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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 12:03 AM
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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

>
>James Harris




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