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Topic: Units in algebraic integers
Replies: 18   Last Post: Sep 29, 2004 9:12 AM

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Andrzej Kolowski

Posts: 353
Registered: 12/6/04
Re: Units in algebraic integers
Posted: Sep 27, 2004 7:45 PM
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akolowski@hotmail.com (Andrzej Kolowski) wrote in message news:<200409270403.i8R43fY08660@proapp.mathforum.org>...
> On 26 Sep 2004, James Harris wrote:
> >I've given the example of
> >
> >x^2 + (2s_1 + 3s_2)x + s_1 s_2 6 = (x + 2s_1))(x + 3s_2)
> >
> >where s_1 s_2 = 1, and 2s_1 + 3s_2 is an integer
> >
> >to highlight the oddity that if s_1 and s_2 are rational, then
> >trivially you can have both be units as both can equal 1, but if s_1
> >and s_2 are irrational, then they cannot be units in the ring of
> >algebraic integers.
> >
> >It gets weirder.
> >
> >I picked 2 and 3 because they're the first two primes--and therefore
> >coprime to each other--and are small, but let's replace them with 'a'
> >and b, non-unit coprime algebraic integers:
> >
> >
> >x^2 + (as_1 + bs_2)x + s_1 s_2 ab = (x + as_1))(x + bs_2)
> >
> >where s_1 s_2 = 1, as_1 + bs_2 is an integer, and ab is an integer.
> >
> >Question: Can s_1 and s_2 be irrational and algebraic integers?
> >
>
> Consider P(x) = x^2 + 55*x + 110.
>
> The roots are r1, r2 = (-55 +/- sqrt(3025 - 12100))/ 2.
>
> = 55*(-1 +/- sqrt(-3))/2.
>
> Letting a = 5 and b = 11, we have
>
> 5*s1 = -55*(-1 + sqrt(-3))/2 and
>
> 11*s2 = -55*(-1 - sqrt(-3))/2.
>
> That is,
>
> s1 = -11*(-1 + sqrt(-3))/2 and
>
> s2 = -5*(-1 - sqrt(-3))/2.
>
> Finally, note that (-1 +/- sqrt(-3))/2 are algebraic
> integers.
>
> Therefore s1 and s2 are irrational and are algebraic
> integers.
>
> But so what? Did you think this couldn't happen??
>
> Andrzej
>

I have to admit as James Harris pointed out that I
overlooked the requirement that s1 * s2 = 1, both in
this note and the subsequent one in which

P(x) = x^2 + 100*x + 100.

I would *guess* with that additional restriction,
there are not going to be any examples. But again,
so what? Why should there be? Is every equation
supposed to have a solution in algebraic integers?

Incidentally, the current version of APF still has
an error in the last section. There should be an
f^2 in the third part of the coefficient of x^3.

Andrzej

> >
> >James Harris




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