For all ring with unit element $A$ let $W(A)$ be the ring of $p$-typical Witt vectors. Denote by $$\phi\;:\;W(A)\to A^{\mathbb{N}}$$ the ghost map, which is given by $$\phi(a_0,a_1,a_2,\ldots)\;=\;(\phi_0,\phi_1,\phi_2,\ldots)$$ where $\phi_n=\phi_n(a_0,\ldots,a_n)$ is defined by $$\phi_n=a_0^{p^n}+pa_1^{p^{n-1}}+p^2a_2^{p^{n-1}}+\cdots+p^na_n\;.$$

I have 3 questions. The first is general, while the second and the third seem more approachable.

ASSUMPTION. $A$ is the ring of integers of a complete ultrametric field $K$ containing the field of $p$-adic numbers $\mathbb{Q}_p$. Denote by $k$ the residual field of $A$, it is a field of positive characteristic $p$.

QUESTION 1: Under the above assumption, I'm interested in understanding the image of the map $\phi$.

Notice that the map $\phi:W(K)\to K^{\mathbb{N}}$ is bijective.

Allow me to add some useful information. Under the assumption, $\phi:W(A)\to A^{\mathbb{N}}$ is injective and it is also known that if there is a ring endomorphism $\sigma:A\to A$ such that $\sigma(a)\equiv a^p \pmod{pA}$ for all $a\in A$, then a sequence $(\phi_0,\phi_1,\phi_2,\ldots)$ lies in $\operatorname{Im}(\phi)$ if and only if for all $n\geq 0$ one has $$\sigma(\phi_n)\equiv\phi_{n+1}\pmod{p^{n+1}A}$$ (for a proof see Bourbaki Alg. Comm. Chapter 9, $\S1$, $N.2$, Prop. 2, page AC IX.3). This criterion is due to B.Dwork.

The criterion provides an answer to Question 1 when $K$ is unramified over $\mathbb{Q}_p$ (i.e. the maximal ideal of $A$ is $pA$), because in this case $A=W(k)$ has a canonical Frobenius. What's going on in the general case? As soon as the ideal $pA$ is not maximal (ramification) the presence of a Frobenius endomorphism of $A$ (i.e. a ring endomorphism of $A$ lifting of the $p$-th power of $k$) seems not enough to guarantee a description of the image by means of congruences as above.

I also have the following important related questions, that might be more approachable.

QUESTION 2: Maintain the above assumption. Can we describe the set of elements $a\in A$ such that there exists $x=(x_0,x_1,\dotsc)\in W(A)$ with $$\phi(x_0,x_1,\dotsc)=(a,a,a,\dotsc)\quad ?$$ In other words, what is the set of $a\in A$ such that we can solve in $A$ the system of conditions \begin{align*} x_0&{}=a \\ x_0^p+px_1&{}=a \\ x_0^{p^2}+px_1^p+p^2x_2&{}=a \\ \cdots&\qquad? \end{align*} Of course, if $a$ lies in $\mathbb{Z}_p=W(\mathbb{F}_p)$ the ring of $p$-adic integers, then $a$ is a solution of my problem (by the above congruences criterion). Moreover, if $A=W(k)$, then Dwork's criterion proves that the only solutions $a\in A$ are the elements of $\mathbb{Z}_p$.

In general, the set of such elements $a$ is a sub-ring $B$ of $A$ containing $\mathbb{Z}_p$.

Question 2 is relevant for some convergence properties of certain Artin–Hasse exponentials that I do not mention here. However, it is also related to the following interesting question.

QUESTION 3: The ring $A^{\mathbb{N}}$ is naturally an $A$-algebra via the diagonal map $A\to A^{\mathbb{N}}$ sending $a\mapsto (a,a,a,\dotsc)$. Now, when does this $A$-algebra structure induce an $A$-algebra structure on $W(A)$ such that $\phi:W(A)\to A^{\mathbb{N}}$ is $A$-linear?

Alternatively, can we describe the maximal sub-ring $B\subseteq A$ such that $W(A)$ has a $B$-module structure making $\phi:W(A)\to A^{\mathbb{N}}$ a $B$-linear ring homomorphism?

Any comment will be useful and really appreciated.


1 Answer 1


I don't have a full answer, however I also don't have enough reputation to just comment, so I will post this as an answer.

In what follows, we will overline all projections modulo $p$.

Let $A$ be any commutative ring with no $p$-torsion. Then the ghost map $\phi$ is injective. Let $a\in A$ be such that $\left(a,a,\ldots\right)$ is the image of an element $w\in W\left(A\right)$ by the ghost map. We have $\phi\left(F\left(w\right)\right)=\phi\left(w\right)$, so $F\left(w\right)=w$. Modulo $p$, this implies that for all $n\in\mathbb{N}$ we have $\overline{w_{n}}^{p}=\overline{w_{n}}$. In particular, $\overline{w}=\overline{w}^{p}=\overline{\left[{w_{0}}^{p}\right]}$ where the last operator is the Teichmüller representative.

Let $S^{0}\left(A\right)=A$. For all $n\in\mathbb{N}^{*}$, consider the following subring of $A$: \begin{equation*} S^{n}\left(A\right)=\left\{a\in S^{n-1}\left(A\right)\mid\overline{a}^{p}=\overline{a}\right\}\text{.} \end{equation*}

Notice that for any such $n$ we have $pS^{n-1}\left(A\right)\subset S^{n}\left(A\right)$.

Put: \begin{equation*} B=\bigcap_{n\in\mathbb{N}^{*}}S^{n}\left(A\right)\text{.} \end{equation*}

The canonical morphism $\overline{B}\to\overline{A}$ is injective (thanks to Andrea for his remark). Indeed, if $b\in B$ satisfies $b=pa$ for some $a\in A$, then if $a\notin B$ there exists $n\in\mathbb{N}$ such that $a\in S^{n}\left(A\right)$ and $a\notin S^{n+1}\left(A\right)$. In particular, $b$ is not divisible by $p$ in $S^{n+1}\left(A\right)$ because $A$ has no $p$-torsion. But $b^{p}=p\times p^{p-1}a^{p}$, with $p^{p-1}a^{p}\in S^{n+1}\left(A\right)$ by the above remark. This means that $b^{p}-b$ should not be divisible by $p$ in $S^{n+1}\left(A\right)$, which contradicts the hypothesis $b\in B$. So we must have $a\in B$, and $\overline{B}\to\overline{A}$ is indeed injective.

It follows that the identity on $B$ is a lift of the Frobenius morphism of $\overline{B}$. In particular, we get an injective morphism of rings $\varphi\colon B\to W\left(B\right)$ such that for all $b\in B$ we have $F\left(\varphi\left(b\right)\right)=\varphi\left(b\right)$, and for $\phi\colon W\left(B\right)\to B^{\mathbb{N}}$ we have $\phi\left(\varphi\left(b\right)\right)=\left(b,b,b,\ldots\right)$. Also, ${\varphi\left(b\right)}_{0}=b$.

Consider $B$ as a subring of $W\left(A\right)$ by composing with $W\left(B\right)\to W\left(A\right)$. Any element of $w\in W\left(A\right)$ such that $\phi\left(w\right)$ is a constant sequence will satisfy $\overline{w}=\overline{b}$ for some $b\in S\left(A\right)$ by the above discussion. Since we can factor $\overline{S\left(A\right)}\to\overline{A}$ through the injective morphism $\overline{B}\to\overline{A}$, we can actually suppose that $b\in B$.

For any $x\in W\left(A\right)$, if $F\left(b+px\right)=b+px$, then $F\left(x\right)=x$. So under the further assumption that $W\left(A\right)$ is $p$-adically separated, then $B$ is set of elements answering to both questions 2 and 3.

  • $\begingroup$ Thanks a lot for the answer! By "Frobenius of A" I mean a ring endomorphism of $A$ lifting the $p$-th power of the residual field $k$ of $A$. Dwork's congruence criterion requires a lifting of the Frobenius of $A/pA$. For instance, if $A$ is the ring of integers of a totally ramified field extension of $\mathbb{Q}_p$ , the identity is a Frobenius of $A$, but not a lifting of the $p$-th power of $A/pA\neq \mathbb{F}_p$. EXAMPLE: Let $\pi$ be a non zero solution of $\pi^p+p\pi=0$ (Lubin-Tate). Then $A=\mathbb{Z}_p[\pi]$ and $\pi^p=0$ in $A/pA$. Hence, the Frobenius of $A/pA$ is not the identity. $\endgroup$ Jun 5, 2021 at 7:37
  • $\begingroup$ I think there is a mistake in your argument. The identity is not a lifting of the Frobenius of $B/pB$. You are confusing $B/pB$ and its image in $A/pA$, but the arrow $B/pB\to A/pA$ is not injective. This implies that you (possibly) do not have $\varphi:B\to W(B)$. For instance, if $k$ is perfect and $A=W(k)$ then Dwork's criterion gives $B=\mathbb{Z}_p$, while your proof gives $B'=\mathbb{Z}_p+pA$ (and you see in this example that the map $B'/pB' \to A/pA$ is not injective). The idea is however good and possibly a little changing might fix the proof. $\endgroup$ Jun 5, 2021 at 19:53
  • $\begingroup$ Dear Andrea, thank you for finding this mistake! As you say, only a little changing fixes the proof: the idea is to repeat the process infinitely many times. I have edited my answer. Without it, it would actually fail almost every time. $\endgroup$ Jun 6, 2021 at 18:49
  • $\begingroup$ What exactly is the definition of $S^n(A)$? I don't understand what you mean by "write $S^n(A)$ for the "composition" ". $\endgroup$ Dec 16, 2022 at 6:00
  • $\begingroup$ Okay I have updated the definition in the answer, I hope it is clearer now. $\endgroup$ Dec 19, 2022 at 13:42

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