P-adic numbers
Define $v_p(x)$ to be the largest power of $p$ that divides $x$. Define $v_p(0)=\infty$. Clearly $$ v_p(xy)=v_p(x)+v_p(y),\quad v_p(x+y)\geq\min{v_p(x),v_p(y)} $$
Definition
Analytic definition
Definition. The completion of $\mathbb{Q}$ with respect to $|\cdot|p=s^{-v_p(x)}$, where $s\in\mathbb{R}{> 1}$ is the field of p-adic numbers, denoted $\mathbb{Q}_p$.
Definition. Let $\mathbb{Z}_p\subset\mathbb{Q}_p$ denote the valuation ring $$ {x\in\mathbb{Q}_p\mid|x|\leq 1}. $$ It is called the ring of integer p-adics.
Algebraic definition
For $i>1$, define the natural map $\phi:\mathbb{Z}{p^i}\to \mathbb{Z}{p^{i-1}}$.
Definition. The ring of integer p-adics $\mathbb{Z}p$ is the limit (in the category of unital rings) $$ \lim\leftarrow (\mathbb{Z}_{p^i},\phi_i), $$ i.e. $\mathbb{Z}_p$ is a sequence $$ {{}ix\in\mathbb{Z}{p^i}\mid 1\leq i\leq\infty} $$ such that for all $i>1$ we have $\phi({}ix)={}{i-1}x$.
Consider the ${}ix$ as fitting the following diagram: $$ \cdots\longrightarrow\mathbb{Z}{p^3}\overset{\phi_3}{\longrightarrow}\mathbb{Z}_{p^2}\overset{\phi_2}{\longrightarrow}\mathbb{Z}_p $$
In this way we see that for $p=5$, a number 31 would be represented as a sequence $(31,\dots,31,6,1)$.
If ${}ix$ is determined, then there are $p$ possibilities of what ${}{i+1}x$ could be, corresponding to the $p$ residue classes a number modulo $p^i$ has modulo $p^{i+1}$.
Remark. Consider the absolute value $|x-y|=s^{-v_p(x-y)}$ described above. Intuitively, $x$ and $y$ are close if their difference is close to $0$. Since a coordinate of $x-y$ being 0 means automatically that all lower coordinates are 0, we see that $x-y$ become closer if they share the first $n$ coordinates, where $n$ is large. But this means that $v_p(x-y)$ gets bigger and bigger. So for this intuitive closeness to be reflected in the absolute value, we want to raise $s$ to the negative power, so that a larger $v_p(x-y)$ results in a smaller distance between $x$ and $y$.
Definition. Let the field of $p$-adics, denoted $\mathbb{Q}_p$, be the field of fractions of $\mathbb{Z}_p$.
Let $\mathbb{Q}_p$ be the field of fractions of $\mathbb{Z}_p$. We can extend $v_p$ to $\mathbb{Q}_p\setminus{0}$ by $v_p(z=\frac{x}{y})=v_p(x)-v_p(y)$. Then
- $v_p(z)$ does not depend on the choice of $x,y\in\mathbb{Z}_p$
- $v_p:\mathbb{Q}_p^\ast\to\mathbb{Z}$ is a group homomorphism.
- This extension defines a nonarchimedean absolute value on $\mathbb{Q}_p$ as it did above.
Equivalence of definitions
TODO
Group of units
Definition. Let $U=\mathbb{Z}_p^\ast$ denote the group of $p$-adic units. For every $n\geq 1$, write $U_n=1+p^n\mathbb{Z}_p$.
The group of units $U$ completely describes the multiplicative qualities of $\mathbb{Z}_p$, see for example (1) in properties of units below.
Properties
Of integer p-adics
- $\mathbb{Z}_p$ carries a natural structure of a ring with identity.
- It is an Integral domain
- The map $\mathbb{Z}\to\mathbb{Z}_p$ is injective.
- Multiplication by $p$ (hence also $p^n$) is an isomorphism from $\mathbb{Z}_p$ to itself.
- Let $\epsilon_i$ be the projection $\mathbb{Z}p\to\mathbb{Z}{p^i}$ to the $i$th component. The following sequence is exact: $$0\longrightarrow\mathbb{Z}_p\overset{(-)\cdot p^i}{\longrightarrow}\mathbb{Z}_p\overset{\epsilon_i}{\longrightarrow}\mathbb{Z}/(p^i)\longrightarrow 0.$$
- TFAE:
- $u\in\mathbb{Z}_p$ is invertible
- ${}_1u\neq 0$ in $\mathbb{Z}_p$
- $p\nmid u$ in $\mathbb{Z}_p$.
- Every nonzero element $z\in\mathbb{Z}_p$ can be uniquely represnted as an invertible element and a power of $p$; i.e. there exists a unique $u\in\mathbb{Z}p^\ast$ and $n\in\mathbb{Z}{\geq 0}$ such that $z=p^nu$.
- The $n$ above is equal to $v_p(z)$.
- We denote the group of invertible elements of $\mathbb{Z}_p$ by $U$.
- $\mathbb{Z}$ is dense in $\mathbb{Z}_p$.
- Hensel’s lemma a (simple!) root modulo $p$ can be lifted to a root modulo any higher power of $p$. By taking the limit, it follows that the root modulo $p$ can be lifted to a root over the $p$-adic integers.
- Suppose $\alpha\in\mathbb{Z}p$ and $\alpha\equiv 1\mod p$. The function \begin{gather} \exp\alpha:\mathbb{Z}_p\to\mathbb{Z}_p \ z=\alpha^z={{}iz\in\mathbb{Z}/(p^i)}\mapsto \lim{i\to\infty}{\alpha^{\tilde{{}_iz}}} \end{gather} where $\tilde{{}iz}\in\mathbb{Z}{\geq 0}$ is equivalent to ${}_iz\mod p^i$ defines a group homomorphism from the additive group of $\mathbb{Z}_p$ to the subgroup $U_1$ of its group of units $U$.
Proofs
1
Suppose $x,y\in\mathbb{Z}_p$ are both nonzero. By part (5), we can write $xy=p^mu_1p^nu_2$. Suppose $xy=0$. Since the $u_i$ are invertible, this would mean $p^{m+n}=0$, which is impossible.
3
Suppose $px=0$. We need to show that $x=0$. To that end, consider that $px=0$ implies $px_{n+1}\equiv 0\mod p^{n+1}$ for all $n$. Then $px_{n+1}=p^{n+1}y_{n+1}$ for some $y_{n+1}\in \mathbb{Z}/(p^{n+1})$. Dividing both sides by $p$, we can write $x=p^ny_{n+1}$. Under the natural homomorphism $\phi_{n+1}:\mathbb{Z}/(p^{n+1})\to\mathbb{Z}/(p^n)$, clearly $\phi(x)=0$ since it has $p^n$ as a factor. Thus $x_n$ is zero (for all $n$).
The corresponding claim for multiplication by $p^n$ follows because it is the $n$-fold composition of maps that are multiplication by $p$, which are injective.
4
By (3), multiplication by $p^i$ is an isomorphism. Clearly $\epsilon_i$ is surjective. So it suffices to show that $\ker\epsilon_i=p^n\mathbb{Z}_p$.
Clearly $p^i\mathbb{Z}_p\subset\text{ker}\epsilon_i$ since the $i$th coordinate is multiplied by $p^i$ so it vanishes. Conversely, let $x\in\ker\epsilon_i$. Since the $i$th coordinate vanishes, all lower order coordinates vanish as well. So just define $y\in\mathbb{Z}p$ to be the renumbering of the coordinates of $x$ starting with $x{i+1}=y_1$. Then clearly $x=p^i\cdot y$.
5
The only nontrivial thing is to show that 3 implies 1. Suppose $p\nmid x$. To show that $x$ is invertible, it suffices to show that each component is invertible. Consider $x_m$. Its image in $\mathbb{Z}/(p)$ is invertible hence there exists $y,z\in\mathbb{Z}/(p^m)$ such that $xy+pz=1$. We claim then that $x_m$ is invertible in $\mathbb{Z}/(p^m)$:
\begin{align}
x(y+ypz+\cdots+yp^{n-1}z^{n-1}) =& xy(1+pz+\cdots+p^{n-1}z^{n-1}) \
=& (1-pz)(1+pz+\cdots+p^{n-1}z^{n-1}) \
=& 1.
\end{align}
6
Similar to our discussion in (4). If $x\in\mathbb{Z}p$ is nonzero, then there exists a largest integer $n$ such that $x_n=0$. Take $u$ to be the renumbering of $x$ starting with $u_1=x{n+1}$. All its components are nonzero hence $u$ is invertible.
7
We need to show that every $x\in\mathbb{Z}_p$ is the limit of a sequence $(y_n)$ where $y_n\in\mathbb{Z}$. Just take $y_n\equiv x_n\mod p^n$.
Of p-adics
Of units
- $\epsilon_n\mid_U:U\to(\mathbb{Z}/(p^n))^\ast$ is a surjective group homomorphism with kernel $U_n$.
- $U/U_1\cong\mathbb{F}_p^\ast$, hence is cyclic of order $p-1$.
- By property (5) of integer p-adics, $\mathbb{Q}_p^\ast=p^\mathbb{Z}\times U$ where $U$ is the set of units.
- Let $U_n=1+p^n\mathbb{Z}_p\subset U$. Then $U_n$ is a subgroup and $$\epsilon_n\mid_U:U\to(\mathbb{Z}/(p^n))^\ast$$ from property (3) of integer p-adics is a surjective group homomorphism, and its kernel is $U_n$.
- (Teichmuller decomposition) $U=T\times U_1$, where $T={x\in U\mid x^{p-1}=1}$ is a cyclic group of $p-1$ elements.
Proofs
1
The unrestricted $\epsilon_n$ is already shown to be a surjective ring homomorphism by (4) of integer p-adics. So in particular it sends invertible elements in $\mathbb{Z}_p$ (and therefore invertible elements in $U$) to invertible elements in $(\mathbb{Z}/(p^n))^\ast$. It remains to show the other direction, that if $\epsilon_n(x)$ for $x\in\mathbb{Z}_p$ then $x$ is invertible, hence $x\in U$. An element in $(\mathbb{Z}/(p^n))^\ast$ is invertible if and only if it is relatively prime to $p^n$. So $x$ is invertible if and only if $p\nmid\epsilon_n(x)$. By part (5) of integer p-adics, this shows that $x$ is invertible.
$U_1$ is the kernel of $\epsilon_1:U\to(\mathbb{Z}/p\mathbb{Z})^\ast$. Since $\epsilon_1$ is clearly surjective, $U/U_1\cong(\mathbb{Z}/p\mathbb{Z})^\ast\cong\mathbb{F}_p^\ast$.
Product formula
Theorem. Consider the special set of absolute values on $\mathbb{Q}$
- $|\cdot|_\infty=|\cdot|$
- $|\cdot|p=p^{-v_p(\cdot)}$. Suppose $x\in\mathbb{Q}$. Then $$ |x|\infty\prod_p|x|_p=1. $$