From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: 17th roots of 1
Date: 1 Jul 1998 20:37:59 GMT
In article , Lee Davidson wrote:
>Let R+ = sqrt(17+4sqrt(17)), R- = sqrt(17-4sqrt(17)). Then this
>expression occurs in the step from quartics to quadratics:
>
>sqrt( 1/4 * ( R+- +- 1 )( R+- +- sqrt(17) ) )
>
>+- signs depend on the roots you're looking at.
>
>I tried looking for some way to write the expression within the outer
>sqrt as a perfect square, but couldn't find any. Is this the simplest
>way of writing this? It's pretty simple already, actually, considering.
You want to know whether the numbers
( 1/4 * ( R+- +- 1 )( R+- +- sqrt(17) ) )
are perfect squares -- but in what ring did you mean this?
Note that ((R+-)^2 - 17)/(+-4) = sqrt(17)
so that sqrt(17) already lies in the rings Q[ R+ ] and Q[ R- ]. Moreover,
(R+)(R-) = sqrt(17^2 - 4^2*17) = sqrt(17), so that each of these rings
contain the generator of the other, that is Q[ R+ ] = Q[ R- ].
Now you are asking whether certain elements of this ring are squares.
I take it you mean, are they the squares of other elements in this ring.
Well, if x = y^2 then norm(x)= norm(y)^2, but if I've calculated
correctly, these elements x have norm 17, not a square. So x is
not a square in this ring.
(You can calculate the norm in stages: it's norm = norm1 o norm2 where
these new functions are the norms to and from the intermediate field
Q[sqrt(17)]. Since each is a quadratic extension, each norm(X) is computed
by multiplying X * X', where X' is obtained from X by changing
the sign of the generator R+ or sqrt(17), respectively. In maple:
( 1/4 * ( Ra + ep1 )*( Rb + ep2* sqrt(17) ) );
"*subs({Ra=-Ra,Rb=-Rb},");
expand(");
simplify(",{ep1^2=1,ep2^2=1});
2 2
1/16 (Ra - 1) (Rb - 17)
simplify(",{Ra^2=17+4*ep1*sqrt(17),Rb^2=17-4*ep2*sqrt(17)});
1/2
-17 ep1 ep2 - 4 ep2 17
#(That's now norm2(x) lying in the intermediate field. Now repeat.)
"*subs(sqrt(17)=-sqrt(17),");
expand(");
simplify(",{ep1^2=1,ep2^2=1});
17
dave