# Cardinality of a $\mathbb Q$ basis for $\mathbb C$, assuming the continuum hypothesis

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 $$|E|=\aleph_0 |F|$$

Proof:

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.

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.
• 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 Dec 19, 2016 at 18:04
• 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".) Dec 19, 2016 at 18:20
• $Q(S)$ 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? Dec 19, 2016 at 18:26
• 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 $\Bbb Q(\{s_1,\ldots,s_n\})$. Dec 19, 2016 at 18:43