Recent content by pablis79

Thanks morphism. I'm not particularly up on group/ring theory etc. However, I am learning! From what I understand from your response, we seek a map from \mathbb{Z}_{(p)} to \mathbb{Z}/p\mathbb{Z} = \left\{0,1,\ldots,p-1\right\}. By kernel I think you mean the subset of \mathbb{Z}_{(p)} that...

I defined \mathbb{Z}_{(p)} to be \mathbb{Z}_{(p)} = \left\{\frac{a}{b}\in\mathbb{Q}:p\nmid b\right\}.

In the field of rationals \mathbb{Z}_{(p)} (rationals in the ring of the p-adic integers), how is it possible to prove the residue field \mathbb{Z}_{(p)}/p\mathbb{Z}_{(p)} is equal to \mathbb{Z}/p\mathbb{Z} ? I've narrowed it down to \mathbb{Z}_{(p)}/p\mathbb{Z}_{(p)} = \left\{ a/b\in\mathbb{Q}...

The sequence you give has the Taylor series coefficient function 4/3Sin(Pi*n/3)^2, which may simplify further given n is an integer. If you want a general form for the coefficients of a generating function, you can use an contour integral from complex analysis which extracts the coefficient...

Ok, thanks micromass. So put another way it means that x and y are congruent modulo p in Zp. Cheers!

Hi all, If Zp is the ring of p-adic integers, what does the notation a = b (mod pZp) mean ? I understand congruence in Zp, i.e., a = b (mod p) implies a = b +zp, where z is in Zp (and a, b in Zp). However, I don't get what is meant by (mod pZp) ... does this mean a = b (mod p^k) for all k >= 1...