Adjoining the Sixth Roots of 1

Complex Extensions, Adjoining the Sixth Roots of 1

Adjoining the Sixth Roots of 1

For notational convenience, let q be the square root of -3.  In the world of complex numbers, the sixth root of 1 is (1+q)/2.  Adjoin this to Z and see what happens.

The adjoined element is not a pure imaginary number, hence the vector that is based at the origin is not perpendicular to the x axis.  In other words, the cells of the lattice are no longer rectangles.  They are pushed over rectangles, or parallelograms.  It is tempting to cut each parallelogram in half, making an infinite lattice of perfect equilateral triangles, but that disguises some of the algebra.  There are really two vectors, 1 and (1+q)/2, and they span a lattice of parallelograms, just like any other pair of independent vectors in the plane.

The norm, a2+3b2, is still a valid multiplicative map into the positive integers, but this time a and b can be half integers.  Watch the lattice though; they are either both half integers or they are both whole integers.  Verify that the norm is still an integer, even if a and b are half integers.  Then show there are six different units, namely the six roots of 1 in the complex plane.

Next consider the gcd algorithm.  Let the shorter vector establish a lattice of parallelograms, (in this case rhombuses), find the cell that contains the longer vector, and measure its distance to the four corners.  The distance is always less than the side of the rhombus.  The gcd algorithm runs to completion, and we have a unique factorization domain (ufd).

If q is the square root of -7 or -11, and we adjoin (1+q)/2, the parallelograms grow taller, but the distance between the center and the nearest corner is still less than the short side of the parallelogram.  The gcd algorithm works, and these are all ufds.

Associates

Every prime, and every ring element for that matter, has an associate that does not require half integer coefficients.  If x has coefficients a/2 and b/2, ask whether a = b mod 4.  If they are equal, multiply x by (1-q)/2; otherwise multiply x by (1+q)/2.  The result uses whole coefficients.

Conversely, any prime other than 2 has associates with half integer coefficients.  The norm is odd, so a and b have opposite parity.  Multiply by (1ħq)/2 and you get half integers.  The exception is 2, with norm 4, and associates 1ħq.