Residue field of p-adic integers

[pablis79]

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} : p\nmid a, p \nmid b \right\}, but can't seem to make the last step...

Or maybe I'm barking up the wrong tree. Hmm...

 

[micromass]

How did you define \mathbb{Z}_{(p)}?

 

[pablis79]

I defined \mathbb{Z}_{(p)} to be

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

 

[micromass]

That aren't the p-adic integers...

Anyway, you could prove that the quotient is a field that contains p elements. That shows it.

 

[morphism]

There's a really obvious map from Z_(p) onto Z/pZ whose kernel is pZ_(p). Hint: if p doesn't divide b, then b is a unit in Z/pZ.

 

[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 maps to the zero element in \mathbb{Z}/p\mathbb{Z}. So the kernel is p\mathbb{Z}_{(p)}, i.e. the set of all rationals in \mathbb{Z}_{(p)} such that p divides the numerator. I think one of the things I'm finding difficult is to understand how \mathbb{Z}_{(p)}/p\mathbb{Z}_{(p)} can equal \mathbb{Z}/p\mathbb{Z} (the set with p elements) since \mathbb{Z}_{(p)}/p\mathbb{Z}_{(p)} actually contains not p elements but a whole load of rationals such that p does not divide numerator or denominator. So how can we say they are equal when one contains fractions and the other p integers?

I'm beginning to think that \mathbb{Z}_{(p)}/p\mathbb{Z}_{(p)} = \mathbb{Z}/p\mathbb{Z} because if we take all elements in \mathbb{Z}_{(p)} and modulo (congruence?) them to p\mathbb{Z}_{(p)}, the very basic set of resulting elements that results is \mathbb{Z}/p\mathbb{Z}. Is this along the right/wrong lines?

 

[morphism]

Z_(p)/pZ_(p) contains only p elements. Think about Z/pZ: Z is infinite, but Z/pZ only has p elements in it.

I think your problem is stemming from the fact that you're trying to show that Z_(p)/pZ_(p) and Z/pZ are equal, when they're not (well, depending on your definition of Z/pZ). They're "isomorphic".