$\text{dim} (L(\mathcal{S})|K(\mathcal{S}))\leq \text{dim} (L|K)$

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 $K_j$.
• I have some questions about your answer. When you use $m$ in $L=K(\alpha_1, ..., \alpha_m)$ do you mean $n$? Second, I how do I know that $L(S)=K(S)(\alpha_1, ..., \alpha_n)$? If I assume this as true, isn't it already enough to conclude directly that $[L(S):K(S)]\leq n =[L:K]$? – rmdmc89 Sep 26 '16 at 20:51
• Depends on what you mean by "concluding directly" - the above proof is pretty direct, in my opinion. The equality of fields is just definition chasing: $L(S)$ is an extension of $K(S)$ containing all of the $\alpha_i$ (since these are all contained in $L$), so you get one inclusion, and $K(S)(\alpha_1, \ldots, \alpha_m)$ is an extension of $L$ containing all elements in $S$, which gives the reverse inclusion. – Starfall Sep 26 '16 at 21:06
• Why is $L$ the field of fractions of $K[\alpha_1, ..., \alpha_m]$? – rmdmc89 Sep 27 '16 at 0:19
• Well, because $L = K[\alpha_1, \ldots, \alpha_m]$, of course. (If $\alpha$ is an algebraic element over a field $K$, then $K[\alpha]$ is always a field.) – Starfall Sep 27 '16 at 0:42