An exercise from Rotman's advanced modern algebra

Question

I have been stuck in an exercise from Rotman's advanced modern algebra for a while,which is stated below:

Let $K/k$ be a field extension. If $A \subseteq K$ and $u \in k(A)$, prove that there are $a_1,...,a_n \in A$ with $u \in k(a_1,...,a_n)$.

I have no idea how to deal with it, since $K/k$ may be algebraic or not, so is $k(A)$.

Can anyone give a complete proof or just a hint? Any help would be greatly appreciated!

Answer 1

Hint:

Write $\; A=\bigcup\limits_{\substack{B\text{ finite}}\\B\,\subset\, A}B$ and order the finite subsets of $A$ by inclusion. The set of finite subsets of $A$ is directed.

Show that if $B, C$ are finite such that $B\subset C$, there corresponds an inclusion morphism $\;i:k(B)\hookrightarrow k(C)$, and that if $B\subset C\subset D$, then the composition $k(B)\hookrightarrow k(C)\hookrightarrow k(D)$ is the same as $k(B)\hookrightarrow k(D)$ (resulting from the inclusion $B\subset D$).

This proves that $k(A)$ is the direct limit of the inductive system of fields $\;\bigl(k(B),i\bigr)_{\substack{B\text{ finite}}\\B\,\subset\, A}$.

Any element in the direct limit is the image of an element in some $k(B)$, thus proving the initial assertion.

Answer 2

u is a finite sum de element of k and A with coeficients in k then you only need to attach those elemtens of A that belong in that sum.