p-adic分析note| Hensel’s Analogy

p-adic分析note| Hensel’s Analogy

p-adic分析

p-adic分析note| Hensel’s Analogy

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
    $$
    f(X)=\frac{X}{X-1},
    $$
p-adic分析note| Hensel’s Analogy

hensel原理的类比


这 $p$ 进数由德国数学家 Kurt Hensel 首次提出。Hensel 的出发点是塰数环之间的关比 $\mathbb{Z}$, 连同它的分数域 $\mathrm{Q}$, 和环 $\mathrm{C}[X]$ 复数多项式及其分数域 $\mathrm{C}(X)$. 他从他的博士导师 Leopold Kronecker 那里学到了这个类比,他甚至试图发展一个函盖这两种情兄的单一 理论。
具体来说,让找们使用 $X$ 作为一个不确定的,储葍 $x$ 代表一个数字。一个元素 $f(X) \in \mathbb{C}(X)$ 是一个”有理函数”,即两个项式的商:
$$
f(X)=\frac{P(X)}{Q(X)},
$$
和 $P(X), Q(X) \in \mathbb{C}[X], Q(X) \neq 0$; 我们总是可以要求 $Q(X)$ 是 monic,即它的 前导系数是 1 。 荚似地, 任何有理数 $x \in \mathbb{Q}$ 是两个整数的商:
$$
x=\frac{a}{b},
$$
和 $a, b \in \mathbb{Z}, b \neq 0$; 我们总是可以要求 $b>0$. 此外,这两个环的性质非常相似: 都是 $\mathbb{Z}$ 和 $C[X]$ 是环在哪里
存在唯一分解: 任何整数都可以唯一表示为 $\pm 1$ 乘以素数的乘积,并且任何多项式都可 以唯一地表示为
$$
P(X)=a\left(X-\alpha_{1}\right)\left(X-\alpha_{2}\right) \ldots\left(X-\alpha_{n}\right),
$$
在哪里 $a$ 和 $\alpha_{1}, \alpha_{2}, \ldots \alpha_{n}$ 是書数。这给了我们亨塞尔探索的类比的要点: 素数 $p \in \mathbb{Z}$ 粂似于线生㝖项式 $X-\alpha \in \mathbb{C}[X]$.
夈㩺延伸到方程的解。给定一个多项式,其系数为 $\mathbb{Z}$ ,任何根都是代数数;如果一个函 数是一个多抲的根,其系数为 $\mathbb{C}[X]$ ,它是一个代数函数。所以 $\sqrt{2}$ ,这是 个根
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
假设给定一个多项式 $P(X) \in \mathbb{C}[X]$ 和一个特定的 $\alpha \in \mathbb{C}$. 然后可以 (例如,使用泰勒
$$
P(X)=a_{0}+a_{1}(X-\alpha)+a_{2}(X-\alpha)^{2}+\cdots+a_{n}(X-\alpha)^{n} \quad=\sum_{i=0}^{n} a_{i}(X-\alpha)^{i}
$$
和 $a_{i} \in \mathbb{C}$. 这提供了关于多项式如何在附近表现的非常精觔的信息 $\alpha$.


我们可以为整数做类似的事情吗?


我们可以为笅数做这样的事情吗? 对于正整数,我们可以,而且实际上我们每天都在写 下们隹
$$
321=1+2 \times 10+3 \times 10^{2}
$$
是那种形式。顺人的是 10 不是素数,而 $(X-\alpha)$ 是一个責数 $C[X]$. 但我们可以解决这 个问题: 选择一个素数 $p$ 并在基数中写下㧴们的号码 $p$ : 给定一个正整数 $m$, 我们可以写
$$
m=a_{0}+a_{1} p+a_{2} p^{2}+\cdots+a_{n} p^{n}=\sum_{i=0}^{n} a_{i} p^{i}
$$
和 $a_{i} \in \mathbb{Z}$ 和 $0 \leq a_{i} \leq p-1$. 例如,如果 $p=7$ 我们可以写
$$
320=5+3 \times 7+6 \times 7^{2} .
$$
我们甚至可以将其记录为一电数字”635″(如以十为底,我们倒着记录数字: 最低幕 2 最 后来)。为了保持区别
在标准 (以十为基数) 符号和㫷数之间 $p$, 让㧴们用红色作为启者,所以 $320=635$ 只 要我攸知道我们正在合怍 $p=7$.
我们如何找到扩展? 好吧,要找到 320 的最后一个基数七位数,我们使用除以余数:
$320=45 \times 7+5$. 然后我们取商 45 并将其除以: $45=6 \times 7+3$. 最后
这种扩展已经很有㻓,因为它们提供了”本地”信息: $(X-\alpha)$ 将显示,例如,如果
$P(X)$ 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
$$
72=0+0 \times 3+2 \times 3^{2}+2 \times 3^{3}=2200,
$$
这立即表明 72 可以被 $3^{2}$
现在,对于多项式及其商,实际上可以将这一点推得更远。服用 $f(X) \in \mathbb{C}(X)$ 和
$$
\begin{aligned}
&\alpha \in \mathbb{C} \text {, 总有一个戔开 } \
&f(X)=\frac{P(X)}{Q(X)}=a_{n_{0}}(X-\alpha)^{n_{0}}+a_{n_{0}+1}(X-\alpha)^{n_{0}+1}+\ldots \quad=\sum_{i \geq n_{0}} a_{i}(X-\alpha)^{i}
\end{aligned}
$$
这只是复分析中的 Laurent 展开,但在我们的例子中,它可以很容易地通过对 $P(X)$ 和
$Q(X)$ 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
    $$
    f(X)=\frac{X}{X-1},
    $$
p-adic分析note| Hensel’s Analogy
微积分note| Laws of vector theorem

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