Is the tr.deg of Q_p over Q 1? and what about C over Q?
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9$\begingroup$ If K is a subfield of L, and K is infinite, and #L > #K, then tr.deg(L/K) = #L. $\endgroup$– Bjorn PoonenCommented Feb 16, 2010 at 5:08
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2 Answers
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In both cases the transcendence degree is the cardinality of the continuum. CH is not needed.
This is a corollary of the following result: let $K$ be any infinite field, and let $L/K$ be any extension. Then
$\# L = \operatorname{max} (\# K, \operatorname{trdeg}_K L)$.
To prove this, in turn it suffices to establish the following two results (each of which is straightforward):
1) If $K$ is infinite and $L/K$ is algebraic, then $\# L = \# K$.
2) If $K$ is any infinite field, $T = \{t_i\}_{i \in I}$ is an arbitrary set of indeterminates and $K(T)$ is a purely transcendental function field in the indeterminates $T$, then $ \# K(T) \leq \# T + \# K$.
answered Feb 16, 2010 at 5:08
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$\begingroup$ Thank you. It's really clear when one considers the cardinality...Actually I was wondering if Q_p could be the function field of some variety besides Spec(Q_p) or Spec(Z_p), as some kind of analogy of fields of tr.deg 1 corresponds to curves. I don't know if it's a "right" question though... $\endgroup$– naturaCommented Feb 16, 2010 at 5:20
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3$\begingroup$ @basic: Your "actual" question is actually interesting. More generally, one can ask whether a field $K$ is a "nontrivial function field", i.e., whether there exists a subfield $F$ of $K$ such that $K/F$ is finitely generated and tr. deg. of $K/F$ is positive and finite. I believe such a field must be "Hilbertian", which rules out $Q_p$. $\endgroup$ Commented Feb 16, 2010 at 5:28
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$\begingroup$ Very nice, Pete: +1. Bourbaki gives a formula analogous to your max formula for the cardinality of L (in his first exercise for §14, Chap.V of Algebra), only it is false for finite fields [which you wisely excluded] ! How do you downvote him ? :-) $\endgroup$ Commented Feb 21, 2010 at 17:19
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The transcendence degree of either $\mathbb{C}$ or $\mathbb{Q}_p$ over $\mathbb{Q}$ is exactly the cardinality of the continuum. Certainly it can't be countable, since any field with countable transcendence degree over a countable field is countable. On the other hand, either transcendence degree is at most that of the continuum. So we're already done assuming CH. But I think the result holds even without CH; can anyone confirm / deny? (Edit: Yes, as the other answer shows.)
answered Feb 16, 2010 at 5:04