[FOM] Only one proof

Timothy Y. Chow tchow at alum.mit.edu
Mon Sep 21 12:02:44 EDT 2009

...the very existence of the algebraic numbers seems to depend on topology.

> Certainly one can, by purely algebraic means, form from the field C
> of complex numbers and a polynomial p an extension C' of C containing
> a root of p.  But to prove the Fundamental Theorem of Algebra that
> way you need the extra step of showing how to collapse C' to C.
> How do you do that without appealing to the completeness of C, or
> some topological counterpart thereof?

> Vaughan, you seem to be conflating the two statements
> 1. the field of complex numbers is algebraically closed; and
> 2. there exists an algebraic closure of the rationals.

>Tim, you seem to be viewing the elements of splitting fields as numbers. 
>  In my first post I said "algebraic *numbers*", not "roots of 
>polynomials."  Granted this could be clarified, but how did my second 
>post fail to do so?

I think that Melvyn and I were just trying to point out that your use of 
the phrase "the existence of the algebraic numbers" is nonstandard.  The 
most natural interpretation of this phrase, in the context of your 
original sentence, is "the existence of an algebraic closure of the 
rationals."  But you meant something like, "the fact that the set of 
algebraic complex numbers is an algebraically closed subfield of C."  
This usage is a little odd, and it was not until just a moment ago that I 
finally figured out that this is what you were trying to say.  Even if we 
formally define "algebraic number" to mean "a complex number that is a 
root of a polynomial equation with rational coefficients," the "existence 
of the algebraic numbers" is a triviality and not an important theorem.


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