What role does ring $R$ play in formal group?

Question

This is related to Iwasawa's Local Class Field Theory, Chpt 4, sec 1's Lemma 4.2 proof.

Let $R$ be a commutative $k-$algebra s.t. $char(k)=0$. Let $G_a(X,Y)=X+Y$ be a commutative formal group over $R$. Then one can show that $End(G_a)\cong R$ by $R\to End(G_a)$ through $r\to rX$.(This is not hard to achieve. Consider $f:G_a\to G_a$ s.t. $f(x+y)=f(x)+f(y)$ with $f\in (X)R[[X]]$. Take derivative against $y$ and set $x=0$. One gets $f'(y)=f(x)=a\in R$. Matching degrees on both sides to deduce vanishing of higher order terms. Thus $f(X)=aX$ only.)

$\textbf{Q:}$ What role does this $R$ play here? Consider a finitely generated Abelian group $G$. Clearly $End_Z(G)$ is a ring. In particular, $G$ is $End_Z(G)$ module. $G_a$ is merely abstraction of the additive map $G\times G\to G$. This $G_a$ is really a pair, $(G_a,R)$ where $R$ is the coefficient ring. How should I interpret ring $R$? Or should I interpret $G_a$ as a family of groups $G$'s addition law s.t. each $G_a$ is equipped with its corresponding $End_Z(G)$ for each $G$ where $R$ becomes $End_Z(G)$?

Answer 1

I think this is an oft-told tale, with many variants, as befits traditional story-telling, and I’m sure you can find out more pretty easily. Here I can give you only my own very partial account, and others will emphasize other aspects of the story.

The additive formal group law $\mathbf G_{\mathrm a}$ is just one of many formal group laws. While thinking of it, you might always keep in mind the other well-known group law over $\Bbb Z$, namely the multiplicative formal group law $\mathbf G_{\mathrm m}(X,Y)=X+Y+XY=(1+X)(1+Y)-1$. With the second formula, you see that if $n\in\Bbb Z$, there’s the $[n]$-endomorphism, amounting to combining $X$ with itself $n$ times (if $n\ge0$) using the group law. And explicitly, you get $[n](X)=(1+X)^n-1$, even when $n<0$.

Now, it’s a theorem that if your ring $R$ is a $\Bbb Q$-algebra, that is, when every $1/n\in R$, then every formal group law $G$ is isomorphic to $\mathbf G_{\mathrm a}$, and in particular the endomorphism ring will be isomorphic to $R$, by the map $\text{End}_R(G)\to R$, $f\mapsto f'(0)$.

But this is emphatically not true if $R$ is not so special; for instance, if $R$ is the integer-ring of a number field, and $G$ is the multiplicative formal group law, then $\text{End}_R(G)=\Bbb Z$, no bigger no matter how big the number field is.

How you should interpret a formal group law, even one so special as $\mathbf G_{\mathrm a}$, is up to you: different outlooks produce different theorems. Good luck.

Answer 2

It's easier to describe what's going on non-formally, so let me switch to the ordinary additive group scheme $\mathbb{G}_a$, not the formal one. Explicitly, over an arbitrary base ring $R$ the additive group scheme is

$$\mathbb{G}_{a, R} = \text{Spec } R[x]$$

equipped with the group operation $\mathbb{G}_{a, R} \times \mathbb{G}_{a, R} \to \mathbb{G}_{a, R}$ given by the polynomial $(x, y) \mapsto x + y$. In this case I think a more conceptual description of $\mathbb{G}_{a, R}$ is via its functor of points: namely, its functor of points sends an $R$-algebra $S$ to the underlying abelian group

$$\mathbb{G}_{a, R}(S) = (S, +)$$

of $S$. I think the cleanest way to think about this construction is that $\mathbb{G}_{a, R}$ itself is a "scheme-theoretic version of the free $R$-module $R$ on one generator," in the sense that we can also write $\mathbb{G}_{a, R}(S) = R \otimes_R S$, so the functor of points can be defined by extension of scalars starting from $R$ itself. This definition generalizes to $R$ being replaced by any finitely generated projective $R$-module $M$, but for more general modules we don't necessarily get a representable functor.

The functor of points makes precise the sense in which $\mathbb{G}_{a, R}$ is a family of abelian groups, namely it is the family of abelian groups $\mathbb{G}_{a, R}(S) = S$ given by the underlying additive groups of all $R$-algebras.

Now we can ask:

What is $\text{End}(\mathbb{G}_{a, R})$?

On the one hand, we can think of endomorphisms of $\mathbb{G}_{a, R}$ explicitly in terms of morphisms of $R$-algebras $R[x] \to R[x]$; this gives that an endomorphism is an additive polynomial over $R$, meaning a polynomial $f(x) \in R[x]$ satisfying $f(x + y) = f(x) + f(y)$ in $R[x, y]$. On the other hand, via the functor of points, an endomorphism of $\mathbb{G}_{a, R}$ is a natural family of endomorphisms

$$f_S : \mathbb{G}_{a, R}(S) \to \mathbb{G}_{a, R}(S)$$

or equivalently a natural family of endomorphisms $f_S : S \to S$ on the underlying additive groups of all $R$-algebras $S$ (we do not require that $f_S$ is $R$-linear). These endomorphisms are obtained from the polynomial $f$ by substituting values $x, y \in S$; to go from this to the explicit polynomial $f$ we use the Yoneda lemma.

The "obvious" additive polynomials are the linear ones $f(x) = rx, r \in R$; this gives an injection $R \hookrightarrow \text{End}(\mathbb{G}_{a, R})$. In terms of functors of points this is just the observation that by definition an $R$-algebra $S$ is in particular an $R$-module, and by definition morphisms of $R$-algebras are in particular morphisms of $R$-modules.

This perhaps already addresses your question. However it may be amusing to know that in general there can be more elements of $\text{End}(\mathbb{G}_{a, R})$. Setting $y = \varepsilon \in R[\varepsilon]/\varepsilon^2$, or equivalently quotienting by $y^2$, gives

$$f(x + \varepsilon) = f(x) + \varepsilon f'(x) = f(x) + f(\varepsilon) = f(x) + f(0) + \varepsilon f'(0)$$

so we get that $f(0) = 0$ and $f'(x) = f'(0)$ is a constant, say $r \in R$. However it does not follow that $f(x) = rx$! We get this conclusion if $R$ has characteristic $0$, but if $R$ has positive characteristic $n$ then, for example, $f(x)$ could have terms of the form $x^n$.

I'm not sure what happens if $R$ is an arbitrary commutative ring, but if $R$ is an algebra over a field $F$ of characteristic $p$ then the conclusion is that $f(x) = rx + g(x^p)$ where $g(x) \in R[x]$ is a polynomial satisfying $g(x^p + y^p) = g(x^p) + g(y^p)$, hence which is another additive polynomial. Continuing in this way we can show that $f(x)$ is a "polynomial in Frobenius"

$$f(x) = \sum_{k=0}^n r_k x^{p^k}$$

so in this case $\text{End}(\mathbb{G}_{a, R})$ is a polynomial ring $R[F]$ generated by Frobenius. This recently came up here.

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It sounds to me from your phrasing ("$\mathbb{G}_a$ is merely abstraction of the addition map") like you would like to interpret $\mathbb{G}_{a, R}$ as being something like an abstraction of the addition map on all $R$-modules $M$, not necessarily just the underlying additive groups of $R$-algebras.

There is an object carrying an addition operation which deserves to be called the "abstract addition map on all $R$-modules," but it is not a group scheme; instead it is just the forgetful functor $F : \text{Mod}(R) \to \text{Ab}$ from $R$-modules to abelian groups (while $\mathbb{G}_{a, R}$ was the forgetful functor from $R$-algebras to abelian groups). In the same way that a group scheme is a group object in the category of schemes, this functor $F$ is an abelian group object in the category of functors $\text{Mod}(R) \to \text{Set}$.

Now we can again ask: what is $\text{End}(F)$? An endomorphism of the functor $F$ is, similar to the above, a natural family of endomorphisms $f_M : M \to M$, but now defined on all $R$-modules. There is again an "obvious" choice given by scalar multiplication by $r \in R$, which again gives an inclusion $R \to \text{End}(F)$.

This time, this inclusion is always an isomorphism, for an arbitrary base ring $R$ (not necessarily commutative). This follows from the fact that $F$ is representable, by the free module $R$ on one generator, together with the Yoneda lemma again.

Going back to algebraic geometry, the conclusion I draw from all this is that the additive group scheme $\mathbb{G}_{a, R}$ is not an abstraction of the addition on arbitrary $R$-modules; it is really an abstraction of the addition on $R$-algebras specifically. Other group schemes have functors of points which make clearer use of the multiplication on $R$-algebras, such as the multiplicative group scheme

$$\mathbb{G}_{m, R}(S) = (S^{\times}, \cdot).$$

Also it is possible to understand arbitrary formal groups in this language but their functor of points are a bit more annoying to describe; I think one way to do it is to only define the functor of points on nilpotents.