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Re: Units in algebraic integers
Posted:
Sep 27, 2004 9:22 AM
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On 27 Sep 2004, Andrzej Kolowski wrote:
>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??
>
Please ignore above. I did a wrong calculation. Here
is (I think) a valid example:
Let P(x) = x^2 + 100*x + 100.
Roots are -50 +/- 20*sqrt(6).
Let a = 2, b = 5. Then
s1 = 25 - 10*sqrt(6) and
s2 = 10 + 4*sqrt(6).
Both irrational, both algebraic integers.
Again: why were you asking?
Andrzej
> Andrzej
>
>>
>>James Harris
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