Is the transcendence degree of a domain over a subfield the same as that of the fraction field of that domain?

Consider the inclusion $k\subset A$ of the field $k$ in the domain $A$ and the fraction field $K=Frac(A)$ of $A$.
Obviously if a family $(a_i)_{i\in I}$ of elements $a_i \in A$ is algebraically independant over $k$ it will remain algebraically independant in $K$.
Consider however a family $(\alpha _i) _{i \in I}$ of elements $\alpha _i \in K$ algebraically independent over $k$.
To my puzzlement, I can't construct from it an algebraically independent family $(a_i)_{i\in I}$ of elements $a_i \in A$. Although my real question is whether it is possible to actually construct such a family in a natural way, I'll ask something more precise:

Precise question Given the $k$- algebraically independent set $(\alpha _i) _{i \in I}$ in $K$, does there exist in $A$ some $k$- algebraically independent set $(a_i)_{i\in I}$ ( with the same index set $I$) ?

The answer is "yes" if $A$ is finitely generated over $k$., thanks to E.Noether's normalization theorem. Interestingly the proof of that theorem is not purely field-theoretic, since it makes use of Krull dimension.

NB I'm not sure (despite the title of the question!) that I know what the transcendence degree of $A$ is: the "correct" definition might follow from the answers to this question!

Edit a-fortiori has proved in his comment that the anwer to the "Precise question" is yes, and that as a consequence the only reasonable definition of transcendence degree of $A$ is that it equals the transcendence degree of $K$.
I now think that it is impossible to naturally associate to the $k$- algebraically independent family $(\alpha _i) _{i \in I}$ in $K$ a $k$- algebraically independent family $(a_i)_{i\in I}$ in $A$, even though we now know thanks to a-fortiori that such a family exists. For example, if $X, Y$ are algebraically independent over $k$, and we take the family of just one element $\alpha=\frac {X}{Y}\in k(X,Y)$, which transcendental element in $k[X,Y]$ should we choose?! It would be great if someone could come up with a rigorous statement of the impossibility of a natural choice.

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If you are willing to well-order your set $I$, I think that you can first reduce to the extension of a single element of $K$ and then observe that at least one among "numerator" or "denominator" of the next element must be transcendental over the domain you have so far constructed. –  M P Sep 12 '11 at 13:30
The usual strengthening of the existence theorem for transcendence bases states: If $K$ is algebraic over $k(A)$ for some subset $A\subset K$, there is a transcendence basis contained in $A$. –  user2035 Sep 12 '11 at 13:53
You are absolutely right,a-fortiori, thanks a lot! And I can't even plead ignorance: I knew the result you mention! I'll edit my question. –  Georges Elencwajg Sep 12 '11 at 14:55
a-fortiori, I (and certainly other users too) would be very happy to upvote you if you transformed your comment above into an answer. –  Georges Elencwajg Sep 12 '11 at 15:57

As already said in the comments, there is a general statement (analogous to extending bases in linear algebra) that for a field extension $K/k$ and subsets $A'\subseteq A\subseteq K$ such that $A'$ is algebraically independent and $K$ is algebraic over $k(A)$ there is a transcendence basis $A'\subseteq B\subseteq A$.
However, as your example $\alpha=X/Y\in K=k(X,Y)$ shows, it is possible that there are $k$-automorphisms of $K$ leaving $\alpha$ and $A$ invariant ($X,Y\mapsto \lambda X,\lambda Y$ for $\lambda\in k^\times$) such that there are no invariant transcendental elements of $A$ (say, $\mathrm{char}(k)=0$).