# Residue field of p-adic integers

1. Jan 13, 2012

### 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...

Last edited: Jan 13, 2012
2. Jan 13, 2012

### micromass

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

3. Jan 13, 2012

### pablis79

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

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

4. Jan 13, 2012

### micromass

Staff Emeritus

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

5. Jan 13, 2012

### 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.

6. Jan 14, 2012

### 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?

7. Jan 15, 2012

### 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".