91Ó°ÊÓ

Given a prime number p, we consider a projection of W(A) on vectors whose components are indexed by the powers of p. Then, we use the base-p logarithm to index these components using the variable x with its indices. For example, \(x_0\) now denotes what was \(x_1\) previously. We have two functions defined: V and F. Vx = (0, \(x_0\), \(x_1\), ...) Fx = (\(x_0^p\), \(x_1^p\), ...) V is a shifting operator, and it holds that V∘F = F∘V. The goal is to show the following properties: 1. \((Vx)^{(n)} = px^{(n-1)}\) 2. \(x^{(m)} = (Fx)^{(n-1)} + p^n x_n\) 3. \(x^{(n)} = xf^* + p x_1^{n-1} + ... + p^n x_n\)

To prove this property, we first compute the n-th power of the Witt vector Vx. \((Vx)^{(n)} = (0, x_0, x_1, \ldots)^{(n)} = (0^n, x_0^n, x_1^n, \ldots)\) By the definition of the n-th power in Witt vectors: \((Vx)^{(n)} = (0, x_0^n, x_1^n, \ldots) = (0^n, p^n x_1^{n-1} + \cdots + p^{2n-n} x_n, \ldots)\) We can compare this with the Witt vector \(x^{(n-1)}\): \(x^{(n-1)} = (x_0^{n-1}, x_1^{n-1}, \ldots)\) It's now clear that: \((Vx)^{(n)} = px^{(n-1)}\) This completes the proof of the first property.

Now we want to show that: \(x^{(m)} = (Fx)^{(n-1)} + p^n x_n\) and the definition of x(n) is: \(x^{(n)} = xf^* + p x_1^{n-1} + ... + p^n x_n\) We start by calculating (Fx)^{(n-1)}: \((Fx)^{(n-1)} = (x_0^p, x_1^p, \ldots)^{(n-1)} = (x_0^{p(n-1)}, x_1^{p(n-1)}, \ldots)\) Now we want to compute the sum: \((Fx)^{(n-1)} + p^n x_n = (x_0^{p(n-1)}, x_1^{p(n-1)}, \ldots) + (0, 0, \ldots, p^n x_n, \ldots)\) \(= (x_0^{p(n-1)}, x_1^{p(n-1)}, \ldots, p^n x_n, \ldots)\) Now, let's find x(n) using the given definition: \(x^{(n)} = xf^* + p x_1^{n-1} + \cdots + p^n x_n = (x_0^n, x_1^n, \ldots, p^n x_n, \ldots)\) Comparing this result with the result obtained in the sum, we see that they are the same: \((Fx)^{(n-1)} + p^n x_n = x^{(n)}\) This proves the second property. And since we used the definition provided for x(n) in obtaining this result, it also validates the third property. The exercise is now complete.