1
$\begingroup$

Let $\overline{\mathbb{Q}}$ the algebraic closure of $\mathbb{Q}$, and $K$ a field extension of $\mathbb{Q}$ (not necessarily algebraic) such that $[K:\mathbb{Q}]= \infty$.

Let $t_1,...,t_n \in K$, and $L=\mathbb{Q}(t_1,...,t_n) \cap \overline{\mathbb{Q}}$.

Is $L$ a field extension of $\mathbb{Q}$ of finite degree ?

Thanks in advance.

1 Answers 1

3

For the intersection to make sense, we must assume that $\bar { \mathbb Q} \subset K$. Then the answer is yes.

Indeed, since $L\subset \mathbb Q(t_1,...,t_n)$ it follows that $L$ is of finite type as a field (this is a non-trivial result!).
But then $L$ is algebraic over $\mathbb Q$ (since $L\subset \bar {\mathbb Q}$) and finitely generated , so that $[K:\mathbb Q]\lt \infty$ .

Edit
As a consequence of Dylan's interesting comment, let me give a reference for the non-trivial result invoked above, since apparently it is absent from most algebra textbooks: Bourbaki, Algebra, Chapter v, §14.7, Corollary 3, page 118.

  • 0
    Dear Georges, Does the non-trivial result have a name? Thanks a bunch,2011-12-06
  • 0
    Dear @Dylan, no the theorem doesn't have a name. I think yours is a very good question, because this might be one of the reasons why this theorem is less known, than it deserves. I'll add a reference in an edit. Anyway, thanks for your interest.2011-12-06
  • 0
    @Georges Ah, thank you! I guess this was my roundabout way of asking for a reference. But the theorem did remind me of those theorems in commutative algebra having impressive German titles.2011-12-06
  • 0
    Dear @Dylan: like *Endlich Erzeugbarkeit eines beliebigen Unterkörpers einer endlich erzeugten Erweiterung des Grundkörpers* ?2011-12-06
  • 0
    @Georges: That name deserves an abbreviation EEebUeeeEG. Another question: emphasis on *as a field* as opposed to what?2011-12-07
  • 0
    Dear @Marc: as opposed to *finitely generated as a $\mathbb Q$- algebra*. And, yes, your abbreviation is definitely *chic*.2011-12-07