite while $[K_1:K][K_2:K]$ is countably infinite? – user9077 Jun 3 '16 at 1:49
It is true that defining $K_1K_2$ as the smallest subfield of $L$ containing $K_1$ and $K_2$ does not make it easy to work with. A general element of $K_1K_2$ is of the form $$\frac{r_1s_1+\cdots+r_\ell s_\ell}{u_1v_1+\cdots+u_mv_m},$$ where $r_i,u_i\in K_1$ and $s_i,v_i\in K_2$, with $u_1v_1+\cdots+u_mv_m\neq0$ in $L$. Luckily, your field extension $L/K$ is finite. This implies that the smallest sub-$K$-algebra $A$ of $L$ containing $K_1$ and $K_2$ is equal to $K_1K_2$. Indeed, $A$ is a domain as a subring of $L$, and is of finite dimensions over $K$, hence is a field. Since $A\subseteq K_1K_2$, one has $A=K_1K_2$ by minimality of $K_1K_2$. Now, life gets a lot easier since an arbitrary element of $A$ is of the form $$r_1s_1+\cdots+r_\ell s_\ell,$$ where $r_i\in K_1$ and $s_i\in K_2$.
It is easy to proceed with your idea of taking a $K$-basis $x_1,\ldots,x_m$ of $K_1$ and a $K$-basis $y_1,\ldots,y_n$ of $K_2$, and show that the family $x_iy_j$ generates $K_1K_2$. From which one deduces that indeed $$[K_1K_2:K]\leq[K_1:K][K_2:K].$$
• Thank you. I understand the elements of $K_1K_2$ better now. – user9077 Jun 3 '16 at 1:43