most recent 30 from http://mathoverflow.net 2013-05-22T14:06:01Z http://mathoverflow.net/feeds/question/75219 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75219/is-the-transcendence-degree-of-a-domain-over-a-subfield-the-same-as-that-of-the-f Georges Elencwajg 2011-09-12T13:05:32Z 2011-09-13T07:53:12Z

<p>Consider the inclusion $k\subset A$ of the field $k$ in the domain $A$ and the fraction field $K=Frac(A)$ of $A$.<br> 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 $K$.<br> Consider however a family $(\alpha _i) _{i \in I}$ of elements $\alpha _i \in K$ algebraically independent over $k$.<br> 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: </p> <p><strong>Precise question</strong> 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 $I$) ?</p> <p>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. </p> <p><strong>NB</strong> 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!</p> <p><strong>Edit</strong> 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 $K$.<br> I now think that it is impossible to <em>naturally</em> 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.</p>

http://mathoverflow.net/questions/75219/is-the-transcendence-degree-of-a-domain-over-a-subfield-the-same-as-that-of-the-f/75285#75285 a-fortiori 2011-09-13T07:53:12Z 2011-09-13T07:53:12Z

<p>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 $A'\subseteq B\subseteq A$.</p> <p>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, $\mathrm{char}(k)=0$).</p>