Re: Synopsis, algebraic integers

From: John Roberts-Jones (johnrj_at_globalnet.co.uk)
Date: 09/19/04 

Date: Sun, 19 Sep 2004 23:33:49 +0100

On 19 Sep 2004 06:16:16 -0700, jstevh@msn.com (James Harris) wrote:

>I've claimed a problem with the ring of algebraic integers. Here very
>quickly I'd like to outline the problem.
>
>Consider
>
>2x^2 + (6q_2 - q_1)x - 3q_1 q_2 = 0
>
>which has
>
>x = q_1/2 and x=-3q_2
>
>as the roots, and in general let
>
>q_1 q_2 = 1, and 6q_2 - q_1 be an integer.
>
>If both roots are rational then note that q_1 may be coprime to 2 in a
>ring where coprimeness holds.
>
>For instance, that can be seen directly with q_1=1, q_2=1, which
>fulfils all the conditions given, and shows coprimeness in the ring of
>integers.
>
>There are an infinity of rational solutions for which q_1 is coprime
>to 2 in the ring of integers.

It is only meaningful to say that q1 is coprime to 2 in the ring of
integers if q1 is an integer. If q1 and 6q2 - q1 = 6/q1 - q1 are
both integers then q1 is a divisor of 6 in the ring of integers.
But in the ring of integers the only divisors of 6 coprime to 2
are 1, -1, 3 and -3.

Four is hardly an infinity.
>
>But, surprisingly, if q_1 and q_2 are irrational, then standard
>methodology indicates that q_1 has non-unit algebraic integer factors
>of 2 forced upon it in the ring of algebraic integers.
>
>That puzzling result follow from the fact that q_2 cannot be an
>algebraic integer if q_1 and q_2 are irrational.

Well done! At last I see what you think is going wrong in the ring
of algebraic integers, A. It all comes down to that pesky theorem,
that if z is an algebraic integer and it is a zero of its minimum
polynomial P(t) in Z[t] then in A it has a nonunit common factor
with each rational prime dividing the constant term of P(t).
But if P(t) is not irreducible in Z[t] then it seems easy to find
counterexamples; why should reducibility make a difference?

It seems particularly blatant when P(t) is quadratic, doesn't it?

But note that word _minimum_ in the theorem. The minimum
polynomial of an algebraic integer x is the unique polynomial of
least degree of which x is a zero.

If the zeros of a monic quadratic polynomial P(t) are irrational
then P(t) is irreducible in Z[t] and it is the minimum polynomial
in Z[t] of both zeros.

Otherwise, the zeros are rational, so they are rational integers,
j and k, say. In this case P(t) is reducible in Z[t], as
P(t) = P1(t) P2(t), where P1(t) = t - j and p2(t) = t - k.

The minimum polynomial of j is P1(t) and the theorem obtains: if p
is a rational prime dividing j then in A there is a nonunit common
factor, p, of j and p that is a divisor of p. Note that the
theorem has nothing to say about any common factors of j and the
constant term of P2(t).

Similarly, the minimum polynomial of k is P2(t), the theorem
obtains, but it says nothing about common factors of k and the
constant term of P1(t).

If P(t) is reducible over Z then it cannot be the minimum
polynomial of anything; the theorem does not apply.

John Roberts-Jones

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