Prove that $\operatorname{tr.deg}(\mathbb{C/Q})=\mathfrak{c}$ (where $\mathfrak{c}$ is the cardinality of $\mathbb{R}$) using the continuum hypothesis.

$Proposition$ If $E/F$ is an algebraic extension then $|E|=\aleph_0 |F|$


Let $S$ be a trancendental basis of the extension $\mathbb{C/Q}$.Then $|S| \leqslant \mathfrak{c}$.Suppose that $|S|< \mathfrak{c}$. Then, by the continuum hypothesis, $S$ is finite or countable.

Let $S=\{s_1,s_2, \ldots \}$ a countably infinite set.

Then let $A_n=\mathbb{Q}(s_1,,,s_n)$

$(\dagger)$ We have that every $A_n$ is a countable set and $\bigcup_{n=1}^{\infty}A_n=\mathbb{Q}(S)$ which is countable as a countable union of countable sets.

Also by definition of the trancendental basis we have that $\mathbb{C}/\mathbb{Q}(S)$ is an algebraic extension thus by the above proposition $|\mathbb{C}|=\aleph_0 |\mathbb{Q}(S)|=\aleph_0 \aleph_0=\aleph_0$ - deriving a contradiction.

Is this proof correct?

I'm not sure about if the equation of the sets in $(\dagger)$ is correct.


1 Answer 1


Yes, the proof you suggest is correct. To see why, note that $\mathbb{Q}(S)$ is obtained by taking the rational functions in $|S|$ variables. If $S$ is finite, then there are only countably many such rational functions.

If you know basic cardinal arithmetic, you can in fact show that $|\mathbb Q(S)|=|S|+\aleph_0$, and therefore the continuum hypothesis is not needed for the proof.

  • $\begingroup$ thank you for your help.the continuooum hypothesis is a hint that our instructor gave me.my main concern is if the set equation in $(\dagger)$ is correct $\endgroup$ Dec 19, 2016 at 18:04
  • $\begingroup$ That the union is equal I think is correct, but this is no longer a set theory question, but rather a field theory question. How do you define $\Bbb Q(S)$? (As for the CH part, yes, I was just making a general remark. It's easier if you just know there is "countable" and "uncountable".) $\endgroup$
    – Asaf Karagila
    Dec 19, 2016 at 18:20
  • $\begingroup$ $Q(S)$ is the set of rational functions with $|S|$ variables.A typical element of this set is a $u=f(s_1,,,,s_n)/g(s_l1,,,,s_lm) $for some elements in $S$? I know the form of the elements if S is finite,but what form does an element have in Q(S) is S is infinite? $\endgroup$ Dec 19, 2016 at 18:26
  • 1
    $\begingroup$ The same form. It's a ratio between two finite sequences in finitely many variables and rational coefficients. Moreover, if $x\in\Bbb Q(S)$, then by definition only finitely many symbols appear in $x$ in either $f$ or $g$. So $x$ lies in some $\Bbb Q(\{s_1,\ldots,s_n\})$. $\endgroup$
    – Asaf Karagila
    Dec 19, 2016 at 18:43

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