\classheader{2013-02-20}


When we complete the ring $\C[T]$ at the prime ideal $(T)$, we saw last time that we get the ring $\C[[T]]$, the formal power series in the variable $T$. Its fraction field is denoted $\laurentseries{\C}{T}$, and elements of $\laurentseries{\C}{T}$ are called Laurent series in $T$:
\[\laurentseries{\C}{T}=\left\{\sum_{n=n_0}^\infty a_nT^n\;\middle\vert\;a_i\in\C\text{ for some }n_o\in\Z\right\}.\]
We can also form the fields $\laurentseries{\C}{T-1}$, $\laurentseries{\C}{T-2}$, $\ldots$ but there is one more field, which is $\laurentseries{\C}{\frac{1}{T}}$. We say that this is the Laurent expansion at $\infty$. The field $\C(T)$ is like $\Q$, the fields $\laurentseries{\C}{T}$, $\laurentseries{\C}{T-1}$, etc. are like the $p$-adic fields $\Q_2$, $\Q_3$, and $\R$ is like the expansion at $\infty$. This analogy is very mysterious.
\begin{center}
\begin{tikzpicture}[scale=2]
\node at (0,0) {$\C(T)$};
\node at (90:1) {$\laurentseries{\C}{\frac{1}{T}}$};
\node at (0:1) {$\laurentseries{\C}{T}$};
\node at (320:1) {$\;\;\;\;\;\;\laurentseries{\C}{T-1}$};
\node at (270:1) {$\laurentseries{\C}{T-2}$};
\node[rotate=90] at (90:0.5) {$\subset$};
\node[rotate=0] at (0:0.5) {$\subset$};
\node[rotate=315] at (315:0.5) {$\subset$};
\node[rotate=270] at (270:0.5) {$\subset$};
\node[rotate=315] at (215:1) {$\cdots$};
\end{tikzpicture}\qquad\qquad\qquad\begin{tikzpicture}[scale=2]
\node at (0,0) {$\Q$};
\node at (90:1) {$\R$};
\node at (0:1) {$\Q_2$};
\node at (315:1) {$\Q_3$};
\node at (270:1) {$\Q_5$};
\node[rotate=90] at (90:0.5) {$\subset$};
\node[rotate=0] at (0:0.5) {$\subset$};
\node[rotate=315] at (315:0.5) {$\subset$};
\node[rotate=270] at (270:0.5) {$\subset$};
\node[rotate=315] at (215:1) {$\cdots$};
\end{tikzpicture}
\end{center}

The Greeks wanted to understand the world using mathematics. But once they realized $\Q$ and $\R$ are different, it was no longer so clear to them how to do that. The student who told others about the existence of irrational numbers was killed by the gods, or perhaps just thrown out of the boat by the other students.

Nowadays, we worry if the weight of our body becomes too small or too large, but we don't worry about whether it is a rational number or an irrational number.

Why does $\Q$ want to grow to $\R$ or $\Q_2$ or $\Q_3$? Its heart has holes, for example at $\sqrt{2}$ and $\sqrt{3}$. This is similar to mankind; we can grow to be big boys or big girls, but there is still some sadness in our hearts, and we grow to love another person.

\begin{theorem}[Hensel's lemma]
Let $A$ be a commutative ring, and let $I$ be an ideal of $A$. Let $f\in A[T]$, and assume that $a\in A$ has the property that $f(a)=0\bmod I$, and that $f'(a)\bmod I$ is invertible in $A/I$. Then there exists a unique $b\in \widehat{A}:=\projlim A/I^n$ such that $f(b)=0$ and $b\equiv a\bmod I$. In other words, there exist unique $b_n\in A/I^n$ such that $f(b_n)\equiv 0\bmod I^n$ and $b_n\equiv a\bmod I$, and then we have $b=(b_n)_{n\geq 1}$.
\end{theorem}
\begin{proof}
We proceed by induction on $n\geq 1$. Assume the unique existence of $b_n\in A/I^n$. Fix a choice of $\widetilde{b_n}\in A/I^{n+1}$, a lifting of $b_n$. Then by the general fact that
\[g(T+x)\equiv g(T)+g'(T)x\bmod x^2\]
for any polynomial $g$, we have that
\[f(\widetilde{b_n}+x)=f(\widetilde{b_n})+f'(\widetilde{b_n})x+\underbrace{(\text{a multiple of }x^2)}_{\in I^{2n}\subset I^{n+1}}.\]
Because $f'(\widetilde{b_n})\in (A/I^{n+1})^\times$, there is a unique choice of $x$, namely
\[x=-f'(\widetilde{b_n})^{-1}f(\widetilde{b_n}),\]
such that $f(\widetilde{b_n}+x)\equiv 0\bmod I^{n+1}$. We then set $b_{n+1}=\widetilde{b_n}-f'(\widetilde{b_n})^{-1}f(\widetilde{b_n})$, which is the unique good choice in $A/I^{n+1}$. Note that we also have that $b_{n+1}\equiv b_n\bmod I^n$, because $f(\widetilde{b_n})\in I^n/I^{n+1}$.
\end{proof}

For example, suppose that we want to find $\sqrt{-1}$ in $\Z_5$. We know that we can choose $\sqrt{-1}=2$ in $\Z/5\Z$ (the other choice is 3). We hope to find the $y$ such that
\[(2+5y)^2\equiv -1\bmod 25,\]
and this $y$ is the $x$ in the proof. Solving, we get that $y\equiv 1\bmod 5$, and we note that $2+5\cdot 1=7$ has the property that $7\equiv 2\bmod 5$ and $7^2\equiv -1\bmod 25$.

Here is another application. If $p$ is an odd prime and $m\in\Z$ is an integer with $p\nmid m$, then we can show that if $\legendre{m}{p}=1$, then $x^2=m$ has a solution $\Q_p$ (which is in fact in $\Z_p$; one way to see this is that $\Z_p$ is a PID, hence normal). We can see this by applying Hensel's lemma to $f(T)=T^2-m$; there is an $a\in\Z$ such that $a^2-m\equiv 0\bmod p$, and $f'(a)=2a$ which is invertible in $\Z/p\Z$.

We can use Hensel's lemma and the $p$-adics to better understand number theory.

It may happen that tomorrow, when you wake up, taking three steps returns you close to your starting point, and taking nine steps returns you even closer.