分类： 数学

今天我们讨论多项式

交换环上

1. 
   Units, nilpotents, and idempotents lift from A/\mathfrak{N} to A.Proof: Units and nilpotents are obvious. In fact they lift to any of their representatives.
   For idempotents, if x^2=x\in A/\mathfrak{N}, then (1-x)x=0 \in A/\mathfrak{N}, so (1-x)^kx^k=0\in A for sufficiently large k. And (1-x)^k+x^k=1-x+x=1\in A/\mathfrak{N}, so lifts to a unit (1-x)^k+x^k. Moreover, its inverse u=1\in A/\mathfrak{N}. So (ux)^k(u(1-x))^k=0,ux^k+u(1-x)^k=1\in A and ux=x,u(1-x)=1-x\in A/\mathfrak{N}.
   This can be interpreted by sheaf theory, which is to be discussed in later posts. Continue reading “《交换代数》习题研讨 (I)”

Problem: If the decimal expansion of a contains all 10 digits (0,...,9), then the number of length n strings (shorted as n-strings) is greater than n+8.

If you’ve established the simple lemma, the solution is instant. Otherwise very impossible.

Lemma: The number C_n of n-strings is strictly greater than C_{n-1}, that of n-1-strings.
Actually,  we always have C_n \ge C_{n-1}: Every n-1-string corresponds to an n-string by continuing 1 more digit. The map is clearly injective. If C_n=C_{n-1}, it is bijective, meaning we have a way to continue uniquely, which means rationality. Rigidly, at least one of the n-1-strings occurs infinitely, but all digits after some n-1-string is totally determined by it. So if an n-1-string appears twice, it must appear every such distances, and so do the digits between. Continue reading “一个简单的组合问题及相关的思考”

今天我们讨论生成函数

Problem 1: Give a finite set of positive integers T. Let \mathfrak{T}_n be the collection of sequences (t_1,t_2,...,t_m), such that \sum_{i=1}^m t_m=n, and each t_i\in T. Let a_n=|\mathfrak{T}_n| for n\ge1 and a_0=1, and f(x)=\sum_{n=0}^{\infty}a_n x^n. Find out what is f(x) explicitly.

Solution: It is not hard to note the recursive relation a_n=\sum_{t\in T} a_{n-t} for n\ge1, if we set a_i=0 for negative i. So that f(x)=1+\sum_{t\in T} x^t f(x) and f(x)=1/(1-\sum_{t\in T} x^t), which is a rational function.

Variantion 1: If T is infinite, what would happen? Would f(x) still be rational?

We first analyze simple cases. If T=\mathbb{N}^+, it is expected that f(x)=1+\sum_{t=1}^{\infty} x^t f(x)=1+f(x) x/(1-x), so that f(x)=(1-x)/(1-2x)=1+\sum_{n=1}^{\infty} 2^{n-1} x^n. Indeed, in this case, counting the sequences amounts to divide a sequence of n objects arbitrarily. You can choose to divide between the ith and i+1th for 1\ge i\ge n-1, so in all 2^{n-1} choices, justifying the expansion.

I think it is equivalent to f being rational.

Theorem: \mathbb{Q}_p(t)\cap\mathbb{Q}[[t]]=\mathbb{Q}(t)

It is very interesting that this theorem is used for the rationality of \zeta-functions for algebraic varieties, which is part of the Weil conjectures.

2013 年 7 月 10 日