Hyperalgebra of a group scheme

Question

The following definitions are from Mumford's book on Abelian Varieties:

A group scheme $G$ is a finite type scheme over an algebraically closed field $k$ with morphisms $m:G\times G\rightarrow G$, $e:\operatorname{Spec}(k):\rightarrow G$ and $i:G\rightarrow G$ such that for any finite type scheme $S$ over k, the maps $m$, $e$ and $i$ induce a group structure on $G(S)= Hom_{\operatorname{Spec}(k)}(S, G)$.

For any closed $x\in G$ we define $$\textbf{H}_x= Hom_{cont}(\mathcal{O}_{x,G}, k)$$ Now, the hyperalbegra $\textbf{H}$ of $G$ is defined as: $$\textbf{H}=\bigoplus_{\operatorname{closed}x\in G}\textbf{H}_x$$ where $Hom_{cont}$ is the are the maps $L:\mathcal{O}_{x,G}\rightarrow k$ which are continuous in the sense that $L(m_x^{n}) = (0)$ for some natural number $n$ (i.e. continuous in the respective m-adic topologies).

My question is the following. What sort of continuous maps are being considered? Are they morphisms between vector spaces or morphism between groups?

I believe the answer is vector spaces because the following is stated immediately after:

$$\textbf{H} = \varinjlim_{Z\in \Sigma}\Gamma(\mathcal{O}_Z)^*$$ where $\Sigma$ is the collection of $0$-dimensional subschemes of $G$ and $W^*$ is the dual of any $k$-vector space $W$.

If my guess is incorrect, could someone please help me with the proof of the equality stated above?

Edit: I believe I see why my question might now be clear enough! My doubts would be answered if someone could help prove the equality mentioned above: $$\textbf{H} = \varinjlim_{Z\in\Sigma}\Gamma(\mathcal{O}_Z)^*$$

Thank you!

Answer 1

An element of $\mathbf{H}_x$ is a continuous morphism of $k$-vector spaces from $\mathcal{O}_{G,x}$ to $k$. To see that $\mathbf{H}=\bigoplus_{x\in G_{(0)}}\mathbf{H}_x$ is isomorphic to the direct limit stated in the question, we start from the obvious equality $$ \mathbf{H}=\varinjlim_{S\subset G_{(0)}\text{ finite}}\varinjlim_{n\geq 1}\bigoplus_{x\in S}\operatorname{Hom}_{k\text{-vect}}(\mathcal{O}_{G,x}/\mathfrak{m}_{G,x}^n,k). $$ Setting $Z_{S,n}=\operatorname{Spec} \prod_{x\in S}\mathcal{O}_{G,x}/\mathfrak{m}_{G,x}^n \subset X$, we can rewrite this as $$ \mathbf{H}=\varinjlim_{S\subset G_{(0)}\text{ finite}}\varinjlim_{n\geq 1}\Gamma(Z,\mathcal{O}_Z)^\vee. $$ Now it suffices to show that $\{Z_{S,n}\mid S\subset G_{(0)}\text{ finite, }n\geq 1\}$ is cofinal in the set of $0$-dimensional closed subschemes of $G$. This is not difficult, so I leave it as an exercise.

Answer 2

Any local ring $A$ with maximal ideal $I$ can be endowed with the $I$-adic topology, the topology where the zero element $0$ has as fundamental basis of neighborhoods the powers of the ideal $I$. This topology is also called "maximal ideal topology".

$\mathcal{O}_{x,G}$ is a local ring with maximal ideal $m_x$ and $k$ is also a local ring with maximal ideal $(0)$. Hence $\mathcal{O}_{x,G}$ and $k$ can both be endowed with their respective maximal ideal topologie, and $L:\mathcal{O}_{x,G}\rightarrow k$ said to be continuous if it is continuous for these topologies.

This amounts, as $L$ is a ring morphism, to be only continuous at the point zero $0$ of $k$, which amount to ask that the inverse image of the open set $(0)$ of $k$ is open as well in $\mathcal{O}_{x,G}$, hence contains a power $\mathcal{O}_{x,G}^n$ of $\mathcal{O}_{x,G}$, which implies that $L(m_x^{n}) = (0)$.