\classheader{2013-02-18}

Last time, we started talking about completion and the $p$-adic fields.

\begin{definition}
Assume we are given sets $X_n$ and maps $f_n$, as follows:
\begin{center}
\begin{tikzcd}
\cdots \ar{r}{f_3} & X_3\ar{r}{f_2} & X_2 \ar{r}{f_1} & X_1
\end{tikzcd}
\end{center}
Then the projective (a.k.a. inverse) limit of this system, denoted $\projlim X_n$, is defined to be
\[\projlim X_n:=\{(x_n)_{n\in\N}\mid x_n\in X_n\text{ and }f_n(x_{n+1})=x_n\}.\]
\end{definition}
\begin{definition}
Let $A$ be a commutative ring, and let $I$ be an ideal of $A$. The $I$-adic completion of $A$, denoted $A_I$, is defined to be
\[A_I:=\projlim A/I^n,\]
where the map $f_n:A/I^{n+1}\to A/I^n$ is just the quotient map.
\end{definition}
\begin{example}
If $A=R[T]$ and $I=(T)$, then we have
\[A/I^n=R[T]/(T^n)=\{a_0+a_1T+\cdots+a_{n-1}T^{n-1}\mid a_i\in R\},\]
and thus
\[\projlim A/I^n=R[[T]]=\left\{\sum_{n=0}^\infty a_nT^n\;\middle\vert\; a_n\in R\right\},\]
the formal power series ring in the variable $T$. More generally, if $A=R[T_1,\ldots,T_n]$ and $I=(T_1,\ldots,T_n)$, then
\[\projlim A/I^n=R[[T_1,\ldots,T_n]]=\left\{\sum_{m_1,\ldots,m_n\geq 0} a_{m_1,\ldots,m_n}T_1^{m_1}\cdots T_n^{m_n}\;\middle\vert\; a_{m_1,\ldots,m_n}\in R\right\}.\]
If $A$ is a local ring and $\frak{m}$ is its maximal ideal, then $\projlim A/\frak{m}^n$ is called the completion of $A$. This is often denoted $\widehat{A}$.
\end{example}

Let $A$ be a commutative ring, and $\frak{p}\in\Spec(A)$. The idea of localization is to make things simple; completion makes things even more simple.
\begin{center}
\begin{tikzcd}[row sep=0in]
A \ar{r} & A_\frak{p} \ar{r} &\widehat{A}_\frak{p}\mathrlap{\,=\projlim A_\frak{p}/(\frak{p}A_\frak{p})^n}\\
& \text{simple} & \text{simpler}
\end{tikzcd}
\end{center}
If $\frak{p}\in\max(A)$, then $\widehat{A}_\frak{p}=\projlim A/\frak{p}^n$ because we have a canonical isomorphism $A/\frak{p}^n\cong A_\frak{p}/(\frak{p}A_\frak{p})^n$. This is because elements of $A\setminus\frak{p}$ are invertible in $A/\frak{p}^n$.

\begin{example}
Let $a\in\C$. Then
\begin{center}
\begin{tikzcd}[row sep=0.1in]
\C[T]_{(T-a)} \ar[white]{r}[description]{\text{\normalsize\color{black}{$\subset$}}} & \widehat{\C[T]}_{(T-a)}\\
\left\{\frac{f}{g}\;\middle\vert\;f,g\in\C[T],\,g(a)\neq 0\right\} & \C[[T-a]] \mathrlap{\;\;\;\text{``Taylor expansions at $a$''}}
\end{tikzcd}
\end{center}
\end{example}


In 1896, Hensel defined $\Z_p$ and $\Q_p$ where $p$ is a prime number.
\[\Z_p:=\projlim \Z/p^n\Z=\projlim\Z_{(p)}/p^n\Z_{(p)}.\]
The ring $\Z_p$ is an integral domain, and then we define
\[\Q_p:=\text{fraction field of }\Z_p.\]
Thus, we have $\Z\hookrightarrow\Z_{(p)}\hookrightarrow\Z_p$, and $\Q\hookrightarrow\Q_p$. In $\Q_p$, we have the following relations:
\[\Z_{(p)}=\Q\cap \Z_p,\qquad \Q_p=\Q+\Z_p,\qquad \Z=\Z[\tfrac{1}{p}]\cap \Z_{(p)},\qquad \Q_p=\Z[\tfrac{1}{p}]+\Z_p=\Z_p[\tfrac{1}{p}].\]

\subsection*{Picture of $\Z_3$}

\begin{center}
\begin{tikzpicture}[commutative diagrams/every diagram]
\node (a) at (0,4) {$\Z/3\Z$};
\node (b) at (6,4) {$\Z/9\Z$};
\node (c) at (12,4) {$\Z/27\Z$};
\path[commutative diagrams/.cd, every arrow, every label]
(b) edge (a)
(c) edge (b);
\node (ac) at ($(a)+(0,-3)$) {};
\node (ac1) at ($(ac)+(90:1.5)$) {};
\node (ac2) at ($(ac)+(210:1.5)$) {};
\node (ac3) at ($(ac)+(330:1.5)$) {};
\draw[thick] (ac1) circle (0.9);
\draw[thick] (ac2) circle (0.9);
\draw[thick] (ac3) circle (0.9);
\node (bc) at ($(b)+(0,-3)$) {};
\node (bc1) at ($(bc)+(90:1.5)$) {};
\node (bc2) at ($(bc)+(210:1.5)$) {};
\node (bc3) at ($(bc)+(330:1.5)$) {};
\draw[gray!60,thick] (bc1) circle (0.9);
\draw[gray!60,thick] (bc2) circle (0.9);
\draw[gray!60,thick] (bc3) circle (0.9);
\node (bc11) at ($(bc1)+(90:0.45)$) {};
\node (bc12) at ($(bc1)+(210:0.45)$) {};
\node (bc13) at ($(bc1)+(330:0.45)$) {};
\draw[thick] (bc11) circle (0.35);
\draw[thick] (bc12) circle (0.35);
\draw[thick] (bc13) circle (0.35);
\node (bc21) at ($(bc2)+(90:0.45)$) {};
\node (bc22) at ($(bc2)+(210:0.45)$) {};
\node (bc23) at ($(bc2)+(330:0.45)$) {};
\draw[thick] (bc21) circle (0.35);
\draw[thick] (bc22) circle (0.35);
\draw[thick] (bc23) circle (0.35);
\node (bc31) at ($(bc3)+(90:0.45)$) {};
\node (bc32) at ($(bc3)+(210:0.45)$) {};
\node (bc33) at ($(bc3)+(330:0.45)$) {};
\draw[thick] (bc31) circle (0.35);
\draw[thick] (bc32) circle (0.35);
\draw[thick] (bc33) circle (0.35);
\end{tikzpicture}
\end{center}

You can think of an element of $\Z_3$ as being a choice of one of the three petals at each stage. You can feel that you approach some point, some limit; this is the element of $\Z_3$.

This picture also shows us that
\[\Q_p=\coprod_{\substack{a\in\Z[\frac{1}{p}]\\0\leq a<1}}\Z_p+a=\bigcup_{n\geq 0}p^{-n}\Z_p.\]
\begin{center}
image
\end{center}
Moreover, $\Q_p/\Z_p\cong\Z[\frac{1}{p}]/\Z$.

The formal definition of the topology on $\Q_p$ is that $x_n$ converges to $a$ when, for any fixed $n\geq 0$, $x_\lambda-a\in p^n\Z_p$ for all $\lambda\gg 0$.

\begin{remark}
Arithmetic in $\R$ and in each $\Q_p$ is simpler than in $\Q$. We can understand $\Q$ be studying the problem in $\R$ and $\Q_p$ first, and then glueing these pieces of local information together. This is known as the Hasse principle.

For example, in $\Q$, although $\sqrt{1}=1$ exists in $\Q$, $\sqrt{1.1}$ and even $\sqrt{1.01}$ do not exist in $\Q$. In $\R$, $\sqrt{1.1}$ exists, and it is near to $\sqrt{1}$. In $\Q_5$, $\sqrt{1}$ exists, and $\sqrt{1-\frac{5}{4}}=\frac{\sqrt{-1}}{2}$ exists and is near to 1. We know that $\sqrt{-1}$ exists in $\Z_5$ because we can find it in each $\Z/5^n\Z$:
\begin{center}
\begin{tikzcd}[row sep=0.1in]
\Z/125\Z \ar{r} & \Z/25\Z \ar{r} & \Z/5\Z\\
57 \ar[mapsto]{r} & 7 \ar[mapsto]{r} & 2\\
68 \ar[mapsto]{r} & 18\ar[mapsto]{r} & 3
\end{tikzcd}
\end{center}
\end{remark}

If we drop money, we are usually very sad if the money is big. But for example, if we drop $3^{10}$ dollars, we can relax, because this is very small in the 3-adics. This world is dominated by the real numbers, though, not by the $p$-adics; we don't live in a $p$-adic world. That is strange. 

$\R$ is like the sun, and the $p$-adics are like the stars. The sun blocks out the stars during the day, and humans are asleep at night and don't see the stars, even though they are just as important.
\begin{center}
\begin{tikzpicture}
\node at (-2.5,4) {$\R$};
\node at (4.5,4) {daytime sky};
\fill [color=yellow!70] (0,4) circle (0.8);
\node[mypoint,gray,minimum size=3pt] at (-0.35,4.2) {};
\node[mypoint,gray,minimum size=3pt] at (0.35,4.2) {};
\draw[gray,thick] ($(0.35,4.25)+(45:0.14)$) arc (45:135:0.14 and 0.11);
\draw[gray,thick] ($(-0.35,4.25)+(45:0.14)$) arc (45:135:0.14 and 0.11);
\draw[gray,thick] ($(0,3.65)+(190:0.2)$) arc (190:350:0.2 and 0.2);
\foreach \t in {0,30,...,330} {
\draw[gray,thick] ($(0,4)+(\t:1)$) --++ (\t:0.5);
}
\foreach \i in {-2,-1,0,1} {
\foreach \j in {0,72,144,216,288} {
\draw[thick,blue!70,round cap-round cap] ($(\i,1)+(\j+90:0.2)$) -- ($(\i,1)+(\j+234:0.2)$);
}}
\node at  (2,1) {$\cdots$};
\node at (-2,0.5) {$\Q_2$};
\node at (-1,0.5) {$\Q_3$};
\node at (0,0.5) {$\Q_5$};
\node at (1,0.5) {$\Q_7$};
\node at (4.5,1) {nighttime sky};
\end{tikzpicture}
\end{center}