# Why does $\mathbb{C}$ have transcendence degree $\mathfrak{c}$ over $\mathbb{Q}$?

It's pretty well known that $\text{trdeg}(\mathbb{C}/\mathbb{Q})=\mathfrak{c}=|\mathbb{C}|$.

As a subset of $\mathbb{C}$, of course the degree cannot be any greater than $\mathfrak{c}$. I'm trying to understand the justification why it cannot be any smaller. The explanation in my book says that if $\mathbb{C}$ has an at most countable (i.e. finite or countable) transcendence basis $z_1,z_2,\dots$ over $\mathbb{Q}$, then $\mathbb{C}$ is algebraic over $\mathbb{Q}(z_1,z_2,\dots)$. Since a polynomial over $\mathbb{Q}$ can be identified as a finite sequence of rationals, it follows that $|\mathbb{C}|=|\mathbb{Q}|$, a contradiction.

I don't see why the polynomial part comes in? I'm know things like a countable unions/products of countable sets is countable, but could someone please explain in more detail this part about the polynomial approach? Since $\mathbb{C}$ is algebraic over $\mathbb{Q}(z_1,z_2,\dots)$, does that just mean that any complex number can be written as a polynomial in the $z_i$ with coefficients in $\mathbb{Q}$? For example, $$\alpha=q_1z_1^3z_4z_6^5+q_2z_{11}+q_3z^{12}_{19}+\cdots+q_nz_6z_8z^4_{51}?$$

Is the point just that the set of all such polynomials are countable?

Thanks,

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And besides, you have to rule out cardinals strictly between $\aleph_0$ and $\mathfrak{c}$. –  GEdgar Jan 28 '12 at 4:08
The result is correct, the proof you allude to has a problem. It shows that any transcendence basis is uncountable. But (unless we assume the Continuum Hypothesis) that does not force size $c$. –  André Nicolas Jan 28 '12 at 4:09
Suppose that $B$ is a subset of the complex numbers, with infinite cardinality $\kappa$. Then the set of polynomials with coefficients in $B$ has cardinality $\kappa$, and therefore so does the set of complex numbers algebraic over $(\mathbb{Q}\cup B)$. So if $\kappa<c$, this algebraic closure cannot be all of $\mathbb{C}$. (From polynomials to algebraic numbers is easy, any non-zero polynomial has only finitely many roots.) –  André Nicolas Jan 28 '12 at 4:17

(Of course I assume the Axiom of Choice...) Choose a transcendence basis $X = \{x_i\}_{i \in I}$ for $\mathbb{C}$ over $\mathbb{Q}$. Then $\mathbb{C}$ is an algebraic extension of $\mathbb{Q}(X)$. Now here are two rather straightforward facts:

1: If $F$ is any infinite field and $K/F$ is an algebraic extension, then $\# K = \#F$.

2: For any infinite field $F$ and purely transcendental extension $F(X)$, we have $\# F(X) = \max (\#F, \# X)$.

Putting these together we find

$\mathfrak{c} = \# \mathbb{C} = \# \mathbb{Q}(X) = \max (\aleph_0, \# X)$.

Since $\mathfrak{c} > \aleph_0$, we conclude $\mathfrak{c} = \# X$.

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Thanks. Let me see if I understand these two facts. For 1., $\# K\geq\# F$ of course, since $K\supset F$. Since $K$ is algebraic over $F$, there is a surjective map $F[x]\to K$ mapping $p(x)$ to one of its roots in $K$. Since $F$ is infinite, $\# F[x]=\# F$, so $\# F\geq \# K$ and so $\# F=\# K$. –  jain Jan 28 '12 at 8:01
If you have time, do you mind proving why the 2nd fact is? I don't see why $\# F(X)$ should be the maximum of the two. Thank you. –  jain Jan 28 '12 at 8:02
@jain: For any infinite field $F$ and a set $X$ of indeterminates, if $X$ is finite, then $\# F(X) = \# F$: do you see how to prove this? Moreover, if $X$ is infinite, then $F(X) = \bigcup_{Y \subset X} F(Y)$, where $Y$ ranges over the finite subsets of $X$. –  Pete L. Clark Jan 29 '12 at 17:32
@PeteL.Clark Thanks for supplying these thoughts. I do have a question -- Why must we assume the axiom of choice? –  Nik Kumar Apr 21 '13 at 21:58
@Nik: The axiom of choice is necessary in order for every field extension to admit a transcendence basis, in a similar way to how it is necessary for every vector space to admit a basis. –  Pete L. Clark Apr 21 '13 at 22:10

Note that, if $K$ is a countable field and $x$ is transcendental over $K$, then $K[x]$ is also countable (separate polynomials by degree), hence $K(x)$ is countable (the elements can be identified with pairs of elements of $K[x]$). Letting $K_0=\mathbb{Q}$ and $K_n=K_{n-1}(z_n)$, we have that $$\mathbb{Q}(z_1,z_2,\ldots)=\bigcup_{n=0}^\infty K_n$$ is a countable union of countable sets, and hence is countable.

If a field $L$ is countable and $F$ is algebraic over $L$, then $F$ is countable, because $L[x]$ is countable and we can cover $F$ by a countable number of finite sets $S_f$, one for each $f\in L[x]$, where $S_f=\{a\in F\mid f(a)=0\}$.

Thus, if $\mathbb{C}$ had a countable transcendence basis $z_1,z_2,\ldots$ over $\mathbb{Q}$, then $\mathbb{C}$ is algebraic over $\mathbb{Q}(z_1,z_2,\ldots)$, and it would follow that $\mathbb{C}$ is countable, a contradiction.

(This does not explain why $\mathbb{C}$ has transcendence degree $\mathfrak{c}$, though.)

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The field $F=\mathbb Q(z_1,z_2,..)$ is countable as well as the polynomial ring $F[x]$. Since $\mathbb C$ is algebraic over $F$ you can define a surjective map $\phi:F[x]\to \mathbb C$ by sending $p(x)$ to one of its roots.
Edit: This just shows that the transcendence degree is uncountable but you can use the same argument for any base of size less than $\mathfrak c$.

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