# Why does this generalized ring of Witt vectors not depend on a choice of a prime element?

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.

• 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.
– KCd
Apr 15, 2022 at 23:04
• I don't know how to characterize Witt ring in terms of universal property..
– Pont
Apr 15, 2022 at 23:16
• @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 $\pi$.
– KCd
Apr 16, 2022 at 0:53
• 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 $\prod_{n \geq 0} F$.
– KCd
Apr 16, 2022 at 0:56
• I think that you are discussing the ring of Witt vectors. As others pointed out earlier the Witt ring is something else. Apr 16, 2022 at 6:25

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).