Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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,

share|improve this question
1  
And besides, you have to rule out cardinals strictly between $\aleph_0$ and $\mathfrak{c}$. –  GEdgar Jan 28 '12 at 4:08
2  
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
2  
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

3 Answers 3

up vote 7 down vote accepted

(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 $\mathfrak{c} = \# X$.

share|improve this answer
    
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
1  
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

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 $\mathfrak c$.

share|improve this answer

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.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.