algebraic numbers
Jim Propp
but that nevertheless has the property that the only rational numbers in
Z[alpha] are the rational integers?
(Note: If instead of Z[alpha] one considers Z[alpha,alpha_1,alpha_2,...],
where alpha_1,alpha_2,... are the Galois conjugates of alpha, then the
desired property cannot hold. This is equivalent to the characterization
of algebraic integers as roots of monic polynomials with rational integer
coefficients.)
Jim Propp
Department of Mathematics
University of Wisconsin
Gerry Myerson
> Does there exist an algebraic number alpha that is NOT an algebraic integer,
> but that nevertheless has the property that the only rational numbers in
> Z[alpha] are the rational integers?
I think so. Let alpha be either root of 2x^2 - 3x + 2 = 0. Alpha is an
algebraic number & not an algebraic integer. It has denominator 2, that
is, 2 alpha is an algebraic integer.
Suppose r is a non-integer rational in Z[alpha]. Then there's a polynomial p
in Z[x] such that p(alpha) = r. Let deg p = n. Then 2^n p(alpha) is an integer,
so 2^n r is an integer, so r = s/2^t for some odd integer s and some positive
integer t.
Now 2^t p - s is a polynomial with integer coefficients and alpha as a root,
so 2^t p - s = (2x^2 - 3x + 2) times q for some polynomial q in Q[x] of degree
n - 2.
Now it seems unlikely to me that the form on the right could give rise to
a polynomial with an odd constant term but all other coefficients even.
Maybe some Gauss' lemma argument could finish this off.
Gerry Myerson (ge...@mpce.mq.edu.au)
Antoine Chambert-Loir
>Does there exist an algebraic number alpha that is NOT an algebraic integer,
>but that nevertheless has the property that the only rational numbers in
>Z[alpha] are the rational integers?
Doesn't (1+2i)/3 do the job ?
Antoine Chambert-Loir
--
Antoine
michael reid
> Does there exist an algebraic number alpha that is NOT an algebraic integer,
> but that nevertheless has the property that the only rational numbers in
> Z[alpha] are the rational integers?
alpha = 1 / (3 + sqrt(2)).
alpha is integral at P = (3 - sqrt(2)) , but not at P' = (3 + sqrt(2)).
it follows that Z[alpha] is integral everywhere except at P',
and thus Z[alpha] intersect Q is integral everywhere, including at 7.
mike
Gerry Myerson
(Antoine Chambert-Loir) wrote:
=> In article <73uuno$1...@pfaff.mit.edu>, Jim Propp <pr...@math.mit.edu> wrote:
=> >Does there exist an algebraic number alpha that is NOT an algebraic integer,
=> >but that nevertheless has the property that the only rational numbers in
=> >Z[alpha] are the rational integers?
=>
=> Doesn't (1+2i)/3 do the job ?
No. 3 (alpha)^2 - 2 alpha = -5/3.
Gerry Myerson (ge...@mpce.mq.edu.au)
Robert Israel
|> Now 2^t p - s is a polynomial with integer coefficients and alpha as a root,
|> so 2^t p - s = (2x^2 - 3x + 2) times q for some polynomial q in Q[x] of degree
|> n - 2.
|>
|> Now it seems unlikely to me that the form on the right could give rise to
|> a polynomial with an odd constant term but all other coefficients even.
|> Maybe some Gauss' lemma argument could finish this off.
It's simpler than that. The constant term in (2 x^2 - 3 x + 2) q is
2 times the constant term in q, and therefore is even.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Gerry Myerson
Israel) wrote:
= In article <gerry-01129...@abinitio.mpce.mq.edu.au>,
= ge...@mpce.mq.edu.au (Gerry Myerson) writes:
=
= |> Now 2^t p - s is a polynomial with integer coefficients and alpha as a
= |> root, so 2^t p - s = (2x^2 - 3x + 2) times q for some polynomial q in
= |> Q[x] of degree n - 2.
= |>
= |> Now it seems unlikely to me that the form on the right could give rise to
= |> a polynomial with an odd constant term but all other coefficients even.
= |> Maybe some Gauss' lemma argument could finish this off.
=
= It's simpler than that. The constant term in (2 x^2 - 3 x + 2) q is
= 2 times the constant term in q, and therefore is even.
No, q is in Q[x], not Z[x] (until proven otherwise).
(Posted & emailed)
Gerry Myerson (ge...@mpce.mq.edu.au)
Bill Dubuque
|
| ... Now pq in Z[x] for p = (2x^2 - 3x + 2) and some q in Q[x] ...
|
| Now it seems unlikely to me that the above product could give rise to
| a polynomial with an odd constant term but all other coefficients even.
pq in Z[x], primitive p in Z[x], q in Q[x] => q in Z[x] by Gauss' Lemma.
So, indeed, your pq has even constant term - completing your proof.
-Bill Dubuque
Timothy Chow
michael reid <re...@math.brown.edu> wrote:
>Jim Propp asks:
>
>> Does there exist an algebraic number alpha that is NOT an algebraic integer,
>
>
>alpha is integral at P = (3 - sqrt(2)) , but not at P' = (3 + sqrt(2)).
>it follows that Z[alpha] is integral everywhere except at P',
>and thus Z[alpha] intersect Q is integral everywhere, including at 7.
This is a nice argument that is worth explaining in more detail. Among
other things, it illustrates the power of localization.
The key point is that if v_p(x) < 0 for some x in Q and some rational
prime p (here v_p is the p-adic valuation), then v_P(x) < 0 for *every*
prime P lying over p.
Let F = Q(alpha) and let J be the ring of integers in F. We want to
have v_p(x) >= 0 for every x in Z[alpha] and every rational prime p.
We claim that to achieve this, it suffices to choose, for each rational
prime p, one prime P in J lying over p, and ensure that v_P(alpha) >= 0
for all these P. For then v_P(x) >= 0 for all x in Z[alpha], and the
above "key point" will force v_p(x) >= 0 should x happen to be in Q.
On the other hand, we don't want to make v_P(alpha) >= 0 for *all*
primes P in J, for then alpha would be an algebraic integer. So now
it's clear what we need to do: we need to choose a prime p that splits
in J and choose alpha such that v_P(alpha) < 0 for some prime lying
over p and v_P'(alpha) >= 0 for some other prime P' over the same p.
To find the simplest example, we take F = Q(sqrt(2)), which is a UFD,
and note that 7 splits into (3+sqrt(2))(3-sqrt(2)) in J. Then it is
easy to check that alpha = 1/(3+sqrt(2)) satisfies v_P(alpha) >= 0 for
all primes P in J except P = (3+sqrt(2))J, so it does the trick.
--
Tim Chow tchow AT alum DOT mit DOT edu
Where a calculator like the ENIAC today is equipped with 18,000 vacuum tubes
and weighs 30 tons, computers in the future may have only 1,000 vacuum tubes
and perhaps weigh only 1 1/2 tons. ---Popular Mechanics, March 1949, p.258