From: Arturo Magidin (magidin@math.berkeley.edu)
Subject: Re: Two Algebra Questions (answer?) 
Newsgroups: sci.math
Date: 2001-07-14 18:11:45 PST 
Summary: Integer polynomials can be factored completely over algebraic integers

In article <qyr*1E20o@news.cam.virata.com>,
John Rickard  <John.Rickard@virata.com> wrote:

   [.snip.]

>Can you give an example of a non-constant polynomial with integer
>coefficients that cannot be factored as a product of linear terms with
>algebraic integer coefficients?

I asked an Arithmetic Geometer friend of mine, and he said he thought
it was impossible, and said it would follow from a "generalization of
Gauss' Lemma." With this information, I think I can prove the
following:

CLAIM: Let f(x) be a polynomial with integer coefficients. Then it can
be factored into a product off linear terms, each with algebraic
integer coefficients.

(Note that even so, the result claimed by James Harris elsewhere is
much stronger than this, since his "integers plus radicals plus i" is
much smaller than the ring of algebraic integers).

In fact, if the argument I propose works, it should also work with
"integer coefficients" replaced by "algebraic integer coefficients",
in any number field.

What follows is more or less my attempt to parallel the presentation
in Birkhoff-MacLane's "Survey of Modern Algebra", Chapter III, Section
9. Assuming I did not make any mistakes (and I will be greateful for
any mistakes that are pointed out), I'm sure it can be cleaned up
considerably.


Let K be a number field, and let R be the ring of integers of K. Given
any polynomial f(x) with coefficients in K, we define the content of f
to be the "ideal" generated by the coefficients of f (recall that
since R is a Dedekind domain, we have an ideal group; so the ideal I
mentioned is really a product of prime ideals of R, with possibly
negative exponents; it's a "quotient" of an ideal of R by another
ideal of R).

Let's say that f is "primitive" if its content is R.

Lemma 1: Let K<F be number fields, with rings of integers R and S,
respectively. Let f(x) be a polynomial with coefficients in R. Then f
is primitive as a polynomial in R[x] if and only if it is primitive as
a polynomial in S[x].

Proof: Assume that f is not primitive in R[x]. Then there is a prime
ideal P such that every coefficient of f(x) lies in P; let Q be a
prime of S lying over P. Then every coefficient of f(x) lies in Q, so
f is not primitive in S[x] either.

Conversely, if there is a prime ideal Q of S such that every
coefficient of f lies in Q, then every coefficient of f also lies in
Q\cap R=P, which is a prime of R; hence f cannot be primitive in R[x]
either. QED

DEF. Given a polynomial f(x) with algebraic integer coefficients, we
say f(x) is "primitive" if and only if it is primitive in R[x], where
R is the ring of integers of any number field K such that f is defined
over R.

Lemma 2 (Gauss' Lemma) Let f and g be two polynomials with algebraic
integer coefficients, and assume they are both primitive. Then their 
product is also primitive.

Proof. Let K be a number field, with ring of integers R, such that
both f and g are defined over R. Write

   \Sum(c_k x^k)  =  (\Sum a_i x^i)(\Sum b_j x^j)
and assume this product is not primitive. Then there is some prime
ideal P of R such that every c_k lies in P. But let a_i and b_j be the
first coefficients of (\Sum a_ix^i) and (\Sum b_jx^j) respectively
which do NOT lie in P (which is possible since both f and g are
primitive). Then the coefficient c_{i+j} is equal to

c_{i+j} = a_0 b_{i+j} + ... + a_{i-1}b_{j+1} +
            + a_ib_j
            + a_{i+1} b_{j-1} +...+ a_{i+j}b_0.

Then each summand on the first and third rows lie in P, but the
summand a_ib_j is not; so c_{i+j} cannot lie in P either, giving a
contradiction. Thus, the product is primitive. QED

Lemma 3: Let f(x) be a polynomial with algebraic coefficients (not
necessarily algebraic integers). Then there exists a number field K,
with ring of integers R, such that f(x) can be wrirtten as: 
f(x) = c_f f*(x)
where c_f is a constant, and f* is a primitive polynomial with 
algebraic integer coefficients. Moreover, c_f and f* are unique up to
units of R.

Proof. Let K be any number field over which f is defined, and let R
be its ring of integers. We may write

   f(x) = (a_n/b_n)x^n + ... + (a_0/b_0)
where a_i,b_i lie in R. By going to an extension of K if necessary,
we may assume that a_n and b_n are relatively prime (we may need to go
to an extension where all involved primes of R' are principal to
factor out the terms). 

Let c= 1/b_0...b_n Then f(x)=cg(x), where g has coefficients in R.  By
going to an extension of K if necessary, we may assume that the
content of g is principal, generated by an element c'. Then
f*(x)=(1/c')g(x) is primitive, and f(x)=(cc')f*(x). This gives
existence, by setting c_f = cc'.

To give uniequeness, it suffices to show that f* is unique up to units
of R. Suppose that f*=cg*, where f* and g* are primitive, and c lies
in K. Write c =u/v, with u and v in R, and such that (u,v)=R. Then
ug*=vf*. Suppose that v is not a unit. Let P be a prime ideal of R
with v in P. Then every coefficient of ug* lies in P, but since
(u,v)=R, that means that each coefficient of g* lies in P, but g* is
primitive, givign a contradiction. Thus, v is a unit. By symmetry, u
is a unit, so u/v is a unit of R. QED

Note that (c_f) (the ideal generated by c_f) is the content of f.

Lemma 4: Let f(x) and g(x) be two polynomials with algebraic integer
coefficients. Then the content of fg is equal to the product of the
contents of f and g, in any number field over which both f and g are
defined.

Proof. By Lemma 3, we may write f=c_f f^*, g=c_g g^* with f* and g*
primitive, for some number field K. Then their product is
c_fc_gf*g*, and by Gauss' Lemma, f*g* is primitive. Thus the content
is (c_fc_g)=(c_f)(c_g). To see that this holds in any number field,
simply take the ideals lying below (c_fc_g), (c_f), and (c_g). QED

THEOREM: Let f(x) be a polynomial with algebraic integer
coefficients. If it can be factored into a product of polynomials with
algebraic coefficients (not necessarily algebraic integers), then it
can be factored into a product of polynomials of the same degree with
algebraic integer coefficients.

Proof. Suppose we may write f(x)=g(x)h(x). Then by Lemma 3 we may
rewrite as
f(x) = c_g c_h g*(x)h*(x) = c_f f*

Since g*h* is primitive, g*h* ~ f* (they are associates), and (c_gc_h)
= (c_f). Thus, by going to a larger number field if necessary, c_gc_h
= uc_f, where u is a unit. Thus, the factorization is actually a
factorization into a product of polynomials with algebraic integers. QED

COROLLARY: Every polynomial with integer coefficients may be factored
into a product of linear factors, each with algebraic integer
coefficients.

Proof: Apply the Theorem to the factorization into linear terms, each
with algebaric coefficients. QED

I think that does it.

======================================================================
"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