Completed and uncompleted operations for Morava $E$-theory

Question

Let $E = E_n$ be the $n$-th Morava $E$-theory with coefficient ring $$ E_* = \mathbb{W}(\mathbb{F}_{p^n})[\![u_1,\ldots,u_{n-1}]\!][u^{\pm 1}]. $$ It is usual to consider the completed co-operations $$ E^\vee_* E := \pi_*L_{K(n)}(E \wedge E) $$ rather than the 'ordinary' co-operations $E_*E$. The latter is shown by Hovey (although this was a folklore result) to be isomorphic to $\operatorname{Hom}^c(\mathbb{G}_n,E_*)$ the ring of continuous functions from the $n$-th Morava stabilizer group to $E_*$ under the $\mathfrak{m} = (p,u_1,\ldots,u_{n-1})$-adic topology.

Let $L_0$ be the $L$-completion functor (for example see Appendix A of Hovey-Strickland). It is known that $E^\vee_*E = L_0(E_*E) = (E_*E)^\wedge_{\mathfrak{m}}$. There is a map $\phi:E_*E \to E^\vee_*E$ which can be identified with the usual $\frak{m}$-adic completion map.

Is the map $\phi:E_*E \to E^\vee_* E$ injective?

Standard commutative algebra tells us that the kernel of this map is isomorphic to $\bigcap_{n=0}^\infty \mathfrak{m}^n E_*E$, although it seems difficult to approach this problem in this way.

Here is a closely related question. Let $E(n)$ be the Johnson-Wilson cohomology theory with coefficient ring $$ E(n)_* = \mathbb{Z}_{(p)}[v_1,\ldots,v_{n-1},v_n^{\pm 1}]. $$ Now there is an isomorphism $L_0(E(n)_*E(n)) = E(n)_*^\vee E(n) \simeq \operatorname{Hom}^c(\mathbb{G}_n,\widehat{E(n)}_*)$, where $$\widehat{E(n)}_* \simeq \mathbb{Z}_{(p)}[v_1,\ldots,v_{n-1},v_n^{\pm 1}]^\wedge_I$$ for $I = (p,v_1,\ldots,v_{n-1})$. In this case by work of Johnson it is known that there is an injection of $E(n)_*E(n)$ into $\operatorname{Hom}^c(\mathbb{G}_n \times\mathbb{G}_n,A)$ where $A$ the ring of integers in an unramified degree $n$ extension of the p-adic numbers. It is known that the latter is $L$-complete, and since $E(n)_*E(n) \to L_0(E(n)_*E(n))$ is initial amongst maps with $L$-complete target, in this case it follows that $E(n)_*E(n) \to E(n)^\vee_*E(n)$ is injective.

Answer 1

No, this map is not injective.

To see this, put $W=\mathbb{W}(\mathbb{F}_{p^n})$, which is a free module of finite rank over $\mathbb{Z}_p$. It is standard that $\mathbb{Z}_p\otimes\mathbb{Z}_p$ contains a rational vector space of uncountable dimension, so the same is true of $W\otimes W$. (Here tensor products are implicitly taken over $\mathbb{Z}$ or $\mathbb{Z}_{(p)}$; you get the same anwswer either way.) This rational vector space can only map trivially to the group $E^\vee_*E$. Thus, it will suffice to check that $E_*E$ contains a copy of $W\otimes W$. The left and right unit maps provide a map $W\otimes W\to E_0E$. In the opposite direction, we can define a map $\phi:E_0\to W$ sending $u_i$ to $0$ for $0<i<n$. The resulting formal group law over $W$ has logarithm $\sum_kx^{p^{nk}}/p^k$, so the coefficients of the FGL lie in $\mathbb{Q}\cap W=\mathbb{Z}_{(p)}$. Because of this, the two resulting FGLs over $W\otimes W$ are the same. Thus, the standard universal property of $BP_*BP$ gives a ring map $$ E_*E = E_*\otimes_{BP_*}BP_*BP\otimes_{BP_*}E_* \to W\otimes W, $$ which is left inverse to our map in the opposite direction.