
Re: Units in algebraic integers
Posted:
Sep 27, 2004 9:22 AM


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 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?? >
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

