
Re: Units in algebraic integers
Posted:
Sep 27, 2004 12:03 AM


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 primesand therefore >coprime to each otherand are small, but let's replace them with 'a' >and b, nonunit 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

