Extension field adjointing a set Let $K/k$ an extension field. Consider $A\subseteq K$ and $u\in k(A)$. Prove that exists elements $a_1,a_2,\ldots,a_n\in A$ such that $u\in k(a_1,a_2,\ldots,a_n)$.
I've proved that $k\subseteq k(A)\subseteq K$ and I think thta exists a basis of $k(A)$, over $k$ such that $\beta\subseteq A$, and thus conclude. Any help? or another argument?
 A: A proof will depend on which definition of $k(A)$ you are using.  My definition is $k(A)$ is the intersection of all subfields of $K$ which contain $k$ and $A$.
Let $F$ be the set of elements of $K$ of the form
$$\frac{f(a_1, ... , a_n)}{g(a_1, ... , a_n)}$$
where $n \geq 1$; $a_1, ... , a_n \in A$; $f, g \in k[T_1, ... , T_n]$; and $g(a_1, ... , a_n) \neq 0$.  
It is easy to see that $F$ is contained in every subfield of $K$ which contains $k$ and $A$.  This is because such a subfield must be closed under addition, multiplication, and division.  Therefore, $F \subseteq k(A)$.  
On the other hand, $F$ is itself a subfield of $K$ containing both $k$ and $A$.  Therefore, $F$ appears as one of the subfields in the intersection defining $k(A)$.  So, $k(A) \subseteq F$.  
Then $k(A) = F$, and from this description of $k(A)$, it is obvious that 
$$k(A) = \bigcup\limits_{E \subseteq A \textrm{ finite}} k(E)$$
In particular, if $u \in k(A)$, then $u$ is is in some $k(E)$, for some $E = \{a_1, ... , a_n\}$.  The usual notation is 
$$k(\{a_1, ... , a_n\}) = k(a_1, ... , a_n)$$
