# Changing uniformizer of $p$-adics

In the theory of $p$-adic fields typically a uniformizer $\pi$ is chosen that generates the maximal ideal, $m$. And a few theorems later it can be shown that every element $x \in O$ of the ring of integers can be expressed as $$x = \sum_{i=0}^\infty c_i \pi^i$$ where the $c_i$ are representatives of $O/m$. But $\pi$ is only unique up to multiplication by a unit.

Question 1: What is the proof that the $p$-adic field does not depend on the choice of uniformizer? (Edit: This was shown in the comments to be a silly question, but I believe the following questions are still valid.)

Question 2: Suppose $\pi_2=u\pi$ for some unit $u$. Is there an explicit form of the mapping $c_i \to d_i$ such that $$\sum_{i=0}^\infty c_i \pi^i = \sum_{i=0}^\infty d_i \pi_2^i$$

As an example, consider $\mathbb{Z}_7$, the 7-adics. The usual uniformizer is, of course, $\pi=7$. But $5$ is a unit in $\mathbb{Z}_7$ so we should be able to take $\pi=35$ or $\pi=7/5$.

Question 2 (example version): Is there an explicit form of the mapping $c_i \to d_i$ such that $$\sum_{i=0}^\infty c_i 7^i = \sum_{i=0}^\infty d_i (7/5)^i$$ Note: The answer is not $d_i = 5^i c_i$ because $0 \le d_i \le 6$.

• What do you mean by Question 1? The $p$-adic field is not usually defined in terms of a uniformizer... Commented Apr 26, 2016 at 18:43
• Good point. I mentally flipped the definition (construction) and the theorem (representation). But I believe Questions 2 are still valid. Or am I missing something else? Commented Apr 26, 2016 at 18:49

As regards question 2, this happens to be exactly what I tackled in my answer to Why does this generalized ring of Witt vectors not depend on a choice of a prime element? There, in an intro, I tried to write a $$5$$-adic integer in base $$10 = 2\cdot 5$$ (as $$2$$ is a $$5$$-adic unit). To quote the crucial part of the general statement:

To translate from "standard base $$p$$" to "base $$\tilde p = u\cdot p$$" with a unit $$u \in \mathbb Z_p^\times$$, looking through those congruences one has to solve, I get (denoting $$[\cdot] : (\mathbb Z_p \rightarrow) \mathbb Z_p/p \rightarrow \{0,1,\dots , p-1\}$$ our representatives):

$$\tilde a_0 = a_0$$ ($$= [a_0 \text{ mod } p]$$)

$$\tilde a_1 = \left[ [u^{-1}] [a_1] \qquad \text{ mod } p \right]$$

$$\tilde a_2 = \left[ [u^{-2}] \left([a_2]+ \dfrac{1-[u^{-1}]u}{p} \cdot [a_1] \right) \qquad \text{ mod } p \right]$$

and it gets unwieldy from here, but if so obliged, one will find formulae for the higher terms. Note that although we start and end with expressions modulo $$p$$, we have to go through computations in $$\mathbb Z_p$$ here.

In other words, we can define a map $$\mathbb F_p^{\mathbb N_0} \rightarrow \mathbb F_p^{\mathbb N_0}$$,  $$(a_0 \text{ mod } p, a_1 \text{ mod } p, ...) \mapsto (\tilde a_0 \text{ mod } p, \tilde a_1 \text{ mod }p, ...)$$, such that the induced map on $$\mathbb Z_p (\simeq W(\mathbb F_p))$$, $$\sum a_n p^n \mapsto \sum \tilde a_n \tilde p^n$$ is a ring homomorphism. Actually, it is the identity, written in an unpleasantly complicated way.

I believe that is what you are looking for.

I want to note that the "right" setup for this seems to be the Witt vector machinery in the follow-up in that answer, which gives general polynomials $$T_0, T_1, ...$$ whose manifestations are the above formulas (and you want to apply it to $$R = O/m$$ which should give $$W(R) \simeq O$$ regardless of whether $$W(\cdot)$$ is defined with $$\pi$$ or $$\tilde \pi = u \cdot \pi$$). One thing that makes the above formulas uglier are those representative choices $$[ \cdot ]$$. Note that for all this, it is tacitly supposed that in both extensions, in your terminology $$\sum c_i \pi^i$$ and $$\sum d_i \pi^i$$, the $$c_i$$ and $$d_i$$ come from the same set of representatives, chosen as one section $$[ \cdot ] : O/m \rightarrow O$$ of the projection $$O \rightarrow O/m$$; and conceptually, this choice of representatives brings an unwarranted arbitrariness into the question. (The Witt vector formulation, on the other hand, once one has those general polyomials $$T_n$$, seemingly just works in the residue field.)