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 independent over $k$ it will remain algebraically independent 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 answer 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.