# Image of the ghost map of $p$-typical Witt vectors and $A$-ring structure of $W(A)$

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.

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.

Consider the following subring of $$A$$: $$\begin{equation*} S\left(A\right)=\left\{a\in A\mid\overline{a}^{p}=\overline{a}\right\}\text{.} \end{equation*}$$

For all $$n\in\mathbb{N}$$, write $$S^{n}\left(A\right)$$ for the "composition" (by convention, $$S^{0}\left(A\right)=A$$). Notice that for any such $$n$$ we have $$pS^{n}\left(A\right)\subset S^{n+1}\left(A\right)$$.

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.

• 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. – PULITA ANDREA Jun 5 at 7:37
• 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. – PULITA ANDREA Jun 5 at 19:53
• 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. – Rubén Muñoz--Bertrand Jun 6 at 18:49