Some special subgroups of formal groups

Question

Let $G$ be a 1-dimensional, commutative formal group over a ring $R$. Give $G$ a coordinate $x$ and let $A\subset R$ be the subring generated by the coefficients of the corresponding formal group law $F(x,y)= \sum_{ij}a_{ij}x^iy^j$. So $G$ is really defined over $A$.

Call a finite subgroup $K\subset G$ special if it is the kernel of a homomorphism $T:G\rightarrow \phi^*G$ for some ring map $\phi:A\rightarrow R$. ($\phi^*G$ is the formal group over $R$ with formal group law $\phi^*F(x,y)=\sum_{ij}\phi(a_{ij})x^iy^j$.)

What's an example of a finite subgroup $K\subset G$ that is not (isomorphic to) a special subgroup? (I guess that would be the same as asking that $G/K$ does not inject into any $\phi^*G$.)

In all the cases I've tried (including the additive, multiplicative, and universal formal group laws) I seem to have convinced myself that all subgroups are special. Which leads me to suspect that maybe special subgroups are not so special.

$\textbf{Edit}$: To be explicit, a subgroup $K$ corresponds to a monic polynomial $f_K(x)\in R[x]$ with nilpotent lower order coefficients and such that $f(F(x,y))\equiv 0\ \text{mod}\ (f(x),f(y))$. That subgroup $K$ is special if furthermore there is some invertible power series $u(x)\in R[[x]]$ and a ring map $\phi:A\rightarrow R$ as above such that $u(F(x,y))f(F(x,y))=\phi^*F(u(x)f(x),u(y)f(y)).$

Answer 1

Let me specialize heavily to the case of formal groups (group laws) of dimension one over a $p$-adic ring $\mathfrak o$, i.e. the ring of integers of a finite extension $k$ of $\Bbb Q_p$.

I still am uncertain about what category you’re thinking of. If we restrict further to formal groups of finite height (the endomorphism $[p]$ being of finite degree $p^h$), then these things become $p$-divisible groups, or, if you like, ind-finite objects. For instance the kernel of $[p^n]$ will be a finite $\mathfrak o$-group-scheme, $K_n=\ker([p^n])=\mathrm{Spec}(\mathfrak o[[x]]/([p^n](x))\,)$, and you have natural maps $K_n\hookrightarrow K_{n+1}$, and you see that $\projlim\mathfrak o[[x]]/([p^n](x))\cong\mathfrak o[[x]]$. In this sense, your $G$, if indeed a formal group of finite height over $\mathfrak o$, is the union of its finite subgroups. This is the viewpoint that I tend to work with.

Now, let’s consider just one fairly simple case, where the formal group law has all its coefficients in an unramified extension $A$ of $\Bbb Z_p$, even in $\Bbb Z_p$ itself, and suppose the height is $h=2$ for simplicity. This means that $[p](x)\equiv px+ux^{p^2}\pmod{x^{p^2+1}}$, where $u$ is a unit of $A$, and the congruence ignores all terms in the power series of degree $>p^2$. Look at the Newton polygon and see that all the $z\in\overline k$ with $v_p(z)>0$ and $[p](z)=0$ have $v_p(z)=\frac1{p^2-1}$, plus of course $0$. So $p^2$ in all, and thus they form an elementary $p$-group of order $p^2$.

Now take any of the cyclic subgroups of $\ker[p]$, call it $\Gamma$. One proves that $$ \pi_\Gamma(x)=\prod_{\gamma\in\Gamma}F(x,\gamma)\,, $$ which is defined over a totally ramified extension $A'$ of $A$ (actually of degree $p+1$), is a morphism into another formal group, which I will abuse language in calling $G/\Gamma$.

I ask you to believe that I have shown you a formal group $G/\Gamma$ that, as far as I can see, will prove to you that $\Gamma$ is not a special subgroup of $G$, once you see that the formal group law of $G/\Gamma$ is not isomorphic to that of $G$, not even with a morphism $\varphi^*$ of the type you allow. (I think, because I’m not sure what properties you allow $\varphi^*$ to have.)

How do I know that $G/\Gamma$ is nothing like $G$? By Newtonian magic, you see that the Newton polygon of $[p]_{G/\Gamma}$ has vertices at $(1,1)$, $(p,\frac1{p+1})$, and $(p^2,0)$. The important fact is that this polygon is not the same as that of $[p]_G$; and since the shape of the Newton polygon of $[p]$ is an invariant, it follows that there is no way for $G/\Gamma$ to be isomorphic to $G$.

In the appropriate category, the map from $G$ to $G/\Gamma$ is onto. You can show, for instance, that if $v_p(\eta)>0$, there is $\xi$ in a finite extension of $k(\eta)$ such that $v_p(\xi)>0$ and $\pi_\Gamma(\xi)=\eta$.

(All of this is in an old and poorly-written paper of mine, Finite subgroups and isogenies…. EDIT:The “Newtonian magic” involves the “Newton copolygon”, also called the valuation function. I’ll bet a nickel that somebody else has explained it better than I can, but it’s in a later paper of mine, Canonical subgroups of formal groups, and I fear that it’s at most a little better-written than the other. Copolygon talk begins on p. 109.)

Answer 2

Consider the case of a formal group $G$ of finite height over a complete local Noetherian ring $R$ of residue characteristic $p>0$. For each $m$ there is a finite $R$-algebra $S$ that classifies finite subgroups of $G$ of order $p^m$ in the sense that $R$-algebra homomorphisms $S\to T$ biject with subgroup-schemes $A<\text{spec}(T)\times_{\text{spec}(R)}G$ such that $\mathcal{O}_A$ is free of rank $p^m$ over $T$. The structure of this classifying ring is described in my paper Finite subgroups of formal groups; there is another version on my home page with additional exposition. In another paper I showed how this ring arises from a calculation in algebraic topology, in the case where $G$ is the universal deformation of a formal group $G_0$ over a finite field $F$. In the case where $F$ is of prime order, every finite subgroup of $G_0$ is the kernel of a power of Frobenius, and we can use the universal deformation property to deduce a kind of specialness property for $G$. Specifically, given $\alpha\colon R\to T$ and a finite subgroup $A<\alpha^*G$ there is another map $\beta\colon R\to T$ with $(\alpha^*G)/A\simeq\beta^*G$. There is a similar but slightly more involved statement in the case where $|F|$ is not prime. In algebraic topology this is all closely bound up with the theory of power operations in $H_\infty$ ring spectra. This is explained in a paper by Charles Rezk. The similar specialness property of Lazard's universal FGL is similarly bound up with the $H_\infty$ structure of the complex cobordism spectrum $MU$, via Quillen's fundamental theorem that the homotopy ring $\pi_*(MU)$ is canonically isomorphic to the Lazard ring.