More on Teichmüller Representatives

By ellipticcurves

As I mentioned before, Teichmüller’s great insight was that one might have better luck if using representatives

t(m) = (m, m^p, m^{p^2}, \ldots)

for m = 0, \ldots, p-1, instead of integers 0, \ldots, p-1 themselves. One of the advantages, for example is that this set of representatives is closed under multiplication. In other words, it is an monomorphism of multiplicative monoids from {\mathbb F}_p to {\mathbb Z}_p.

But is this really a set of representative? In other words, can every p-adic integer be written as a p-series in t(0), \ldots, t(p-1)? Since t(m) \equiv m \pmod{p} for each m, and integers 0,\ldots, p-1 form a set of representatives, the answer to this question is yes.

Let’s offer a demonstration. Say we would like to write 5-adic integer 7 in terms of Teichmüller representatives. Say,

7 = t(a_0) + t(a_1)5 + t(a_2)5^2 + \ldots.

Then

  • 7 \equiv 2 \pmod{5}, so let a_0 = 2.
  • 7 \equiv a_0^5 + 5 a_1 \pmod{5^2}, i.e., 5 a_1 \equiv 7 - 2^5 \equiv 0 \pmod{25}, so we take a_1=0.
  • 7 \equiv a_0^{25} + 5 a_1^5 + 5^2 a_2 \pmod{5^3}, i.e., 25 a_2 \equiv 7 - 2^{25} - 0^5 \equiv 75 \pmod{125}, so we take a_2=3.
  • 7 \equiv a_0^{125} + 5 a_1^{25} + 5^2 a_2^5 + 5^3 a_3 \pmod{5^4}, i.e., 125 a_3 \equiv 0 \pmod{125}, so we take a_3=0.
  • etc.

So,

7 = t(2) + t(0) 5 + t(3) 5^2 + t(0) 5^3 + \ldots.

Therein lies a seed of the inductive proof of the fact that

\{t(m) | m\in 0,\ldots, p-1\}

is a set of representatives. If for a fixed p-adic integer a one can choose a_0, \ldots, a_{n-1} \in 0,\ldots, p-1 so that

a \equiv t(a_0) + t(a_1) p + \cdots + t(a_{n-1}) p^{n-1} \pmod{p^n},

or, equivalently,

a \equiv a_0^{p^{n-1}} + a_1^{p^{n-2}} p + \cdots + a_{n-1} p^{n-1} \pmod{p^n},

then

a - (a_0^{p^n} + a_1^{p^{n-1}} p + \cdots + a_{n-1}^p p^{n-1}) \equiv 0 \pmod{p^n},

and there exists a_n \in 0, \ldots, p-1 such that

a - (a_0^{p^n} + a_1^{p^{n-1}} p + \cdots + a_{n-1}^p p^{n-1}) \equiv a_n p^{n} \pmod{p^{n+1}}.

Then

a \equiv a_0^{p^n} + a_1^{p^{n-1}} p + \cdots + a_{n-1}^p p^{n-1} + a_n p^{n} \pmod{p^{n+1}},

or, equivalently,

a \equiv t(a_0) + t(a_1) p + \cdots + t(a_{n}) p^{n} \pmod{p^{n+1}}.

The short version is that any overshot modulo p^{n} can be corrected modulo p^{n+1} using only multiples of p^n.

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