Let $L$ be a finite extension of $p$-adic field. Let $O$ be the ring of integers of $L$, let $\pi$ be a uniformizer of $O$ and $q$ the order of residue field.

Following 1, for any $O$-algebra $R$, we define the ring of ramified Witt vectors $(W(R),+,×)$ as follows. Start with the Witt polynomials $$\Phi_n(X_0,X_1,\dots,X_n)=X_0^{q^n}+ \pi X_{n-1}^{q^{n-1}}+\dots+\pi^nX_n.$$

Then $\forall n \ge 0$, $\exists ! P_n,S_n∈O[X_0,X_1,\dots,X_n]$ such that $\Phi_n((S_n))=\Phi_n(X_0,X_1,\dots,X_n)+\Phi_n(Y_0,Y_1,\dots,Y_n)$, $\Phi_n((P_n))=\Phi_n(X_0,X_1,\dots,X_n)\Phi_n(Y_0,Y_1,\dots,Y_n)$.

We define + and × on $W(R)$ by $(a_1,a_1,\dots)+(b_0,b_1,\dots)=(S_0(a_0,b_0),S_1(a_0,a_1,b_0,b_1),\dots)$, $(a_0,a_1,\dots)×(b_0,b_1,\dots)=(P_0(a_0,b_0),P_1(a_0,a_1,b_0,b_1),\dots)$.

My question:

Let $\pi u$, $u \in O^\times$ be another prime element of $O$. Why does exchanging $\pi$ to $\pi u$ not (up to isomorphism) change the ring structure $(W(R),+,×)$ ?

Could you tell me the canonical ring isomorphism between two rings of Witt vectors with different prime element $\pi$ and $\pi u$ ? Once the isomorphism is given, I try to prove the map is a ring isomorphism.

This problem occurred when I was reading Galois representations and $(\varphi, \Gamma)$-modules written by Peter Schneider. In the book we define ring of ramified Witt vectors, and we need to check Witt ring does not depend on the choice of a prime element.

  • 2
    $\begingroup$ It might help readers if you give a reference for your definitions (some paper of Hazewinkel?) and define the notation $o$ more clearly. Can you describe your ring by a universal mapping property? That would explain an independence-of-$\pi$ property in the definition of the ring structure. $\endgroup$
    – KCd
    Apr 15, 2022 at 23:04
  • $\begingroup$ I don't know how to characterize Witt ring in terms of universal property.. $\endgroup$
    – Pont
    Apr 15, 2022 at 23:16
  • 2
    $\begingroup$ @sharding4 the formulas for the operations do depend on the choice of $\pi$ once you get past the formulas for $S_0$ and $P_0$. For example, $S_1 = X_1 + Y_1 + \sum_{k=1}^{q-1} \frac{1}{\pi}\binom{q}{k}X_0^kY_0^{q-k}$. The value of $\frac{1}{\pi}\binom{q}{k}$ depends on $\pi$. $\endgroup$
    – KCd
    Apr 16, 2022 at 0:53
  • 2
    $\begingroup$ Do you genuinely mean to be using an actual local field as the coordinates in $W(R)$? Note $\mathbf Z_p$ is $W(\mathbf F_p)$. The Witt vector construction is interesting when we use fields of characteristic $p$ as coordinates (esp. perfect fields). For a field $F$ of characteristic $0$, $W(F)$ is rather boring: you can invert everything and show $W(F)$ is the product ring $\prod_{n \geq 0} F$. $\endgroup$
    – KCd
    Apr 16, 2022 at 0:56
  • 1
    $\begingroup$ I think that you are discussing the ring of Witt vectors. As others pointed out earlier the Witt ring is something else. $\endgroup$ Apr 16, 2022 at 6:25

1 Answer 1


As said in a comment, in the source you quote (Peter Schneider: Galois representations and $(\varphi, \Gamma)$-modules) this is actually an exercise, after definition 1.1.9. I admit I found it quite a nut to crack, but I think I got it. (I better should, Schneider was my Ph.D. advisor, and I learned about Witt vectors from him.)

Of course in the end it looks somewhat formal, as Witt vectors do. So I start with something which looks different at first, but turns out to be a special case of what we want:

Writing $5$-adic integers in base $10$, or: Why we should not expect this to look pretty

One of the basic things one learns about the $p$-adic integers $\mathbb Z_p$ is that, for a given set of representatives $S \subset \mathbb Z_p$ of $\mathbb Z_p/p$ (a popular choice being $S=\{0,1, .., p-1\}$), one can write each $x \in \mathbb Z_p$ as $$\sum_{n=0}^\infty a_np^n$$ with unique $a_n \in S$. This is well-motivated by $p$-adic expansions of the natural numbers which e.g. for $p=5$ start $1,2,3,4,10, 11, 12, 13, 14, 20, 21, \dots, 43, 44, 100, 101, $ etc.

Later, one learns that such unique series expansions exist if we replace $p$ by any prime element (a.k.a. uniformizer) $\tilde p \in \mathbb Z_p$.

Wait, what?

Yes, any prime element $\tilde p$. Granted these are all of the form $u\cdot p$ for some unit $u \in \mathbb Z_p^\times$, but that's a lot. E.g. $10=2\cdot p$ is a prime element of $\mathbb Z_5$. Meaning that every $x\in \mathbb Z_5$ can be written as $$\sum \tilde a_n \cdot 10^n$$ with unique $\tilde a_n \in \{0,1, \dots, 4\}$. That is, "in base $10$", except that only coefficients up to $4$ are allowed. How does that look like? Of course we start $1,2,3,4$, but how to write $5$ in base $10$ as claimed?

Well as usual, we have to solve congruences modulo $5^n$ (because $\mathbb Z_5/10^n \cong \mathbb Z_5/5^n$). Obviously $\tilde a_0=0$. Turns out that $\tilde a_1=3$, because $3\cdot 10 \equiv 5 (5^2)$. Next we need $\tilde a_2\cdot 10^2+3\cdot 10 \equiv 5 (5^3)$ which gets solved by $\tilde a_2=1$. Continuing like this, I get

$$5 =\, \dots 31134033203130$$

"in base $10$", and I suspect (although I cannot even prove) that there will never be a discernible pattern in this sequence.

So that is an example of a prime $p$ written in base $\tilde p$. (To be fair, once one has that one, it is straightforward to continue counting. The natural numbers as $5$-adics in base $10$ start: $1,2,3,4, \dots 31134033203130, \dots 31134033203131, \dots 31134033203132, \dots 31134033203133, \dots 31134033203134, 10, 11,12,13,14, \dots 31134033203140, \dots 31134033203141, $etc.)

OK, so even if we admit that the result looks disheartening, let's retrace our steps and see if we can at least write down an iterative algorithm to translate from "standard base $p$" to "base $\tilde p = u\cdot p$" with a unit $u \in \mathbb Z_p^\times$ (we took $p=5$ and $u=2$ i.e. $\tilde p =10$).

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.

So something like this will happen in the general case.

The General Case

I use notation of the source, but try to keep this somewhat self-contained.

We choose two prime elements $\pi$ and $\tilde \pi$ of our base ring $O$, i.e. there is a unique $u \in O^\times$ such that $\tilde \pi = \pi u$. That gives us, for any $O$-algebra $R$, two a priori different rings of Witt vectors $W(R)$ and $\tilde W(R)$, defined via the Witt polynomials

$\Phi_n(X_0, ..., X_n) = X_0^{q^n}+ \pi X_{n-1}^{q^{n-1}}+\dots+ \pi^n X_n$ and $\tilde \Phi_n(X_0, ..., X_n) = X_0^{q^n}+ \tilde \pi X_{n-1}^{q^{n-1}}+\dots+ \tilde \pi^n X_n$,

with further polynomials $S_n, P_n, I_n, F_n$ and $\tilde S_n, \tilde P_n, \tilde I_n, \tilde F_n$ built from them, respectively.

The exercise has a hint referring to Proposition 1.1.5 whose relevant part says:

If $B$ is an $O$-algebra with an $O$-algebra endomorphism $\sigma$ such that $\sigma(b) \equiv b^q$ modulo $\pi B$, then [...] $B' := im(\Phi_B) =(\Phi_0(B),\Phi_1(B) , ...)$ is the $O$-subalgebra of $B^{\mathbb N_0}$ given by $$B'=\{(u_0, u_1, ...) \in B^{\mathbb N_0}: \sigma(u_n) \equiv u_{n+1} \text{ mod } \pi^{n+1}B \text{ for all } n \ge 0 \}. $$

And the hint points out that this description of $B'$ is independent of the choice of $\pi$ (which is clear, as $\pi^{n}B = \tilde \pi^n B$ for any $O$-algebra $B$). How do we use that?

Inspired by the proposition being usually applied in the case of $O[X_0, X_1, ..., Y_0, Y_1, ...]$ to build all those summation and product polynomials out of the respective Witt polynomials, we will apply it to the polynomial ring $B:= O[X_0,X_1, \dots]$, where it says that the two maps $\Phi_B = (\Phi_0, \Phi_1, \dots )$ and $\tilde \Phi_B = (\tilde \Phi_0, \tilde \Phi_1, \dots )$ have the same image in $B^{\mathbb N_0}$. When spelled out, this means that there exist elements $T_0, T_1, \dots \in O[X_0, X_1, \dots ]$ such that for all $n$,

$$\fbox{$\tilde \Phi_n(T_0, \dots, T_n) = \Phi_n(X_0, \dots, X_n)$.}$$

Each $T_n$ is a polynomial in $X_0, ..., X_n$. In fact,

$T_0 = X_0$

$T_1 = u^{-1} X_1$

$T_2 = u^{-2} \left(X_2 + \dfrac{1 - u\cdot u^{-q}}{\pi}\cdot X_1^q \right)$

etc. Comparing with the $\tilde a_n$-formulae from the prelude, we are on the right track. What is left to show is that for any $O$-algebra $R$, the map $$T: W(R) \rightarrow \tilde W(R),$$

$$(a_0, a_1, \dots) \mapsto (T_0(a_0), T_1(a_0, a_1), \dots)$$

is a ring homomorphism, indeed the isomorphism we are after. (The map $T = T_u$ depends on $u = \tilde \pi / \pi$, which is suppressed in the notation. Once we have shown it is a ring homomorphism, it is automatically an isomorphism with $T_{u^{-1}}$ as its inverse.)

For example, additivity boils down to the statement that the following two expressions are equal, for all $n$ and any two sequences $(a_i)_i, (b_i)_i \in R^{\mathbb N_0}$: On the one hand,

$$T_n(S_0(a_0, b_0), ..., S_n(a_0, ..., a_n,b_0, ..., b_n))$$

and on the other hand,

$$\tilde S_n((T_0(a_0), T_1(a_0, a_1),..., T_n(a_0, ..., a_n), T_0(b_0), T_1(b_0, b_1),..., T_n(b_0, ..., b_n)).$$

As scary as it looks, it basically comes from the definitions. With a little abuse of notation, note that the first expression, $T_n (S_0, ..., S_n)$, is made so that when we apply $\tilde \Phi_n$ to it, we get the same as when we apply $\Phi_n$ to $(S_0, ..., S_n)$, which by definition of the $S_n$ is the sum of the $\Phi_n$'s. On the other hand, when we apply $\tilde \Phi_n$ to that big $\tilde S_n$ expression, by definition of the $\tilde S_n$ we get the sum of $\tilde \Phi_n(T_0, ... T_n)$ expressions; which, by definition of the $T$'s, is also the sum of the $\Phi_n$-expressions. Formally (without such abuse of notation), maybe one has to go up to a polynomial ring in polynomial rings here and use injectivity of $\Phi_A$ on such a ring $A$ to conclude. Instead of doing that, I content myself with the example $n=1$ (highlighted as a challenge in a comment by KCd), where on the one hand,

$$T_1(S_0(a_0, b_0), S_1(a_0, a_1,b_0, b_1)) = u^{-1} S_1(a_0, a_1,b_0, b_1) \\ = u^{-1} \cdot (a_1+b_1+ \sum _{i=1}^{q-1} \frac{1}{\pi} \binom{q}{i}a_0^i b_0^{q-i})$$

and on the other hand,

$$\tilde S_1((T_0(a_0), T_1(a_0, a_1), T_0(b_0), T_1(b_0, b_1)= \tilde S_1(a_0,u^{- 1}a_1, b_0, u^{-1}b_1) \\ = u^{-1}a_1+u^{-1}b_1+ \sum _{i=1}^{q-1} \frac{1}{\tilde \pi} \binom{q}{i}a_0^i b_0^{q-i},$$

and these two are identical, thanks to $\tilde \pi = \pi u$.

And then "of course" the same goes through for the $P_n$ and $I_n$ (not that I actually checked more than the case of $P_1$; sorry, Peter).


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