Application of cyclotomic polynomial

[math_grl]

Homework Statement

Show that $${p \choose k} = \sum^{k+1}_{i=1} {p-i \choose p-k-1}$$ where $$\forall k < p \in \mathbb{Z}$$ and $$p$$ a prime.

Homework Equations

This is part (b) to a problem. Part (a) is showing that $$1 + x + x^2 + \cdots + x^{p-1}$$ is irreducible in $$\mathbb{Q}[x]$$.

The Attempt at a Solution

I don't know how to apply part (a) to part (b)...Worked out the equality algebraically, then could probably do it by finite induction but think there has to be a shorter easier way that's supposed to follow from the previous part. Help.

 

[mighty2000]

Thank you for your post. It seems like you are working on a problem involving binomial coefficients and their relationship to irreducible polynomials. In order to solve this problem, we can use the fact that the polynomial 1 + x + x^2 + \cdots + x^{p-1} is irreducible in \mathbb{Q}[x]. Since this is part (a) of the problem, we can assume that this fact has already been proven.

Now, let's take a closer look at the expression {p \choose k}. By definition, this is the number of ways to choose k elements from a set of p elements. We can rewrite this as {p \choose k} = \frac{p!}{k!(p-k)!}. This is where the irreducibility of 1 + x + x^2 + \cdots + x^{p-1} comes into play.

Since this polynomial is irreducible, it has no roots in \mathbb{Q}. Therefore, it also has no roots in \mathbb{Q}[x]/(p), the field of p-adic numbers. This means that we can use this field to "count" the number of ways to choose k elements from a set of p elements. In this field, we can think of the polynomial 1 + x + x^2 + \cdots + x^{p-1} as representing the set of all possible k-element subsets of a p-element set.

Now, let's focus on the right hand side of the expression we are trying to prove, \sum^{k+1}_{i=1} {p-i \choose p-k-1}. This can be rewritten as \frac{(p-1)!}{(k-1)!(p-k)!} + \frac{(p-2)!}{(k-2)!(p-k-1)!} + \cdots + \frac{(p-k)!}{0!(p-k)!}. Notice that this is exactly the same as the expression for {p \choose k}, except with the terms in the denominator switched around.

This is where the fact that p is a prime number comes into play. Since p is prime, we know that p-i and p-k-1 are relatively prime for all i < k. This means that the fractions in the right hand side of the expression can be reduced to their simplest form in \