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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

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

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

&gt; &gt;
&gt; &gt;James Harris

Date Subject Author
9/26/04 JAMES HARRIS
9/26/04 W. Dale Hall
9/26/04 magidin@math.berkeley.edu
9/26/04 W. Dale Hall
9/26/04 Bill Dubuque
9/26/04 W. Dale Hall
9/27/04 Andrzej Kolowski
9/27/04 Count Dracula
9/27/04 Dik T. Winter
9/28/04 Count Dracula
9/28/04 Dik T. Winter
9/29/04 Jesse F. Hughes
9/29/04 Dik T. Winter
9/29/04 Jesse F. Hughes
9/27/04 JAMES HARRIS
9/27/04 David Hartley
9/27/04 David Hartley
9/27/04 Andrzej Kolowski
9/27/04 Andrzej Kolowski