From: gadde@sm.luth.se (Erland Gadde)
Newsgroups: sci.math
Subject: Re: field of algebraic numbers
Date: Fri, 11 Dec 1998 21:14:51 +0100

In article <74rh11$1hbr$1@news-out2.du.gtn.com>, "gap" <gherz@gmx.de> wrote:

> Hi,
> I have a question. Let F be the field of algebraic numbers. Is there a
> polynominal p in F[X] which has Zeros which are NOT in F. So my question is:
> Is F algebraic closed?

Yes, it is! 
Let p(x)=a0+a1*x+...+an*x^n be a polynomial in F(X), and let r be one of
its zeros. We assume, without loss generality, that p(x) is irreducible
over F.
(Otherwise, just take an irreducible factor of which r is a zero).
With Q as the set of rationals, let Q(a0,a1,...,an) be the smallest
subfield of F which contains Q and a0,a1,...,an. 
This is then a finite algebraic extension of Q, which means that there is
a finite  subset {b0,b1,...,b{m-1} } of F which is a basis of
Q(a0,a1,...,an) 
(regarded as a vector space) over Q. This follows by induction from
the fact that if K,L, and M are fields, and if L is a finite algebraic
extension of K with degree k, and if M is a finite algebraic
extension of L with degree l, then M is a finite algebraic
extension of K with degree k*l: If {c1,...,ck} is a basis of L over
K, and {d1,...,dl} is a basis of M over L, then the set
of products ci*dj (1<=i<=k,1<=j<=l) is a basis of M over K. 
Now, Q(a0,a1,...,an,r) is a simple (hence finite) algebraic extension of 
Q(a0,a1,...,an), (the set {1,r,r^2,...,r^{n-1} } is a basis of  
Q(a0,a1,...,an,r) over Q(a0,a1,...,an). But then, Q(a0,a1,...,an,r)
is a finite algebraic extension of Q of degree m*n. It follows that
the set of the m*n+1 elements {1,r,r^2,...,r^{m*n}} is linearly
dependent over Q, that is: there is a rational polynomial (and hence
an integer polynomial) of degree m*n, of which r is a zero.
Hence r is in F, and F is algebraically closed.

-- 
Erland Gadde
Department of Mathematics
Luleå University of Technology
Sweden
==============================================================================

From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: field of algebraic numbers
Date: 14 Dec 1998 06:30:26 GMT

In article <74v2t8$mti$1@nnrp1.dejanews.com>,  <phunt@interpac.net> wrote:
>You previously stated that F is the field of algebraic numbers,
>and now you're asking other folks: Is the field of algebraic
>numbers closed?
>
>What a charade! ... Perfectly exampling my complaint respecting
>the practice of secondary teaching in public schools.
[etc.]

Lovely rant, but I think you've missed the point. Beginning with the
field  Q  of rational numbers, we form the field  F  of all roots of
all polynomials with coefficients in  Q. It takes some doing to be
sure that this really is a field. Then we consider all polynomials whose
coefficients lie in  F. What's the field containing all of _their_ roots?

The answer is, "again  F",  but that's not so obvious. If  A, B, and  C
are the three roots of the polynomial  X^3 + X + 1 = 0, then is it 
obvious that the roots of  X^3 + A X^2 + B X + C = 0  are also in  F?
If so, what's their minimal polynomial over  Q?

Again, I contend that it's not hard to answer these questions, and indeed
they should be answered quite early in a course on Fields, but I don't
really think it's necessary to get on a soapbox over this particular issue.
Your average mathematician couldn't prove  F  was algebraically closed
either (since your "average mathematician" is not an algebraist!)

dave