From: magidin@math.berkeley.edu (Arturo Magidin)
Subject: Re: Answering the critics
Date: Fri, 3 Aug 2001 14:50:26 +0000 (UTC)
Newsgroups: sci.math
Summary: Factoring polynomials over the algebraic integers

In article <9kdfb2$nla$1@fang.dsto.defence.gov.au>,
The Scarlet Manuka <sacha@maths.uwa.edu.au> wrote:

   [.snip.]

>Also, this only applies to quadratics and does not generalise
>to an arbitrary degree, so does not apply to your proof. The
>closest we have is Arturo Magidin's result that the a_i and b_i
>can be chosen to be algebraic integers if we allow a constant
>out the front (which will be a general algebraic number, if I
>understand his result correctly).

Hmm... No. No constant out in front. A polynomial in one variable with
algebraic integer coefficients can be factored into a product of
linear terms, each eith algebraic integer coefficients (although you
may need to go to an extension of the splitting field to do it).

It is the analogous statement to the usual corollary of Gauss' Lemma:
If f(x)\in Z[x] can be written as f=GH, where G(x),H(x)\in Q[x], then
there are polynomials g(x),h(x)\in Z[x], of the same degrees as G and
H, respectively, such that f=gh.

The statement you write is actually trivial, given the following well
known result:

Lemma: If a is an algebraic number, then there is an nonzero integer c such
that ca is an algebraic integer.

Proof. Since a is algebraic, there is a polynomial (not necessarily
monic) with integer coefficients 

b_n x^n + ... + b_1 x + b_0

such that a is a root. Therefore,

b_n a^n + ... + b_1 a + b_0 = 0

Multiplying by b_n^{n-1}, we have

b_n^n a^n + ... + b_n^{n-1}b_1 a + b_n^{n-1}b_0 = 0

Therefore, b_n*a is a root of the monic polynomial 

x^n + b_{n-1}x^{n-1} + ... + b_n^{n-2}b_1 x + b_n^{n-1}b_0

which has integer coefficients; therefore, b_n*a is an algebraic
integer. let c=b_n, and we're done. QED

Now, we can improve on your statement easily from this: every
polynomial with algebraic coefficients may be written as a product of
linear terms, each with algebraic integer coefficients, times a
rational number.

For let f(x) be a polynomial with algebraic coefficients. Use the
Fundamental Theorem of Algebra to write it

f(x) = g_1(x)...g_n(x)
where each g_i(x) = a_ix + b_i, with a_i and b_i algebraic. Therefore,
there exists an integer c_i such that c_ia_i and c_ib_i are algebraic
integers (find the one for a_i and the one for b_i, and take c_i to be
their lcm). Let D be the lcm of c_1,...,c_n. Then
f(x) = D^n/D^n * g_1(x)...g_n(x)
     = (1/D^n) (Dg_1(x)) * (Dg_2(x)) *...* (Dg_n(x))

and each Dg_i(x) has algebraic integer coefficients, by
construction. So you would get it as a product of linear factors, each
with algebraic integer coefficients, times a rational constant up
front. The result I gave earlier shows that you can actually get rid
of the constant up front.

======================================================================
"It's not denial. I'm just very selective about
 what I accept as reality."
    --- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
magidin@math.berkeley.edu