# Is the formal completion of an affine group necessarily a formal group?

Notation and Setting: let $$\operatorname{Aff}$$ denote the category of affine schemes whose objects are covariant representable functors $$\operatorname{X}:\operatorname{Ring}\rightarrow\operatorname{Set}$$ and $$\operatorname{Spec}:\operatorname{Ring^{op}}\rightarrow\operatorname{Func(Ring, Set)}$$ be the contravariant Yoneda embedding of $$\operatorname{Ring^{op}}$$ in its category of presheaves so that $$\operatorname{Aff}\simeq\operatorname{Ring^{op}}$$.

In addition, let $$\mathcal{O}:\operatorname{Func(Ring, Set)}\rightarrow\operatorname{Ring^{op}}$$ be the functor that sends a functor $$\operatorname{X}:\operatorname{Ring}\rightarrow\operatorname{Set}$$ to the ring of maps $$\operatorname{X}\rightarrow \mathbb{A}^1$$ (where $$\mathbb{A}^1$$ is the forgetful functor) so that $$\operatorname{Spec}$$ and $$\mathcal{O}$$ are inverse of one another.

Let $$\widehat{\operatorname{Aff}}$$ be the indization of $$\operatorname{Aff}$$, i.e. the category whose objects are functors $$\operatorname{X}:\operatorname{Ring}\rightarrow\operatorname{Set}$$ that are small filtered colimits of affine schemes

Let $$\operatorname{G}\in\operatorname{Aff}$$ be a group object (affine group), i.e $$\operatorname{(G(R),m_G)}$$ is a group for every ring R, whereas $$m_G$$ is the multiplication map $$m_G:\operatorname{G(R)}\times\operatorname{G(R)}\rightarrow\operatorname{G(R)}$$ and $$\mathcal{O}_G$$ the ring of maps $$\operatorname{G}\rightarrow\mathbb{A}^1$$.

Let $$I_0\supseteq I_1 \supseteq I_2 \supseteq \cdots$$ be an infinite sequence of ideals in $$\mathcal{O}_G$$, i.e $$I_n\in\mathcal{O}_G$$ for every $$n$$. From Martin's argument we know that the formal colimit $$\operatorname{Y}:=\mathrm{colimit}_{n\in\mathbb{Z}_{\geq0}}\operatorname{Spec(\mathcal{O}_{X}/I_n)}$$ exists and is not an affine scheme (not representable).

My general question: When is $$\operatorname{Y}$$ a formal group, i.e. a group object in $$\widehat{\operatorname{Aff}}$$?

My considerations were as follows: since $$\operatorname{Spec(\mathcal{O}_{G}/I_n)}\subseteq\operatorname{G}$$ is a subfunctor for every $$n$$, the universal property of $$\operatorname{Y}$$ as a colimit gives us a unique map $$\phi:\operatorname{Y}\rightarrow\operatorname{G}$$ which implies that $$\operatorname{Y}$$ is a subfunctor of $$\operatorname{G}$$ (since every two parallel maps to Y which have the same image on G are equal to one another). In particular, $$y\in\operatorname{Y(R)}$$ corresponds to a map $$f:\mathcal{O}_G\rightarrow R$$ such that $$I_n\subseteq\operatorname{Ker(f)}$$ for some $$n$$. For every ring R, we thus have $$\phi:\operatorname{Y(R)}\rightarrow\operatorname{G(R)}, y\mapsto y$$.

This fact, and the example of $$\operatorname{Nil}\simeq\mathrm{colimit}_{n\in\mathbb{Z}_{\geq0}}\operatorname{Spec(\mathbb{Z}[x]/(x)^{n})}$$ which delivers $$\operatorname{(Nil(R),+)}\subseteq\operatorname{(\mathbb{A}^1(R),+)}$$ as a subgroup, lead me to the claim:

$$\operatorname{Y}$$ is a formal group if and only if $$\operatorname{Y}$$ is a sub group of $$\operatorname{G}$$

Now $$\leftarrow$$ is trivial. To show $$\rightarrow$$, we let $$\operatorname{Y}$$ be a formal group with multiplication map $$m_Y$$. Now if it is true (this is the point I am not sure about)
that $$\phi$$ is a map of formal groups and not just a normal natural transforamtion then we have $$\phi(m_{Y}(y,y'))=m_{G}(\phi(y),\phi(y'))$$ which, since $$\phi(y)=y$$, is equivalent to $$m_{Y}=m_{G}$$ and we are done.

my specific questions:

1. Is the universal map $$\phi$$ necceserally a map of formal groups and if so why?

2. Assuming the answer to 1. is negative then:

2.1 In which cases is $$\phi$$ indeed a map of formal groups?

2.2 are there any examples of such situations where $$\operatorname{Y}$$ is not a subgroup of $$\operatorname{G}$$?

Edit: taking S. Carnahan's answer into consideration

Not assuming that $$\phi:\operatorname{Y(R)}\rightarrow\operatorname{G(R)}, y\mapsto y$$ is automatically a map of formal groups, it follows that that $$\phi$$ is a map of formal groups iff there is some $$k$$ such that the kernel of the map given by $$m_G(y,y')$$ lies in $$I_k$$.

Hopf ideals are kernals of maps of Hopf algebras. Thus saying that $$\operatorname{I}$$ is a Hopf ideal is equivalent to saying that the map $$\mathcal{O}_{G}\rightarrow\mathcal{O}_{G}/I$$ is a map of Hopf algebras which is equivalent to the map $$\operatorname{Spec(\mathcal{O}_{G}/I)}\rightarrow\operatorname{Spec(\mathcal{O}_{G})}$$ being a map of affine groups.

My question: How do we show that the property $$m_G(y,y')$$ lies in a certain $$I_k$$ is equivalent to the fact that $$I$$ is a Hopf ideal?

Things to take into consideration:

• We have that $$I_k\subseteq I$$.
• We know (Milne Algebraic Groups Pro. 3.12., p.67) that, given a Hopf ideal $$I$$, any Hopf algebra homomorphism $$A\rightarrow B$$ whose kernal contains $$I$$ factors uniquely through $$A\rightarrow A/I$$
• In the example of $$\operatorname{Nil}$$ we have for $$a\in\operatorname{Nil_n(R)}, b\in\operatorname{Nil_m(R)}, (a+b)\in\operatorname{Nil_{n+m}(R)}$$ and thus $$k=n+m$$. In addition, the map $$\mathbb{Z}[x]\rightarrow\mathbb{Z}[x]/(x)\simeq\mathbb{Z}$$ is a map of Hopf algebras whose kernal $$(x)$$ is hence a Hopf ideal. How to show the equivalence in this example?

Thanks

The universal map is not a map of formal groups without some extra condition. An easy class of counterexamples comes from completions of an affine group along a closed subscheme that does not contain the identity element. In general, when you want to complete an affine group to get a formal group, you set $$I_n = I^n$$ for an ideal $$I$$ that defines a closed subgroup. That is, $$I$$ is a Hopf ideal for the coordinate ring. In particular, $$\Delta(I) \subseteq \mathcal{O}_G \otimes I + I \otimes \mathcal{O}_G$$, so $$\Delta(I_n) \subseteq \sum_{i+j=n} I_i \otimes I_j$$.
To be more specific in the language you use, you need your system of ideals to be compatible with multiplication in the following way: If $$y$$ and $$y'$$ are $$R$$-points, such that the maps $$\mathcal{O}_G \to R$$ have kernel containing $$I_n$$ and $$I_{n'}$$, respectively, then there is some $$k$$ such that the kernel of the map given by $$m_G(y,y')$$ lies in $$I_k$$. You have not specified any structure that forces this condition to hold. For Hopf ideals, our calculation of $$\Delta(I_n)$$ shows that setting $$k \geq n+n'$$ is sufficient.
Here is an explicit example. We complete the additive group $$\mathbb{G}_a = \operatorname{Spec} k[x]$$ (for some nonzero commutative ring $$k$$) at the point $$1$$, so our ideals are $$I_n = (x-1)^n$$. Any $$R$$-point factors through some map $$k[x]/(x-1)^n \to R$$, but the product of such $$R$$-points comes from the composite $$k[x] \to k[y,z] \to k[y,z]/((y-1)^m(z-1)^n) \to R$$, where the first map takes $$x$$ to $$y+z$$, and hence $$(x-2)^{m+n} \mapsto ((y-1)+(z-1))^{m+n}$$. Then, this map necessarily factors through $$k[x]/(x-2)^{m+n}$$. Since $$2 \neq 1$$ in $$k$$, this map does not factor through $$k[x]/(x-1)^r$$ for any positive $$r$$.
• @sagirot $\Delta(I^n) = \Delta(I)^n$ in a Hopf algebra. – S. Carnahan Feb 28 at 13:54