Let $L|K$ be a finite field extension with basis $\{\beta_1, ...,\beta_n\}$ and $M\supset L$ a field. If $\mathcal{S}$ is an arbitrary subset of $M$ and $L(\mathcal{S}), K(\mathcal{S})$ are the fields obtained by adjoining $\mathcal{S}$ to $L$ and $K$ respectively, prove that: $$\text{dim}(L(\mathcal{S})|K(\mathcal{S}))\leq n$$
Here is what I've done so far: I need to prove that an arbitrary element $\frac{F(\alpha_1, ..., \alpha_k)}{G(\alpha_1, ..., \alpha_k)} \in L(\mathcal{S})$ (where $F, G\in L[X_1, ..., X_k]$, $G\neq 0$ and $\alpha_1, ..., \alpha_k \in \mathcal{S}$) can be written in the form:
$$\sum_{i=1}^{n}\frac{r_i(\alpha_1, ..., \alpha_k)}{s_i(\alpha_1, ..., \alpha_k)} \beta_i, \text{ where } \frac{r_i}{s_i} \in K(X_1, ..., X_k)$$
By writing $F(X_1, ..., X_k)=\sum_{i=1}^{n}p_i \beta_i$ and $G(X_1, ..., X_k)=\sum_{i=1}^{n}q_i \beta_i$ where $p_i, q_i \in K[X_1, ..., X_k]$, we need to find elements $r_i, s_i \in K[X_1, ..., X_k]$ such that:
$$\frac{\sum_{i=1}^{n}p_i \beta_i}{\sum_{i=1}^{n}q_i \beta_i}=\sum_{i=1}^{n}\frac{r_i}{s_i}\beta_i$$
I thought about multiplying the above equation by $\sum_{i=1}^{n}q_i \beta_i$, but then I don't know how to deal with the products $\beta_i\beta_j$ which arise from the product distribution.
Is there a simpler way to do this? Thanks!
You know that the extension $ L/K $ is finite. Let $ L = K(\alpha_1, \alpha_2, \ldots, \alpha_m) $, and note that $ L(S) = K(S)(\alpha_1, \alpha_2, \ldots, \alpha_m) $. Denote $ K_j = K(\alpha_1, \ldots, \alpha_j) $. Now, you have
$$ [L(S) : K(S)] = \prod_{j = 0}^{m-1} [K_{j+1}(S) : K_j(S)] \leq \prod_{j=0}^{m-1} [K_{j+1} : K_j] = [L:K] $$
where the inequality $ [K_{j+1}(S) : K_j(S)] \leq [K_{j+1} : K_j] $ follows since the minimal polynomial of $ \alpha_{j+1} $ over $ K_j(S) $ has smaller degree than the minimal polynomial over $ K_j $.