Today we discuss something on polynomials.

Over a Commutative Ring

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 “Discussion on Exercises of Commutative Algebra (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 “A Simple Combinatorial Problem and Related Thoughts”

Today we discuss some generating functions.

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.

Wednesday, July 10, 2013