p-adic分析note| Hensel’s Analogy

p-adic分析

Hensel’s Analogy

The $p$-adic numbers were first introduced by the German mathematician Kurt Hensel. Hensel’s starting point was the analogy between the ring of integers $\mathbb{Z}$, together with its field of fractions $\mathbb{Q}$, and the ring $\mathbb{C}[X]$ of polynomials with complex coefficients, together with its field of fractions $\mathbb{C}(X)$. He learned the analogy from his doctoral adviser, Leopold Kronecker, who even attempted to develop a single theory that covered both cases.

To be specific, let’s use $X$ as an indeterminate, saving $x$ to stand for a number. An element of $f(X) \in \mathbb{C}(X)$ is a “rational function,” i.e., a quotient of two polynomials:
$$f(X)=\frac{P(X)}{Q(X)},$$
with $P(X), Q(X) \in \mathbb{C}[X], Q(X) \neq 0$; we can always require that $Q(X)$ is monic, i.e., its leading coefficient is 1. Similarly, any rational number $x \in \mathbb{Q}$ is a quotient of two integers:
$$x=\frac{a}{b},$$
with $a, b \in \mathbb{Z}, b \neq 0$; we can always require that $b>0$. Furthermore, the properties of the two rings are quite similar: both $\mathbb{Z}$ and $\mathbb{C}[X]$ are rings where

there is unique factorization: any integer can be expressed uniquely as $\pm 1$ times a product of primes, and any polynomial can be expressed uniquely as
$$P(X)=a\left(X-\alpha_{1}\right)\left(X-\alpha_{2}\right) \ldots\left(X-\alpha_{n}\right),$$
where $a$ and $\alpha_{1}, \alpha_{2}, \ldots \alpha_{n}$ are complex numbers. This gives us the main point of the analogy Hensel explored: The primes $p \in \mathbb{Z}$ are analogous to the linear polynomials $X-\alpha \in \mathbb{C}[X]$.

The analogy extends to solutions of equations. Given a polynomial with coefficients in $\mathbb{Z}$, any root is an algebraic number; if a function is a root of a polynomial with coefficients in $\mathbb{C}[X]$, it is an algebraic function. So $\sqrt{2}$, which is a root of $Y^{2}-2$, is an algebraic number, while $f(X)=\sqrt{X^{3}-3 X+1}$, which is a root of $Y^{2}-\left(X^{3}-3 X+1\right)$, is an algebraic function.

Hensel was studying a specific problem about algebraic numbers. Pursuing the analogy, he considered the identical problem in the context of algebraic functions; that problem turned out to be easy to solve, because he could expand the algebraic functions into power series.

Suppose we are given a polynomial $P(X) \in \mathbb{C}[X]$ and a particular $\alpha \in \mathbb{C}$. Then it is possible (for example, using a Taylor expansion) to write the polynomial in the form
\begin{aligned} P(X) &=a_{0}+a_{1}(X-\alpha)+a_{2}(X-\alpha)^{2}+\cdots+a_{n}(X-\alpha)^{n} \ &=\sum_{i=0}^{n} a_{i}(X-\alpha)^{i} \end{aligned}
with $a_{i} \in \mathbb{C}$. This gives very precise information on how the polynomial behaves near $\alpha$.

Can we do something like this for integers?

Can we do something like this for integers? For positive integers, we can, and indeed we do it every day when we write them down:
$$321=1+2 \times 10+3 \times 10^{2}$$
is in that form. The annoying thing is that 10 is not a prime, while $(X-\alpha)$ is a prime in $\mathbb{C}[X]$. But we can fix that: choose a prime number $p$ and write our number in base $p$ : given a positive integer $m$, we can write it in the form
$$m=a_{0}+a_{1} p+a_{2} p^{2}+\cdots+a_{n} p^{n}=\sum_{i=0}^{n} a_{i} p^{i}$$
with $a_{i} \in \mathbb{Z}$ and $0 \leq a_{i} \leq p-1$. For example, if $p=7$ we can write
$$320=5+3 \times 7+6 \times 7^{2} \text {. }$$
We can even record this as a string of digits “635” (as in base ten, we record the digits backwards: lowest powers ${ }^{2}$ come last). To keep the distinction

between standard (base ten) notation and base $p$, let’s use red for the latter, so $320=635$ as long as it’s understood that we are working with $p=7$.
How do we find the expansions? Well, to find the last base seven digit of 320 we use division with remainder: $320=45 \times 7+5$. Then we take the quotient, 45 , and divide it as well: $45=6 \times 7+3$. And finally $6=0 \times 7+6$. The main rule is that the remainder must be one of the numbers $0,1, \ldots$, $p-1$.

Such expansions are already interesting in that they give “local” information: the expansion in powers of $(X-\alpha)$ will show, for example, if $P(X)$ vanishes at $\alpha$, and to what order. Similarly, the expansion “in base $p$ ” will show if $m$ is divisible by $p$, and to what order. For example, expanding 72 in base 3 gives
$$72=0+0 \times 3+2 \times 3^{2}+2 \times 3^{3}=2200,$$
which shows at once that 72 is divisible by $3^{2}$.
Now, for polynomials and their quotients, one can in fact push this much further. Taking $f(X) \in \mathbb{C}(X)$ and $\alpha \in \mathbb{C}$, there is always an expansion
\begin{aligned} f(X)=\frac{P(X)}{Q(X)} &=a_{n_{0}}(X-\alpha)^{n_{0}}+a_{n_{0}+1}(X-\alpha)^{n_{0}+1}+\ldots \ &=\sum_{i \geq n_{0}} a_{i}(X-\alpha)^{i} \end{aligned}
This is just the Laurent expansion from complex analysis, but in our case it can be very easily obtained either by doing long division with the expansions of $P(X)$ and of $Q(X)$ or by using division with remainder as before. Notice that it is a much more complicated object than the preceding expansion:

• We can have $n_{0}<0$, that is, the expansion can begin with a negative exponent; this would signal that $\alpha$ is a root of $Q(X)$ and not of $P(X)$ (more precisely, that its multiplicity as a root of $Q(X)$ is bigger). In the language of analysis, we would say that $f(X)$ has a pole at $\alpha$ of order $-n_{0}$. This is not much of a problem: we first remove the pole by multiplying by $(X-\alpha)^{\left|n_{0}\right|}$, expand the result into powers of $(x-\alpha)$, then divide again at the end.
• The expansion will usually not be finite. In fact, it will only be finite if when we write $f(X)=P(X) / Q(X)$ in lowest terms with $Q(X)$ monic, then $Q(X)$ happens to be a power of $(X-\alpha)$ (can you prove it?). In other words, this is usually an infinite series, and it can be shown that the series $f(\lambda)$ that we get when we replace $X$ by $\lambda \in \mathbb{C}$ will converge whenever $\lambda$ is close enough (but not equal) to $\alpha$. However, since we want to focus on the algebraic structure here, we will treat the series as a formal object: it is just there, and we do not care about convergence.
Here’s an example. Take the rational function
$$f(X)=\frac{X}{X-1},$$

hensel原理的类比

$$f(X)=\frac{P(X)}{Q(X)},$$

$$x=\frac{a}{b},$$

$$P(X)=a\left(X-\alpha_{1}\right)\left(X-\alpha_{2}\right) \ldots\left(X-\alpha_{n}\right),$$

Hensel 正在研究一个关于代数的具体问题。为了进行类比。他在代数函数的背昗下考 虑了相同的问题。结果证明这个问题很容易解决，因为他可以将代数函数扩展为直级

$$P(X)=a_{0}+a_{1}(X-\alpha)+a_{2}(X-\alpha)^{2}+\cdots+a_{n}(X-\alpha)^{n} \quad=\sum_{i=0}^{n} a_{i}(X-\alpha)^{i}$$

我们可以为整数做类似的事情吗?

$$321=1+2 \times 10+3 \times 10^{2}$$

$$m=a_{0}+a_{1} p+a_{2} p^{2}+\cdots+a_{n} p^{n}=\sum_{i=0}^{n} a_{i} p^{i}$$

$$320=5+3 \times 7+6 \times 7^{2} .$$

$320=45 \times 7+5$. 然后我们取商 45 并将其除以: $45=6 \times 7+3$. 最后

$P(X)$ 消失在 $\alpha$ ，以及按什么顺序。同样，扩展”在棊 $p$ ” 将显示如果 $m$ 可以被 $p$ ，以及按 什么顺序。例如，以 3 为底展开 72 给出
$$72=0+0 \times 3+2 \times 3^{2}+2 \times 3^{3}=2200,$$

\begin{aligned} &\alpha \in \mathbb{C} \text {, 总有一个戔开 } \ &f(X)=\frac{P(X)}{Q(X)}=a_{n_{0}}(X-\alpha)^{n_{0}}+a_{n_{0}+1}(X-\alpha)^{n_{0}+1}+\ldots \quad=\sum_{i \geq n_{0}} a_{i}(X-\alpha)^{i} \end{aligned}

$Q(X)$ 或者喼以前一样使用带余数的除法。 请住意，它比前面的展开要复杂得多:

• 我们可以有 $n_{0}<0$, 即展开式可以从负指数开始; 这侍表明 $\alpha$ 是一个根 $Q(X)$ 而不 $f(X)$ 有一个极点 $\alpha$ 有秩序的 $-n_{0}$. 这不是1十/问迫题: 㧴们首先通过乘移除极点 $(X-\alpha)^{n_{0} \mid}$, 将结果展开为的最 $(x-\alpha)$, 然后在最后再除。
• 扩展通常不是有限的。事实上，只有当我们写 $f(X)=P(X) / Q(X)$ 最低限度地 旬话兑，这通常是一个无限级数, 并且可以姃明该级数 $f(\lambda)$ 㧴们更换时得到的 $X$ 经 过 $\lambda \in \mathbb{C}$ 无论何时都会收致 $\lambda$ 足够㢺近 (但不等于) $\alpha$. 但是, 由于㧴们在这里要关 侏。
这是一个列子。取有理函数
$$f(X)=\frac{X}{X-1},$$

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