most recent 30 from http://mathoverflow.net 2013-05-25T03:34:24Z http://mathoverflow.net/feeds/question/67704 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67704/what-is-a-best-transcendence-basis-for-r-q Bruno 2011-06-13T21:02:28Z 2011-06-13T21:26:14Z

<p>It is easy to show, using the axiom of Zorn, that there exists a transcendence basis for $\mathbb{R}/\mathbb{Q}$, i.e. a set $S$, algebraically independent over $\mathbb{Q}$, such that $\mathbb{R}/\mathbb{Q}(S)$ is an algebraic extension.</p> <p>What can we say about $T=\mathbb{R} - \mathbb{Q}(S)$? It is easy to show that $\mathbb{R}/\mathbb{Q}$ is not purely transcendental, so that $T \neq \emptyset$. (Indeed, no element of $\mathbb{R}-\mathbb{Q}$ algebraic over $\mathbb{Q}$ may be contained in $\mathbb{Q}(S)$ - an algebraic relation for such an element over $\mathbb{Q}$ would immediately translate into a relation of algebraic dependence in $S$ over $\mathbb{Q}$.)</p> <p>Thus, all <em>real</em>, <em>non-rational</em> algebraic numbers are contained in $T$. Is it possible, for example, to choose $S$ so that this inclusion is an equality? If not, how "small" can we make $T$?</p> <p>I'm sorry if this turns out to be trivial, or if the answer to my question can be easily found in the literature. I tried!</p> <p>Thank you very much and have a pleasing day.</p>

http://mathoverflow.net/questions/67704/what-is-a-best-transcendence-basis-for-r-q/67705#67705 Pace Nielsen 2011-06-13T21:26:14Z 2011-06-13T21:26:14Z

<p>No, you cannot choose $T$ so nicely. For cardinality reasons, we can choose some $s\in S$ with $s\notin \overline{\mathbb{Q}}$. Now, consider the element $\sqrt{s}$. I'll leave it to you to consider how large this makes $T$.</p>