Quadratic Integers & Associates?
Describe the set of quadratic integers α in Q[√-3] so that α and the conjugate of α are associates.
Ummm... in the book, it says that α is an associate of β iff α/β is a unit.
Help?
Describe the set of quadratic integers α in Q[√-3] so that α and the conjugate of α are associates.
Ummm... in the book, it says that α is an associate of β iff α/β is a unit.
Help?
Comments
Determine the units of Z[ (-1+root(-3))/2 ]. Let a + b(-1+root(-3))/2 be a unit. Taking norms a^2 -ab + b^2 =1, so that 4a^2 - 4ab + 4b^2 = 4, and finally (2a-b)^2 = 4-3b^2. The possible values for b are 0, +/-1. If b = 0 then a = +/-1, if b = 1 then a = 0 or 1, and if b = -1 then a = 0 or -1. The possible units are +/-1, +/- (-1+root(-3))/2, and +/- (-1 - root(-3))/2. The units are all associates, and are generated by {+/- 1}*{ (-1-root(-3))/2}^n for n = 0, 1, 2.
Suppose (a + b(1+root(-3))/2)*unit = a + b(1-root(-3))/2, then by above it suffices to consider the 2 cases unit = -1 and (-1-root(-3))/2
i) unit = -1 implies a = b = 0.
ii) unit = (1-root(-3))/2 implies a(1-root(-3))/2 + b = a + b(1-root(-3))/2, so that a = b, and the numbers are of the form a(3+root(-3))/2.
So the set of quadratic integers that are associates to their conjugates are of the form a*{unit}*{(3 + root(-3))/2}^n where a is in Z and n = 0 or 1.
Another way to show this is as follows. If A and its conjugate are associates then so are any primes dividing A. Otherwise the conjugate of a prime would equal another prime times a unit - impossible for in a UFD since both a prime and it's conjugate divide the same rational prime. So it suffices to work at the prime level in Z[ (-1+root(-3))/2] and find primes in Z that factor as squares time units in Z[ (-1+root(-3))/2 ]. Let p be such a prime. The factorization of the polynomial f(X) = X^2 + X + 1 mod p determines how p splits (factors). The splitting of f (X) is determined by the splitting of 4X^2 + 4X + 4 = (2X+1)^2 + 3 mod p. A square factorization means f(X) is a sqaure mod p. If 4 f(X) factors as (2X+2a)^2 mod p then we see that 3 = 0 mod p, so that only the primes dividing 3 are associates to their conjugates. This is exactly what was determined above, but it's the answer you would give in an algebraic number theory course.
Just as your prior question could have been simplified, so was this one. There I could have simply noted that a + b(-1+root(-3))/2 = a -2b +b(3+root(-3))/2 = a-2b mod (3+root(-3))/2, so that all conguence classes are in Z. Then followed the argument that Norms would force any two rational integers to be congruent mod 3 when in the same class, and deduced that 0, 1 and 2 were the only 3 classes.
I'm not sure what the level of your class is, but my guess is that it's aiming for a slightly more general than specific approach. If you're confused about anything I write please post a comment or email and I'll try my best to clear it up.
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