# How is the $p$-adic Tate module of a formal group defined?

I am familiar with the definition of the $p$-adic Tate module of an elliptic curve defined over a $p$-adic field $k$ (a finite extension of $\mathbb{Q}_p$). But I have also seen some instances where the $p$-adic Tate module of a formal group is talked about. I wanted to know what the precise definition of $T_p(F)$ is, where $F$ is any formal group, like $T_p(\mathbb{G}_m)$. I could not find a satisfactory definition anywhere. Thank you.

The answer of Alex Youcis is perfect, and applies to formal groups of any dimension. Let me simplify for the one-dimensional case. This means your formal group has the shape $F(x,y)\in\mathfrak o[[x,y]]$ satisfying conditions that you evidently know, and where $\mathfrak o$ is probably the ring of integers in a finite extension of $\Bbb Q_p$ for some $p$.

As Alex says, to even have a Tate module, you need your formal group to be $p$-divisible, and for formal group laws of dimension one, this means that the $[p]$-endomorphism has a unit coefficient somewhere, the first such necessarily appearing in degree $p^h$ for some $h\ge1$. This number is called the height of $F$, as again you probably already know.

The important thing is that the first unit coefficient of $[p^n](x)$ will be in degree $p^{nh}$, and this means that according to Weierstrass Preparation, the roots of $[p^n](x)$ in the maximal ideal of the integers of the algebraic closure of $\Bbb Q_p$ are the same as the roots of a polynomial $P_n(x)\in\mathfrak o[x]$, where $\deg(P_n)=p^{nh}$. The roots are all of multiplicity one, as is easily seen, and this means that $[p^n]$ has kernel of order $p^{nh}$. The kernels fit together in just the way you know from the theory of elliptic curves, and you get a Tate module that’s a free $\Bbb Z_p$-module of rank $h$.

• I have just asked a question here math.stackexchange.com/questions/3153491/… about your "this means that the $[p]$-endomorphism has a unit coefficient" - statement above. I would be very thankful if you would answer it! – Layer Cake Mar 19 at 0:10

I've personally only seen this notion defined when $\mathbb{G}$ is $p$-divisible. This means, that if we're working over $A$ (some Noetherian local ring with residue characteristic $p$, e.g. $\overline{\mathbb{F}_p}$ or $\mathbb{Z}_p$) then the map

$$[p^n]^\ast:A[[T_1,\ldots,T_n]]\to A[[T_1,\ldots,T_n]]$$

is finite and free (i.e. makes the RHS into a finite free module over the left) for all $n$ (equiv. just for $n=1$).

So, now suppose that $\mathbb{G}$ is such a formal group. Then, define for all $n\geqslant 1$

$$\mathbb{G}[p^n]=\text{Spec}\left(\frac{A[[ T_1,\ldots,T_n]]}{([p^n]^\ast(T_1),\ldots,[p^n]^\ast(T_n))}\right)$$

which, one shows (although not too difficultly) is a finite flat group scheme over $A$ of rank $p^{nh}$ (if $h$ is the height of your formal group over the residue field of $A$).

These $\mathbb{G}[p^n]$ fit together, in the obvious way, to create a $p$-divisble group denoted $\mathbb{G}[p^\infty]$ of height $h$. This notion is important since Tate showed, in his famous article on $p$-divisible groups, that $\mathbb{G}\mapsto \mathbb{G}[p^\infty]$ is an equivalence between connected $p$-divisible groups over $A$ and $p$-divisible formal groups over $A$ which, as already mentioned, preserves height.

So with all this being said let's now assume that $A=k$ is a field. Then, one defines

$$T_p\mathbb{G}:= T_p\mathbb{G}[p^\infty]:=\varprojlim \mathbb{G}[p^n](\overline{k})$$

as per usual.