0
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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 $.

$\endgroup$
  • $\begingroup$ 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 $[L(S):K(S)]\leq n =[L:K]$? $\endgroup$ – rmdmc89 Sep 26 '16 at 20:51
  • 1
    $\begingroup$ 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. $\endgroup$ – Starfall Sep 26 '16 at 21:06
  • $\begingroup$ Why is $L$ the field of fractions of $K[\alpha_1, ..., \alpha_m]$? $\endgroup$ – rmdmc89 Sep 27 '16 at 0:19
  • $\begingroup$ 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.) $\endgroup$ – Starfall Sep 27 '16 at 0:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.