The degree of a field extension is smaller than the product of the degrees of field extensions of intermediate fields that generate the field Suppose that $L:K$ is a field extension, and $K_1, K_2$ are intermediate fields such that $L =K(K_1, K_2)$, then $[L:K]\leq [K_1:K][K_2:K]$. I prove it as follows:

Proof: Suppose $\{a_1,\ldots,a_n\}$ and $\{b_1,\ldots,b_m\}$ are bases for $K_1$ and $K_2$ respectively (over $K$). Now the set $\{a_ib_j : 1\leq i \leq n, 1\leq j \leq m \}$ spans $L$ (over $K$). Now said set has cardinality $nm = [K_1:K][K_2:K]$ and hence the inequality follows.

Is this proof correct or am I missing something? Some help will be much appreciated!
 A: Based on @JyrkiLahtonen's comment above.

If either $[K_1 : K] = \infty$ or $[K_2 : K] = \infty$, then the inequality is trivially true. So, let both, $K_1/K$ as well as $K_2/K$ be finite extensions. Let $\{ a_1,\dots,a_n \}$ be a basis for $K_1$ over $K$ and let $\{ b_1,\dots,b_m \}$ be a basis for $K_2$ over $K$. Consider $V = \operatorname{span}_K\{ a_i b_j : 1 \leq i \leq n, 1 \leq j \leq m \}$, the vector subspace of $L$ spanned by the vectors $a_i b_j$ over the field $K$. We wish to show that $V$ is a field.
To show that $V$ is a ring, it suffices to show that $a_i b_j \cdot a_k b_l \in V$ for all $1 \leq i, k \leq n$, $1 \leq j, l \leq m$. Now,
$$
a_i b_j \cdot a_k b_l = a_i a_k \cdot b_j b_l
$$
and $a_i a_k \in K_1$, $b_j b_l \in K_2$. So, we can express them as a $K$-linear combinations of $ a_1,\dots,a_n$ and $b_1,\dots,b_m$, respectively. That is,
$$
a_i a_k = \sum_{p = 1}^n c_{ikp} a_p \quad \text{and} \quad b_j b_l = \sum_{q = 1}^m d_{jlq} b_q
$$
for some scalars $c_{ikp}, d_{jlq} \in K$. Hence,
$$
a_i b_j \cdot a_k b_l = \left( \sum_{p = 1}^n c_{ikp} a_p \right) \cdot \left( \sum_{q = 1}^m d_{jlq} b_q \right) = \sum_{p=1}^n \sum_{q=1}^m (c_{ikp}d_{jlq}) a_p b_q \in V.
$$
Hence, $V$ is a ring. In particular, $V$ is an integral domain because it is contained in $L$ which is a field.
Next, we need to show that the multiplicative inverse in $L$ of every non-zero element in $V$ lies in $V$ itself. Let $r \in V$, $r \neq 0$. Since $V$ is spanned over $K$ by a finite set, $V$ is a finite-dimensional vector space over $F$. If $\dim_K V = d$, then the set $\{ 1, r, r^2, \dots, r^d \}$ is a $K$-linearly dependent set. Hence, there exist $c_0,c_1,\dots,c_d \in K$, not all zero, such that
$$
c_0 + c_1 r + c_2 r^2 + \dots + c_d r^d = 0.
$$
Let $k = \min\{ 0 \leq i \leq d : c_i \neq 0 \}$.
Then,
$$
c_k r^k + c_{k+1} r^{k+1} + \dots + c_d r^d = 0.\tag{1}
$$
It cannot be that $c_i = 0$ for all $i \neq k$ because otherwise we would have
$$
c_k r^k = 0 \implies r^k = 0 \implies r = 0,
$$
which is a contradiction. Note that here we are crucially using the fact that $V$ is an integral domain. So, we have concluded that $k < d$. Now, from $(1)$ we get that
$$
\begin{align}
& &c_k r^k + c_{k+1}r^{k+1} + \dots + c_d r^d &= 0 \\
&\implies &r^k(c_k + c_{k+1}r + \dots + c_d r^{d-k}) &= 0 \\
&\implies &c_k + c_{k+1}r + \dots + c_d r^{d-k} &= 0\\
&\implies &r(c_{k+1} + c_{k+2} r + \dots + c_d r^{d-k-1}) &= -c_k\\
&\implies &-c_k^{-1}(c_{k+1} + c_{k+2} r + \dots + c_d r^{d-k-1}) &= r^{-1}.
\end{align}
$$
So, $r^{-1}$ lies in the $K$-span of $\{ 1 , r, r^2, \dots, r^d \}$ which is a subspace of $V$. Therefore, $r^{-1} \in V$ for all nonzero $r \in V$. Thus, $V$ is a field.
Any field containing both $K_1$ and $K_2$ must contain $a_i b_j$ for all $1 \leq i \leq n$, $1 \leq j \leq m$. Hence, it must also contain the $K$-span of $\{ a_i b_j \}$. But we have just shown that this is a field, so it must be the minimal field containing both $K_1$ and $K_2$. In other words, $L = K(K_1,K_2) = V$. Thus, any basis for $L$ over $K$ can contain no more than $nm$ elements. In other words,
$$
[L : K] \leq [K_1 : K] [K_2 : K].
$$
Hence, proved.
A: This is not an answer to the question, but instead expands on another comment of @JyrkiLahtonen, just to shed more light on the proof.

In proving that $V = \operatorname{span}_K\{ a_i b_j : 1 \leq i \leq n, 1 \leq j \leq m \}$ is a field, we crucially use the fact that it is a finite-dimensional vector space over $K$. To see why this is important, let $K = \mathbb{Q}$ and $L = \mathbb{Q}(x,y)$. Then, taking $K_1 = \mathbb{Q}(x)$ and $K_2 = \mathbb{Q}(y)$, we see that $L = K(K_1,K_2)$. Let
$$
\mathcal{B}_1 = \left\{ \frac{f_i(x)}{g_i(x)} : i \in I \right\}
$$
be a basis for $\mathbb{Q}(x)$ over $\mathbb{Q}$, and let
$$
\mathcal{B}_2 = \left\{ \frac{h_j(y)}{k_j(y)} : j \in J \right\}
$$
be a basis for $\mathbb{Q}(y)$ over $\mathbb{Q}$. Consider $V$, the $\mathbb{Q}$-span of the set
$$
\mathcal{B} = \left\{ \frac{f_i(x)h_j(y)}{g_i(x)k_j(y)} : i \in I, j \in J \right\}
$$
in $\mathbb{Q}(x,y)$. We claim that $\frac{1}{x+y} \in \mathbb{Q}(x,y)$ does not lie in $V$, and so the element $x+y \in V$ does not have a multiplicative inverse in $V$.
Suppose, for the sake of contradiction, that
$$
\frac{1}{x+y} = \sum_{\substack{i \in I\\ j \in J}} a_{ij} \frac{f_i(x) h_j(y)}{g_i(x) k_j(y)}
$$
where $a_{ij} \in \mathbb{Q}$ and all but finitely many of them are zero. Then, by clearing out denominators, we can write
$$
\frac{1}{x+y} = \frac{\sum c_{ij}\tilde{f}_i(x)\tilde{h}_j(y)}{\prod d_{ij}\tilde{g}_i(x)\tilde{k}_j(y)},
$$
for some polynomials $\tilde{f}_i,\tilde{g}_i \in \mathbb{Q}[x]$ and $\tilde{h}_j, \tilde{k}_j \in \mathbb{Q}[y]$, $c_{ij}, d_{ij} \in \mathbb{Q}$, and where the sum and product are both finite.
Now, since $x+y$ is an irreducible polynomial in $\mathbb{Q}(x,y)$, it must divide the product in the denominator in the RHS. Since $\mathbb{Q}(x,y)$ is a UFD, we can write the denominator as a product of irreducibles. But $x+y$ cannot divide any of those irreducible factors because the factors are all polynomials purely in the variable $x$ or in the variable $y$. So, we have a contradiction.
A: Here is another way to solve the problem, using some basic results from field theory.

We assume that $K_1/K$ and $K_2/K$ are finite extensions, for otherwise the inequality is trivially true. Since
$$
[L:K] = [L:K_1] \cdot [K_1:K],
$$
it suffices to show that
$$
[L:K_1] \leq [K_2 : K].
$$
Now, every finite extension is algebraic and finitely generated, so we can write $K_1 = K(a_1,\dots,a_n)$ and $K_2 = K(b_1,\dots,b_m)$, for some elements $a_1, \dots, a_n, b_1, \dots, b_m$ that are algebraic over $K$.
It is easy to see that $L = K(K_1,K_2) = K_1(b_1,\dots,b_m)$. Define $L_0 = K_1$ and $L_i = L_{i-1}(b_i)$ for each $1 \leq i \leq m$. So, we have a chain of simple extensions
$$
K_1 = L_0 \subseteq L_1 \subseteq L_2 \subseteq \dots \subseteq L_m = L.
$$
Similarly, define $F_0 = K$ and $F_i = F_{i-1}(b_i)$ for each $1 \leq i \leq m$. So, we have a chain of simple extensions
$$
K = F_0 \subseteq F_1 \subseteq F_2 \subseteq \dots \subseteq F_m = K_2.
$$
Since the extensions are simple (and algebraic), we have that
$$
[L_i:L_{i-1}] = [L_{i-1}(b_i):L_{i-1}] = \deg \min(L_{i-1},b_i),\\
[F_i:F_{i-1}] = [F_{i-1}(b_i):F_{i-1}] = \deg \min(F_{i-1},b_i).
$$
Note that $F_{i-1} \subseteq L_{i-1}$, so $\min(L_{i-1},b_i)$ divides $\min(F_{i-1},b_i)$, which implies that $\deg \min(L_{i-1},b_i) \leq \deg \min(F_{i-1},b_i)$. Therefore,
$$
[L_i : L_{i-1}] \leq [F_i : F_{i-1}]
$$
for each $1 \leq i \leq m$. Hence,
$$
[L_m:L_0] = \prod_{i=1}^m [L_i : L_{i-1}] \leq \prod_{i-1}^m [F_i : F_{i-1}] = [F_m : F_0].
$$
In other words,
$$
[L:K_1] \leq [K_2 : K].
$$
Hence, proved.
