angle for gaussian primes ?

bhelmes · 2019-05-07, 20:16

A peaceful and pleasant evening,

is it possible to calculate an angle for gaussian primes ?

for example 5=(2+i)(2-i) alpha = arc tan (2/1)

Would be nice to get a link or a clear answer,

Greetings from the gaussian primes

Bernhard

paulunderwood · 2019-05-07, 20:41

Quote:

Originally Posted by bhelmes

A peaceful and pleasant evening,

is it possible to calculate an angle for gaussian primes ?

for example 5=(2+i)(2-i) alpha = arc tan (2/1)

Would be nice to get a link or a clear answer,

Greetings from the gaussian primes

Bernhard

See: http://mathworld.wolfram.com/ArgandDiagram.html

LaurV · 2019-05-09, 10:34

What's wrong with polar form?

Dr Sardonicus · 2019-05-09, 11:53

If R = Z[i], p is a prime number, p == 1 (mod 4), then pR = PP', the product of two conjugate prime ideals. If P = (a + b*i)R, it is easily shown that the argument of a + b*i is not a rational multiple of the number pi. (P^k is not a rational integer for any integer k other than 0.)

However, it is also easily shown that, if p₁, p₂, ..., p_k are k distinct prime numbers congruent to 1 (mod 4), P_j = (a_j + i*b_j)R is a prime divisor of p_jR, j = 1 to k, and

x_j = arg(a_j + i*b_j)/pi <-- the circle number,

then the x_j are Z-linearly independent -- a much stronger result. This result follows from unique factorization in R -- the product of integer powers of the P_j is not a rational number unless all the exponents are 0.

Why the above argument does not apply to the prime divisor of 2R is left as an exercise for the reader.

2019-05-07, 20:16	#1
bhelmes Mar 2016 3·5·23 Posts	angle for gaussian primes ? A peaceful and pleasant evening, is it possible to calculate an angle for gaussian primes ? for example 5=(2+i)(2-i) alpha = arc tan (2/1) Would be nice to get a link or a clear answer, Greetings from the gaussian primes Bernhard

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2019-05-09, 10:34	#3
LaurV Romulan Interpreter Jun 2011 Thailand 7×1,373 Posts	What's wrong with polar form?

2019-05-09, 11:53	#4
Dr Sardonicus Feb 2017 Nowhere 4,643 Posts	If R = Z[i], p is a prime number, p == 1 (mod 4), then pR = PP', the product of two conjugate prime ideals. If P = (a + bi)R, it is easily shown that the argument of a + bi is not a rational multiple of the number pi. (P^k is not a rational integer for any integer k other than 0.) However, it is also easily shown that, if p₁, p₂, ..., p_k are k distinct prime numbers congruent to 1 (mod 4), P_j = (a_j + ib_j)R is a prime divisor of p_jR, j = 1 to k, and x_j = arg(a_j + ib_j)/pi <-- the circle number, then the x_j are Z-linearly independent -- a much stronger result. This result follows from unique factorization in R -- the product of integer powers of the P_j is not a rational number unless all the exponents are 0. Why the above argument does not apply to the prime divisor of 2R is left as an exercise for the reader.