---
address:
- 'Departamento de Matemáticas, Universidad de Extremadura, Avenida de Elvas s/n, 06071 Badajoz, Spain'
- 'Departamento de Matemáticas, Universidad de Salamanca, Plaza de la Merced 1-4, 37008 Salamanca, Spain'
- 'Departamento de Matemáticas, Universidad de Extremadura, Avenida de Elvas s/n, 06071 Badajoz, Spain'
author:
- Amelia Álvarez
- Carlos Sancho
- Pedro Sancho
date: 'June, 2007'
title: Functorial Cartier Duality
---

Introduction {#introduction .unnumbered}
============

In this paper we obtain the Cartier duality for $k$-schemes of commutative monoids functorially without providing the vector spaces of functions with a topology (as in [@demazure Exposé VII$_B$ by P. Gabriel, 2.2.1]), generalizing a result for finite commutative algebraic groups by M. Demazure $\&$ P. Gabriel ([@Demazure II, $\S$1, 2.10]).

All functors we consider are functors defined over the category of commutative $k$-algebras. Given a $k$-module $E$, we denote by ${\bf E}$ the functor of $k$-modules ${\bf E}(B):= E \otimes_k B$. Given two functors of $k$-modules $F$ and $H$, ${\bf Hom}_k(F,H)$ will denote the functor of $k$-modules $${\bf Hom}_k(F,H)(B)={\rm
Hom}_{B}(F_{|B}, H_{|B})$$ where $F_{|B}$ is the functor $F$ restricted to the category of commutative $B$-algebras. We denote $F^* := {\bf Hom}_k (F ,{\bf k})$. It holds that ${\bf E}^{**} =
{\bf E}$ ([@Amel 1.10]). Given ${\bf E}$ there exists a $k$-module $V$ such that ${\bf E} = {\bf V}^*$ if and only if $E$ is a projective $k$-module of finite type.

Given a functor of commutative $k$-algebras ${\mathcal A}$, we define ${\rm Spec}\, {\mathcal A}$ to be the functor ${\rm Spec}\,
{\mathcal A}(B) := {\rm Hom}_{k-alg}({\mathcal A}, {\bf B})$. Let $X={\rm Spec}\, A$ be a $k$-scheme and let $X^\cdot$ be its functor of points ($X^\cdot(B):={\rm Hom}_{k-alg}(A,B)$). It is easy to see that $X^\cdot ={\rm Spec}\, {\bf A}$. If ${\bf C}^*$ is a functor of commutative algebras, we prove (Theorem \[1.5\]) that ${\rm Spec}\, {\bf C} ^*$ is a direct limit (of functor of points) of finite $k$-schemes.

If ${\mathcal A} = {\bf A}$, in [@Amel 3.4] we proved that the linear enveloping of ${\rm Spec}\, {\mathcal A}$ is ${\mathcal
A}^*$, that is, $${\rm Hom}_{functors} ({\rm Spec}\, {\mathcal A},
H) = {\rm Hom}_k ({\mathcal A}^*,H)$$ for all dual functors of $k$-modules $H$ (i.e., $H= F^*$). If in addition $G= {\rm Spec}\,
{\mathcal A}$ is an affine $k$-group, in [@Amel 5.4] we proved that the algebraic enveloping of ${\rm Spec}\, {\mathcal A}$ is ${\mathcal
A}^*$ again, that is, $${\bf Hom}_{monoids} ({\rm Spec}\,
{\mathcal A}, {\mathcal B}) = {\bf Hom}_{k-alg} ({\mathcal A}^*,
{\mathcal B})$$ for all dual functors of $k$-algebras ${\mathcal
B}$. As a consequence, we obtained that the category of $G$-modules is equivalent to the category of ${\mathcal A}^*$-modules.

In this paper we prove simultaneously the same results for ${\mathcal A} = {\bf A}, {\bf C}^*$ (Corollary \[2.3\], Proposition \[2.5\], Theorem \[Gmodulos\]). In particular, $${\bf Hom}_{functors} ({\rm Spec}\, {\mathcal A}, {\bf k}) = {\bf Hom}_k ({\mathcal A}^*, {\bf k} ) = {\mathcal A}.$$ Moreover, if ${\rm Spec}\, {\mathcal A}$ is a functor of abelian monoids, then we get $${\bf Hom}_{monoids} ({\rm Spec}\,{\mathcal A}, {\bf k}) = {\bf Hom}_{k-alg} ({\mathcal A}^*,{\bf k}) = {\rm Spec}\, {\mathcal A}^*$$ where ${\bf k}$ is regarded as a functor of monoids with the multiplication operation.

Given a functor of monoids $H$, we say that $H^* = {\bf
Hom}_{monoids} (H,{\bf k})$ is the dual functor of monoids of $H$ and we prove the following theorem.

The category of abelian affine monoids $G={\rm Spec}\, A$ is anti-equivalent to the category of functors ${\rm Spec}\, {\bf
C}^*$ of abelian monoids. The functors giving the anti-equivalence are the ones that assign to each functor of monoids its dual functor of monoids.

In particular, $G^{**}=G$.

Finally, we get three immediate applications of the functorial Cartier duality:

1.  The tangent space to $G^*$ at the identity element is isomorphic to the vector space $V$ of linear functions of $G$.

2.  If ${\rm car}\, k =0$ and $G$ is a unipotent abelian $k$-group, then it is isomorphic to the dual of $V$.

3.  If $G$ is semisimple, then the transposed Fourier transform of $G^*$ is the inverse of the Fourier transform of $G$.

Spectrum of a functor of algebras
=================================

The basic references for reading this paper are [@Amel] and [@carlos].

Let $k$ be a commutative ring with unit. If $E$ is a $k$-module we will say that ${\bf E}$ is a cuasi-coherent $k$-module. The category of $k$-modules is equivalent to the category of cuasi-coherent $k$-modules ([@Amel 1.12]). In particular, ${\rm Hom}_k ({\bf E},{\bf E'}) = {\rm Hom}_k (E,E')$. ${\bf E}^*
= {\bf Hom}_k ({\bf E},{\bf k})$ is the functor of points of the scheme ${\rm Spec}\,S_k^\cdot E$ and we say that it is a $k$-module scheme. As it holds that ${\bf E}^{**} = {\bf E}$, the category of cuasi-coherent modules is anti-equivalent to the category of module schemes ([@Amel 1.12]) .

Throughout this paper we assume that ${\bf A}^*$ is a commutative $k$-algebra scheme (equivalently, $A$ is a cocommutative $k$-coalgebra, [@Amel 4.2]). From Proposition $1.5$ on, we will always assume that $A$ is a projective $k$-module.

Given a functor of $k$-algebras $\mathcal{A}$, the functor ${\rm Spec}\, \mathcal{A}$, “spectrum of $\mathcal{A}$”, is defined to be $$({\rm Spec}\, \mathcal{A})(B) := {\rm Hom}_{k-alg}
(\mathcal{A}, {\bf B})$$ for each commutative $k$-algebra $B$.

If $C$ is a commutative $k$-algebra, then $${\rm Spec}\, {\bf C} =
({\rm Spec}\, C)^\cdot .$$

Let $X$ be a set. Let us consider the discrete topology on $X$. Let $\mathcal{X}$ be the functor, which we will call the constant functor $X$, defined by $${\mathcal X}(B) := {\rm Aplic}_{cont.}
({\rm Spec}\, B, X)$$ for each commutative $k$-algebra $B$.

Let ${\mathcal A}_X$ be the functor of algebras defined by $${\mathcal A}_X(B) := {\rm Aplic}(X,B) = \prod_X B$$ for each commutative $k$-algebra $B$. Let us observe that ${\mathcal A}_X = \underset{X}{\prod} {\bf k}
=(\underset{X}{\oplus} {\bf k})^*$ is a commutative algebra scheme.

Let us prove that ${\rm Spec}\, {\mathcal A}_X = {\mathcal X}$.

The morphisms of functors of $k$-algebras are, in particular, morphisms of functors of $k$-modules $$({\rm Spec}\, {\mathcal
A}_X )(B) = {\rm Hom}_{k-alg} ({\mathcal A}_X, {\bf B} ) \subset
{\rm Hom}_k ( {\mathcal A}_X, {\bf B} )= {\rm Hom}_k ( \underset{x
\in X}{\prod} {\bf k}, {\bf B} )$$ and from [@Amel 4.5] we know that morphisms of functors of $k$-modules from $\underset{X}{\prod} {\bf k}$ to ${\bf B}$ factorize through the projection onto a finite number of factors. Therefore, given a morphism $\underset{X}{\prod} {\bf k} \to {\bf B}$ there exists a subset $Y \subset X$ of finite order, such that we get the factorization $$\underset{X}{\prod} {\bf k} \to
\underset{Y}{\prod} {\bf k} \to {\bf B}.$$ Every continuous morphism ${\rm Spec}\, B \to X$ has finite image, because ${\rm
Spec}\, B$ is compact and $X$ is discrete. We can assume that $X$ is a finite set.

Thus, we get

$$\aligned {\rm Hom}_{k- alg}(\prod_X {\bf k,B})& ={\rm Hom}_{k- alg}(\prod_X k,B)\\
&  = {\rm Cont. Aplic.}_{{\rm Spec}\, k}({\rm Spec}\, B, {\rm Spec}\,
\prod_X k = \coprod_X {\rm Spec}\, k) \\&  = {\rm
Aplic}_{cont.}({\rm Spec}\, B, X) . \endaligned$$

Later we will see that the ring of functions of ${\rm Spec}\,
{\mathcal A}_X$ coincides with ${\mathcal A}_X$, that is, $${\bf Hom}_{functors}({\rm Spec}\, {\mathcal
A}_X , {\bf k}) = {\mathcal A}_X .$$

Let $\mathcal A$ a functor of commutative algebras. Then, ${\rm
Spec}\,{\mathcal A}={\bf Hom}_{k-alg}({\mathcal A},{\bf k})$.

By the adjoint functor formula ([@Amel 1.15]) (restricted to the morphisms of algebras) it holds that $$\aligned {\bf Hom}_{k- alg} ({\mathcal A}, {\bf k}) (B) &
= {\rm Hom}_{B- alg} ({\mathcal A}_{|B}, {\bf B})
= {\rm Hom}_{k-alg} ({\mathcal A}, {\bf B}) \\
& = ({\rm Spec}\, {\mathcal A}) (B) . \endaligned$$

Therefore, ${\rm Spec}\,{\bf A^*}={\bf Hom}_{k-alg} ({\bf A}^*,{\bf
k}) \subset {\bf Hom}_{k}({\bf A}^*,{\bf k}) = {\bf A}$.

If $E$ is a finitely-generated $k$-module, we will say that ${\bf
E}$ is coherent.

\[1.5\] Let ${\bf A}^*$ be a commutative algebra scheme and let us assume that $A$ is a projective $k$-module. By [@Amel 4.12], ${\bf A}^* = \underset{\underset{i}{\longleftarrow}}{\lim}\, {\bf
A}_i$, where ${\bf A}_i$ are the coherent algebras that are quotients of ${\bf A}^*$. It holds that $${\rm Spec}\,{\bf A}^* =
\underset{\underset{i}{\longrightarrow}} {\lim}\, {\rm Spec}\,
{\bf A}_i.$$

By [@Amel 4.5], the image of every $k$-linear morphism ${\bf
A}^* \to {\bf B}$ is a coherent $k$-module; therefore, every morphism of functors of $k$-algebras ${\bf A}^* \to {\bf B}$ factorizes through a coherent algebra ${\bf A}_i$ that is a quotient of ${\bf A}^*$. Then,

$$\begin{split} ({\rm Spec}\, {\bf A^*})(B) & ={\rm
Hom}_{k-alg} ({\bf A}^*, {\bf B}) =
\underset{\underset{i}{\longrightarrow}} {\lim}\, {\rm Hom}_{k-
alg} ({\bf A}_i, {\bf B}) \\ & =
\underset{\underset{i}{\longrightarrow}}{\lim}\,({\rm Spec}\, {\bf
A}_i) (B) .
\end{split}$$

Let ${\mathcal B}$ be a functor of commutative $k$-algebras. Then,

1.  ${\rm Hom}_{functors}({\rm Spec}\, {\bf A},{\rm Spec}\, {\mathcal
    B} ) = {\rm Hom}_{k-alg}({\mathcal B},{\bf A}).$

2.  ${\rm Hom}_{functors}({\rm Spec}\, {\bf A}^*,{\rm Spec}\,
    {\mathcal B} ) = {\rm Hom}_{k-alg}({\mathcal B},{\bf A}^*).$

It holds that ${\rm Hom}_{functors} (({\rm Spec}\, A)^\cdot, F) =
F(A)$ for every functor $F$, by Yoneda’s lemma.

1.  $$\aligned {\rm Hom}_{functors}({\rm Spec}\, {\bf A},{\rm Spec}\, {\mathcal B})
    & = {\rm Hom}_{functors} ( ({\rm Spec}\, {A})^\cdot,{\rm Spec}\, {\mathcal B}) \\
    & ={\rm Hom}_{k-alg} ({\mathcal B},{\bf A}) . \endaligned$$

2.  $$\aligned {\rm Hom}_{functors}({\rm Spec}\, {\bf A}^*,{\rm
    Spec}\, {\mathcal B}) & ={\rm Hom}_{functors}
    (\underset{\underset{i}{\longrightarrow}} {\lim}\, {\rm Spec}\, {\bf
    A}_i,{\rm Spec}\, {\mathcal B}) \\ & =
    \underset{\underset{i}{\longleftarrow}}{\lim}\,{\rm Hom}_{functors}
    ({\rm Spec}\, {\bf A}_i,{\rm Spec}\, {\mathcal B})
    \\ & = \underset{\underset{i}{\longleftarrow}}
    {\lim}\,{\rm Hom}_{k-alg} ({\mathcal B},{\bf A}_i) \\
    & = {\rm Hom}_{k-alg} ({\mathcal B},{\bf A}^*) .
    \endaligned$$

For ${\mathcal A}=\bf A, A^*$, it holds that $${\bf Hom}_{functors}({\rm Spec}\, {\mathcal A}, {\bf k}) = {\mathcal A} .$$

$$\aligned {\bf Hom}_{functors}({\rm Spec}\, {\mathcal A}, {\bf
k}) & = {\bf Hom}_{functors}({\rm Spec}\, {\mathcal A}, {\rm
Spec}\,{\bf k[x]}) \\ & = {\bf Hom}_{k-alg}({\bf k[x],
}{\mathcal A})= {\mathcal A} . \endaligned$$

Explicitly, ${\mathcal A}= {\bf Hom}_{functors} ({\rm Spec}\,
{\mathcal A}, {\bf k})$, $w\mapsto \tilde{w}$, where $\tilde{w}(\phi) = \phi(w)$, for every $\phi \in {\rm Spec}\,
{\mathcal A}= {\bf Hom}_{k-alg} ({\mathcal A}, {\bf k})$.

Closure of dual modules and algebras
====================================

We will say that a functor of $k$-modules $F$ is dual if there exists a functor of $k$-modules $H$ such that $F \simeq H^*$.

Let $F$ be a functor of $k$-modules such that $F^*$ is a reflexive functor. The closure of dual functors of $k$-modules of $F$ is $F^{**}$, that is, it holds the functorial equality $${\rm
Hom}_k(F,G)={\rm Hom}_k(F^{**},G)$$ for every dual functor of $k$-modules $G$.

Let us write $G=H^*$, then $$\aligned {\rm Hom}_k(F,G) & ={\rm Hom}_k(F,H^*)={\rm Hom}_k(H,F^*)={\rm
Hom}_k(H,F^{***}) \\ & ={\rm Hom}_k(F^{**},H^*)={\rm Hom}_k(F^{**},G) . \endaligned$$

Let $F$ be a functor of sets. We will denote by $k[F]$ the functor over the category of commutative $k$-algebras defined by $k[F](B)= \{$formal finite $B$-linear combinations of $f \in F(B) \}$.

\[2.3\] Let $\mathcal A$ be a functor of commutative $k$-algebras. Let us assume that ${\bf Hom}_{functors}({\rm Spec}\, {\mathcal A}, {\bf
k})= \mathcal A$ and that $\mathcal A$ is a reflexive functor of $k$-modules.Then, the closure of dual functors of $k$-modules of $k[{\rm Spec}\, {\mathcal A}]$ is ${\mathcal A}^*$, that is, $${\rm Hom}_{functors}({\rm Spec}\, {\mathcal A}, F) = {\rm Hom}_k(k[{\rm
Spec}\, {\mathcal A}], F) = {\rm Hom}_k({\mathcal A}^*, F)$$ for every dual functor of $k$-modules $F$.

As $k[{\rm Spec}\, {\mathcal A}]^*={\bf Hom}_{functors}({\rm
Spec}\, {\mathcal A}, {\bf k})=\mathcal A$, the closure of dual functors of $k$-modules of $k[{\rm Spec}\, {\mathcal A}]$ is $k[{\rm Spec}\, {\mathcal A}]^{**}={\mathcal A}^*$.

\[2.4\] Let $F$ be a functor of $k$-algebras such that $F^*$ is a reflexive functor of $k$-modules. The closure of dual functors of $k$-algebras of $F$ is $F^{**}$, that is, it holds the functorial equality $${\rm Hom}_{k-alg}(F,G)={\rm Hom}_{k-alg}(F^{**},G)$$ for every dual functor of $k$-algebras $G$. In particular, ${\rm
Spec}\, F= {\rm Spec}\, F^{**}$.

Let us observe that $$\aligned (F \otimes \overset{n}{\ldots} \otimes F)^*  =
& {\rm Hom}_k(F \otimes \overset{n}{\ldots} \otimes F,{\bf k})
= {\rm Hom}_k (F,(F \otimes \overset{n-1}{\ldots} \otimes F)^*)\\
\underset{Induction}= & {\rm Hom}_k (F, (F^{**} \otimes
\overset{n-1}{\ldots} \otimes F^{**})^*) = {\rm Hom}_k
(F^{**},(F^{**} \otimes \overset{n-1}{\ldots} \otimes F^{**})^*)\\
= & (F^{**} \otimes \overset{n}{\ldots} \otimes F^{**})^*
.\endaligned$$ Therefore, given a dual functor of $k$-modules $S^*$, $$\aligned {\rm Hom}_k (F \otimes \ldots \otimes F, S^*) &
={\rm Hom}_k(S, (F \otimes \ldots \otimes F)^*) \\ & = {\rm
Hom}_k(S, (F^{**} \otimes \ldots \otimes F^{**})^*)\\ & = {\rm
Hom}_k (F^{**} \otimes \ldots \otimes F^{**}, S^*) .\endaligned$$

If we consider $S^*=F^{**}$, it follows easily that the structure of algebra of $F$ defines a structure of algebra on $F^{**}$. Finally, if we consider $S^*= G$, we obtain that ${\rm Hom}_{k-alg}(F,G)={\rm
Hom}_{k-alg}(F^{**},G)$.

Let us observe that if $F$ is also a functor of commutative algebras, then so is $F^{**}$: since ${\rm Hom}_k (F \otimes F,
F^{**}) = {\rm Hom}_k (F^{**} \otimes F^{**}, F^{**})$, the morphism $F \otimes F \to F$, $f \otimes f'\mapsto ff'-f'f=0$ extends to a unique morphism $F^{**} \otimes F^{**} \to F^{**}$ (which is $f \otimes f'\mapsto ff'-f'f =0 $).

Let ${\bf A}^*$ and ${\bf B}^*$ be commutative $k$-algebra schemes. By [@Amel 1.8], we have that $({\bf A}^* \otimes_k
{\bf B^*})^* = {\bf Hom}_k ({\bf A^*} \otimes_k {\bf B^*},{\bf k})
= {\bf Hom}_k ({\bf A^*},{\bf B}) = {\bf A} \otimes_k {\bf B}$. Then, ${\rm Spec}\,{\bf A}^* \times {\rm Spec}\,{\bf B}^*= {\rm
Spec}\, ({\bf A}^* \otimes_k {\bf B}^*)
\stackrel{\text{\ref{2.4}}}{=} {\rm Spec}\, ({\bf A} \otimes_k
{\bf B})^* $.

Let $\mathcal A$ be a functor of $k$-algebras and let us assume that $G= {\rm Spec}\, \mathcal A$ is a functor of monoids. The functor of $k$-modules $k[G]$ is obviously a functor of $k$-algebras. Given a functor of $k$-algebras $\mathcal B$, it is easy to check the equality $${\rm Hom}_{monoids} (G, {\mathcal B})
= {\rm Hom}_{k-alg} (k[G], \mathcal B) .$$

\[2.5\] Let $G= {\rm Spec}\, \mathcal A$ be a functor of monoids. Let us assume that ${\bf Hom}_{functors}(G , {\bf k}) = \mathcal A$ and that $\mathcal A$ is a reflexive functor of $k$-modules. Then the closure of dual functors of algebras of $k[G]$ coincides with $\mathcal A^*$, that is,

$${\rm Hom}_{monoids}(G, {\mathcal B}) = {\rm Hom}_{k-alg}(k[G],
{\mathcal B}) = {\rm Hom}_{k-alg}({\mathcal A^*}, {\mathcal B})$$ for every dual functor of $k$-algebras $\mathcal B$.

$k[G]^*=\mathcal A$ is reflexive, then the closure of dual functors of algebras of $k[G]$ is $\mathcal A^*$.

\[Gmodulos\] Let $G= {\rm Spec}\, \mathcal A$ be a functor of monoids. Let us assume that ${\bf Hom}_{functors} ({\rm Spec}\, {\mathcal A}, {\bf
k}) = \mathcal A$ and that $\mathcal A$ is a reflexive functor of $k$-modules. The category of $G$-modules is equivalent to the category of ${\mathcal A}^*$-modules. Likewise, the category of dual functors of $G$-modules is equivalent to the category of dual functors of ${\mathcal A}^*$-modules.

Let $E$ be a $k$-module. Let us observe that ${\bf End}_k ({\bf
E}) = ({\bf E}^* \otimes {\bf E})^*$ is a dual functor. Therefore, $${\rm Hom}_{k-alg} (k[G],{\bf End}_k({\bf E})) = {\rm
Hom}_{k-alg} ({\mathcal A}^*, {\bf End}_k({\bf E})) .$$ In conclusion, endowing $E$ with a structure of $G$-module is equivalent to endowing $E$ with a structure of ${\mathcal
A}^*$-module.

Let us observe that ${\rm Hom}_k (k[G], {\bf E}) = {\rm Hom}_k
({\bf E}^*, {\mathcal A})= {\rm Hom}_k ({\mathcal A}^*, {\bf E})$. Hence, given any two $G$-modules (or ${\mathcal A}^*$-modules) $E$, $E'$, a linear morphism $f: E \to E'$ and $e \in E$, we will have that the morphism $f_1: k[G] \to {\bf E}'$, $f_1(g) := f(ge)
- gf(e)$ is null if and only if the morphism $f_2: {\mathcal A}^*
\to {\bf E}'$, $f_2(a) := f(ae) - a f(e)$ is null. In conclusion, ${\rm Hom}_{G-mod} ({\bf E}, {\bf E'}) = {\rm Hom}_{{\mathcal
A}^*} ({\bf E}, {\bf E'})$.

Let us highlight that the structure of functor of algebras of $\mathcal A^*$ is the one that makes the embedding ${\rm
Spec}\,{\mathcal A}\hookrightarrow \mathcal A^*$ be a morphism of monoids.

The ${\mathbb C}$-linear representations of ${\mathbb{Z}}$ are equivalent to the ${\mathbb C}[{\mathbb{Z}}]$-modules. ${\mathbb C}[{\mathbb{Z}}]={\mathbb
C}[x,1/x]$, $n\mapsto x^n$, is a principal ideal domain. Thus, if $E$ is a finite ${\mathbb C}$-linear representation of ${\mathbb{Z}}$, then $$E= \underset{\alpha\neq 0, n_\alpha>0}{\oplus} {\mathbb
C}[x]/(x-\alpha)^{n_\alpha}$$ such that $n\cdot (\bar
p_{n_\alpha})_{n_\alpha}=(\overline{x^n\cdot
p_{n_\alpha}})_{n_\alpha}$.

\[2.9\] Let $G = {\rm Spec}\, {\mathcal A}$ be a functor of monoids. Let us assume that $ {\bf Hom}_{functors} ( {\rm Spec}\, {\mathcal A},
{\bf k}) = {\mathcal A}$ and that ${\mathcal A}$ is a reflexive functor of $k$-modules. Let ${\mathcal V}$ be a dual functor of $G$-modules (on the left and on the right). Then, $$Der_k
({\mathcal A}^*, {\mathcal V}) = Der(G,{\mathcal V})$$ where $Der(G, {\mathcal V}) := \{ D \in {\rm Hom}_{functors} (G, {\mathcal
V}) : D(g \cdot g') = D(g) \cdot g' + g \cdot D(g') \}$.

$$\aligned Der_k ({\mathcal A}^*, {\mathcal V}) & =
\{ f \in {\rm Hom}_{k-alg} ({\mathcal A}^*, {\mathcal A}^*
\oplus {\mathcal V} \epsilon), \epsilon^2 = 0, f(a)= a \, {\rm mod}
(\epsilon) \} \\ & = \{ f \in  {\rm Hom}_{monoids} (G, {\mathcal A}^*
\oplus {\mathcal V} \epsilon), \epsilon^2 =0,  f(g) =
g \, {\rm mod} (\epsilon) \} \\ & = Der(G, {\mathcal V}) . \endaligned$$

Functorial Cartier Duality
==========================

Given a functor of abelian monoids $G$, where we regard ${\bf k}$ as a monoid with the multiplication operation, $$G^* := {\bf
Hom}_{monoids} (G,{\bf k})$$ will be referred to as the dual monoid of $G$.

If $G$ is a functor of groups, then $G^* = {\bf Hom}_{groups} (G, G_m^{\cdot})$.

\[dual\] Assume that ${\rm Spec}\, {\mathcal  A}$ is a functor of abelian monoids, $\mathcal A$ is a reflexive functor of $k$-modules and ${\bf
Hom}_{functors}({\rm Spec}\, {\mathcal  A},{\bf k})=\mathcal A$. Then

$${\rm Spec}\, {\mathcal  A}^*=({\rm Spec}\, {\mathcal  A})^*$$ (in particular, this equality shows that ${\rm Spec}\,\mathcal
A^*$ is a functor of abelian monoids).

$$\aligned {\rm Spec}\, {\mathcal  A}^* & = {\bf Hom}_{k-alg}
({\mathcal  A}^*,{\bf k}) = {\bf Hom}_{monoids} ({\rm Spec}\,
{\mathcal A},{\bf k})\\ & = ({\rm Spec}\, {\mathcal A})^* .\endaligned$$

\[comentario\] Explicitly, ${\rm Spec}\, {\mathcal A}^* = {\bf Hom}_{monoids} ({\rm
Spec}\, {\mathcal A},G_m^\cdot)$, $\phi \mapsto \tilde{\phi}$, where $\tilde{\phi} (x)$ $ = \phi(x)$, for every $\phi \in {\rm
Spec}\, {\mathcal A}^* = {\bf Hom}_{k-alg} ({\mathcal  A}^*,{\bf
k})$ and $x \in {\rm Spec}\, {\mathcal A} \subset \mathcal A^*$.

Let $G= {\rm Spec}\, {\mathcal A}$. $G^*$ is a functor of abelian monoids ($(f \cdot f')(g) := f(g) \cdot f'(g)$, for every $f,f'\in
G^*$ and $g \in G$), the inclusion $G^* = {\bf Hom}_{monoids}
(G,{\bf k}) \subset {\bf Hom}_{functors} (G,{\bf k})={\mathcal A}$ is a morphism of monoids and the diagram $$\xymatrix{ ({\rm
Spec}\, {\mathcal A})^* \ar@{^{(}->}[r] \ar@{=}[d] & {\bf
Hom}_{functors}({\rm Spec}\,{\mathcal A}, {\bf k}) = \mathcal A
\ar@{=}[d] \\ {\rm Spec}\, {\mathcal A^*} \ar@{^{(}->}[r] & {\bf
Hom}_k({\mathcal A}^*,{\bf k})=\mathcal A^{**}}$$ is commutative.

\[Cartier\] Assume that $G={\rm Spec}\, {\mathcal  A}$ is a functor of abelian monoids, $\mathcal A$ is a reflexive functor of $k$-modules and ${\bf
Hom}_{functors}({\rm Spec}\, {\mathcal  A},{\bf k})=\mathcal A$, ${\bf
Hom}_{functors}({\rm Spec}\, $ ${\mathcal  A}^*, {\bf k}) = \mathcal A^*$. Then:

1.  The morphism $G\overset{**}\to G^{**}$, $g \mapsto
    g^{**}$, where $g^{**} (f) := f(g)$ for every $f \in G^* $, is an isomorphism.

2.  ${\rm Hom}_{monoids} (G_1,G_2) = {\rm Hom}_{monoids}
    (G_2^*,G_1^*)$.

$ $

1.  It is easy to check that the diagram $$\xymatrix{{\rm Spec}\, {\mathcal A^{**}} \ar[r]^-\sim_-{\text{\ref{dual}}} & ({\rm
    Spec} \, {\mathcal A^*})^*
    & ({\rm Spec}\, {\mathcal A})^{**}
    \ar[l]_-\sim^-{\text{\ref{dual}}}\\ & {\rm Spec}\, {\mathcal A}
    \ar@{=}[ul] \ar[ur]^-{**}& }$$ is commutative. Hence the morphism $**$ is an isomorphism.

2.  Every morphism of monoids $G_1 \to G_2$, taking ${\bf
    Hom}_{monoids} (-,{\bf k})$, defines a morphism $G_2^* \to G_1^*$. Taking ${\bf Hom}_{monoids}(-,{\bf k})$ we get the original morphism $G_1 \to G_2$, as it is easy to check. Likewise, every morphism of monoids $G_2^* \to G_1^*$, taking ${\bf
    Hom}_{monoids}(-,{\bf k})$, defines a morphism $G_1 \to G_2$. Taking ${\bf Hom}_{monoids}(-,{\bf k})$ we get the original morphism $G_2^* \to G_1^*$.

The category of abelian affine $k$-monoids $G={\rm Spec}\, A$ is anti-equivalent to the category of functors ${\rm Spec}\, {\bf
A}^*$ of abelian monoids (we assume the $k$-modules $A$ are projective).

In particular, we get the Cartier duality for finite commutative algebraic groups ([@carlos §9.9]). Dieudonné ([@dieudonne Ch. I, §2, 13]) proves the equivalence between the category of $k$-coalgebras in groups and the dual category of linearly compact $k$-algebras in cogroups (where $k$ is a field).

Let $G_1$ and $G_2$ be a pair of functors of abelian monoids. It holds that $$(G_1 \times G_2)^*= G_1^* \times G_2^* .$$

$$\aligned (G_1 \times G_2)^* & ={\bf Hom}_{monoids} (G_1\times G_2, {\bf k})
\\ & = {\bf Hom}_{monoids} (G_1, {\bf k}) \times {\bf Hom}_{monoids} (G_2, {\bf k})
= G_1^* \times G_2^*. \endaligned$$

Let us show some examples.

Let $k$ be a field. Let ${\mathbb{Z}}:= {\rm Spec}\, \underset{{\mathbb{Z}}}{\prod} {\bf k}$. Obviously ${\mathbb{Z}}^* = G_m$, therefore $G_m^* = {\mathbb{Z}}$. In other words, ${\bf Hom}_{grupos} (G_m,G_m)={\mathbb{Z}}$: given $\tau : G_m \to G_m$, there exists $n \in {\mathbb{Z}}$ such that $\tau(\alpha) = \alpha^n$.

Given a quasi-coherent $\prod_X {\bf k}$-module ${\bf E}$, then ${\bf E} = \oplus_{x \in X} {\bf E}_x$ in the obvious way. Given a morphism of algebra schemes $\prod_X {\bf k} \to {\bf A}^*$, then ${\bf A}^* = \prod_{x \in X} {\bf A}^*_x$. Moreover, if the morphism ${\rm Spec}\, {\bf A}^* \to {\rm Spec}\, \prod_X {\bf k}$ is injective, then ${\bf A}^*_x = {\bf k}$ or ${\bf A}^*_x =0$ for each $x \in X$. Therefore, ${\rm Spec}\, {\bf A}^* \subset {\rm Spec}\, \underset{{\mathbb{Z}}}{\prod} {\bf k} = {\mathbb{Z}}$ is a subgroup if and only if ${\rm Spec}\, {\bf A}^* =n \cdot {\mathbb{Z}}$. Dually, the quotient algebraic groups of $G_m$ are the epimorphisms $G_m \to G_m$, $t \mapsto t^n$. Hence, the subgroups of $G_m$ are the $\mu_n = \{
\alpha \in G_m : \alpha^n = 1 \} = {\rm Spec}\, k[x]/(x^n-1)=({\mathbb{Z}}/(n))^*$.

Likewise, $(G_m \times \overset{n}{\ldots} \times G_m)^* = {\mathbb{Z}}\times \overset{n}{\ldots} \times {\mathbb{Z}}$. By the theory of abelian groups (or ${\mathbb{Z}}$-modules), the subgroups of ${\mathbb{Z}}\times
\overset{n}{\ldots} \times {\mathbb{Z}}$ are isomorphic to ${\mathbb{Z}}^r$ through an injective morphism $\phi$ (let us say that its matrix is $(n_{ij})$), $${\mathbb{Z}}\times \overset{r}{\ldots} \times {\mathbb{Z}}\overset{\phi}{\to} {\mathbb{Z}}\times \overset{n}{\ldots} \times {\mathbb{Z}}$$ whose cokernel is isomorphic to ${\mathbb{Z}}^{n-r} \times {\mathbb{Z}}/(n_1) \times
\ldots \times {\mathbb{Z}}/(n_r)$. Dually, the subgroups of $G_m \times
\overset{n}{\ldots} \times G_m$ are isomorphic to $G_m^{n-r}
\times \mu_{n_1} \times \ldots \times \mu_{n_r}$, that is isomorphic to the kernel of the epimorphism $$G_m^n \to G_m^r,\;
(t_1, \ldots ,t_n) \mapsto (t_1^{n_{11}} \cdot \ldots \cdot
t_n^{n_{n1}},\ldots, t_1^{n_{1r}} \cdot \ldots \cdot t_n^{n_{nr}})
.$$

[@borel Ch. III, 8.12] The category of diagonalizable algebraic groups is anti-equivalent to the category of ${\mathbb{Z}}$-modules of finite type (or finitely-generated abelian groups).

Let $M$ be a set with structure of abelian (multiplicative) monoid. The constant functor $\mathcal{M} = {\rm Spec}\, \prod_M
{\bf k}$ is a functor of abelian monoids. The dual functor is the abelian $k$-monoid ${\mathcal M}^* = {\rm Spec}\, \oplus_M {\bf k}
= {\rm Spec}\, k[M]$.

We will say that an abelian monoid $M$ is classic if it is finitely generated and its associated group $G$ is torsion-free and the natural morphism $M \to G$ is injective. It is easy to prove that $M$ is classic if and only if $k[M]$ is an integral finitely generated $k$-algebra.\
[**Theorem.** ]{} The category of abelian monoids (respectively finitely generated, classic) is anti-equivalent to the category of affine $k$-schemes of semisimple abelian monoids (respectively algebraic, integral algebraic). As $G = \mathbb{Z}^n$, the morphism $M \to G$ induces a morphism $G^n_m \to {\mathcal M}^*$. In particular, $G^n_m$ operates on ${\mathcal M}^*$. Furthermore, as $k[G]$ is the localization of $k[M]$ by the algebraically closed system $M$, the morphism $G^n_m \to {\mathcal M}^*$ is an open injection. We will say that an integral affine algebraic variety on which the torus operates with a dense orbit is an affine toric variety. It is easy to prove that there exists a one-to-one correspondence between affine toric varieties with a fixed point whose orbit is transitive and dense, and classic monoids.

Let $G_a={\rm Spec}\, k[x]$ be the additive group. Let us compute $G_a^*$. Assume ${\rm car}\, k=0$. Let $\bf k[[z]]$ be the $k$-algebra scheme defined by ${\bf k[[z]]}(B):=B[[z]]$. Let $\{ \omega_0, \ldots, \omega_n, \ldots \} \subset k[x]^*$ be the dual basis of $\{ 1, x, \ldots, x^n , \ldots \} \subset k[x]$. Let $\phi\colon \bf k[x]^*\to k[[z]]$ be the linear isomorphism defined by $\phi(\omega_i) = \dfrac{z^i}{i!}$. The composition of the natural morphism $G_a\to \bf k[x]^*$ with $\phi$ is equal to the morphism $G_a\to \bf k[[z]]$, $\alpha\mapsto e^{\alpha z}$, which is a morphism of monoids. Therefore, $\phi$ is an isomorphism of functors of algebras.

$$\aligned G_a^*(B) & = {\rm Hom}_{k-alg} ({\bf k[x]}^*,
{\bf B}) = {\rm Hom}_{k-alg} ({\bf k[[z]]}, {\bf B}) \\
& = \underset{\underset{n} {\longrightarrow}}{\lim}\, {\rm
Hom}_{k-alg} ({k[z]/(z^n)},B) = {\rm rad}\, B .
\endaligned$$ Let us follow the notation $G_a^*={\rm rad}\, {\bf k}$. Explicitly, we assign to $n \in {\rm rad}\, {\bf k}$ the morphism $G_a \to G_m$, $\alpha \mapsto e^{\alpha \cdot n}$. Finally, $({\rm rad}\, {\bf k})^*=G_a$; explicitly, $\alpha \in G_a$ defines the morphism ${\rm rad}\, {\bf k} \to G_m$, $u \mapsto e^{\alpha \cdot u}$.

Let $G = {\rm Spec}\, A$ be an abelian $k$-monoid. Let $p \in
G^*(k)$ and let us regard $k$ as a $G$-module via $p$. Then $$T_p
G^* := Der_k ({\bf A}^*, {\bf k})= Der(G, {\bf k}) .$$

Let $q \in G(k)$ and let us regard $k$ as a $G^*$-module via $q$. Then $$T_q G = Der_k ({\bf A}, {\bf k}) = Der(G^*, {\bf k}) .$$

It is a consequence from Proposition \[2.9\], where ${\mathcal V}
= {\bf k}$.

Let ${\bf I}_p^* := \ker p$, then ${\rm Hom}_k ( {\bf I}_p^*/{\bf
I}^{*2}_p , {\bf k}) = Der_k ({\bf A}^*, {\bf k}) = 
Der (G, {\bf k})$.

Structure of commutative $k$-groups in characteristic zero
==========================================================

For simplicity we assume that $k$ is an algebraically closed field. Let ${\bf A}^*$ be a $k$-scheme of commutative algebras and ${\bf
M}^* = \prod_J {\bf k}$ the maximal semisimple quotient of ${\bf
A}^*$. The maximal spectrum of a scheme of algebras is defined to be the set of its maximal bilateral ideal schemes and it is the same as the set of isomorphism classes of simple ${\bf
A}^*$-modules (see [@Amel 6.7]). Then, ${\rm Spec}_{max} {\bf
A}^* = {\rm Spec}_{max} {\bf M}^* = J$. ${\bf A}^*$ is the inverse limit of its cokernels of finite dimension. As every finite commutative $k$-algebra is a direct product of local $k$-algebras, it is easy to prove that ${\bf A}^* = \prod_{j \in J} {\bf
A}^*_j$, where the maximal semisimple quotient of ${\bf A}^*_i$ is ${\bf k}$. Then, there exists an only section of algebra $k$-schemes $s: {\bf M}^* \to {\bf A}^*$ of the epimorphism ${\bf
A}^* \to {\bf M}^*$.

The maximal semisimple quotient of ${\bf A}^* \bar{\otimes} {\bf
A}^* = ( {\bf A} \otimes {\bf A})^*$ is ${\bf M}^* \bar{\otimes}
{\bf M}^* = ( {\bf M} \otimes {\bf M})^*$ ([@Amel 6.23]). If ${\rm Spec}\, {\bf A}^*$ is also a functor of abelian groups, then ${\rm Spec}\, {\bf M}^*$, which is the constant functor $J$, is a functor of abelian subgroups. Furthermore, the morphism $s^*: {\rm
Spec}\, {\bf A}^* \to {\rm Spec}\, {\bf M}^*$ induced by $s$ is a morphism of functors of groups. Therefore, if $1 \in J$ is the identity element, we have the following decomposition of group functors $${\rm Spec}\, {\bf A}^* = {\rm Spec}\, {\bf M}^* \times
{\rm Spec}\, {\bf A}_1^*.$$ In particular, $G = {\rm Spec}\, A$, which is the dual of $G^* = {\rm Spec}\, {\bf A}^*$, is a direct product of a diagonalizable group (every linear representation is a direct sum of simple modules of rank $1$) and a unipotent group (every linear representation has got a filtration of invariant factors).

Given a $k$-vector space scheme ${\bf E}^*$, let us denote by $\bar{S}^n {\bf E}^*$ the representant of the functor $$F({\bf
V}^*)= {\rm Sim}_k ({\bf E}^* \times \stackrel{n}{\ldots} \times
{\bf E}^*, {\bf V}^* ) .$$ $\bar{S}^n {\bf E}^*$ is the quotient of $ {\bf E}^* \bar{\otimes} \stackrel{n}{\ldots} \bar{\otimes}
{\bf E}^* = ({\bf E} \otimes \stackrel{n}{\ldots} \otimes {\bf
E})^*$ by the minimum $k$-vector space subscheme that contains elements of the following type $$\ldots \otimes w \otimes \ldots
\otimes w' \otimes \ldots - \ldots \otimes w' \otimes \ldots
\otimes w \otimes \ldots \; .$$ $\bar{S}^n {\bf E}^*$ coincides with the closure of $k$-vector space schemes of $S^n {\bf E}^*$ and it coincides with $(( {\bf E} \otimes \stackrel{n}{\ldots}
\otimes {\bf E})^{S_n})^*$. In characteristic zero, $(E \otimes
\stackrel{n}{\ldots} \otimes E)^{S_n} = S^n E$, then $\bar{S}^n
{\bf E}^* = (S^n {\bf E})^*$.

If $E = \oplus_X k$, then ${\bf E}^* = \prod_X {\bf k}$ and $\bar{S}^n {\bf E}^* = \prod_{S^n X} {\bf k}$.

We will denote $k[[{\bf E}^*]] := \prod_{n \in {\mathbb{N}}} \bar{S}^n {\bf
E}^*$. If ${\rm car}\, k=0$ and $G = {\bf V}^*$, then $G^* = {\rm
Spec}\, k[[{\bf V}^*]]$. Specifically, the natural morphism ${\bf
V}^* \to k[[{\bf V}^*]]$ assigns $w \in {\bf V}^*$ to $e^w$.

The kernel of the morphism ${\bf A}^* \to {\bf M}^*$ (the morphism of a scheme of algebras onto its maximal semisimple quotient) is said to be the radical of ${\bf A}^*$. We will say that ${\bf A}^*$ is local if ${\bf M}^* = {\bf k}$.

Let $f: {\bf A}^* \to {\bf B}^*$ be a morphism between local $k$-algebra schemes and let ${\bf I}^*_A$ and ${\bf I}^*_B$ be the radical ideals of ${\bf A}^*$ and ${\bf B}^*$, respectively. If the induced morphism ${\bf I}^*_A/{\bf I}^{*2}_A \to {\bf
I}^*_B/{\bf I}^{*2}_B$ is surjective, then $f$ is a surjective morphism. If ${\bf B}^* = k[[{\bf E}^*]]$ and the morphism ${\bf
I}^*_A/{\bf I}^{*2}_A \to {\bf I}^*_B/{\bf I}^{*2}_B$ is injective, then $f$ is an injective morphism.

By abuse of notation, we have regarded ${\bf I}^{*n}$ as the closure of vector space schemes of ${\bf I}^{*n}$ in ${\bf A}^*$ (as we did in [@Amel]). It is convenient to denote by $({\bf I}^{*n})'$ this closure in this paragraph. The natural morphism $S^n
{\bf I}^*/{\bf I}^{*2} \to {\bf I}^{*n}/{\bf I}^{*n+1}$ is surjective and the closure of $k$-vector space schemes of the image of ${\bf
I}^{*n}/{\bf I}^{*n+1}$ in ${\bf A}^*/({\bf I}^{*n+1})'$ is $({\bf
I}^{*n})' / ({\bf I}^{*n+1})'$. Therefore, the morphism $\bar{S}^n
({\bf I}^*/{\bf I}^{*2}) \to ({\bf I}^{*n})' / ({\bf I}^{*n+1})'$ is surjective.

If ${\bf I}^*_A/{\bf I}^{*2}_A \to {\bf I}^*_B/{\bf I}^{*2}_B$ is a surjective morphism, then the morphisms $\bar{S}^n ({\bf
I}^*_A/{\bf I}^{*2}_A) \to \bar{S}^n ({\bf I}^*_B/{\bf I}^{*2}_B)$ are surjective and from the commutative diagram $$\xymatrix{ \bar{S}^n ({\bf I}^*_A/{\bf I}^{*2}_A) \ar[r]^-{epi} \ar[d]^-{epi}
& \bar{S}^n ({\bf I}^*_B/{\bf I}^{*2}_B) \ar[d]^-{epi}
\\ {\bf I}^{*n}_A/{\bf I}^{*n+1}_A \ar[r] &
{\bf I}^{*n}_B/{\bf I}^{*n+1}_B }$$ it is deduced that the morphisms ${\bf I}^{*n}_A/{\bf I}^{*n+1}_A \to {\bf I}^{*n}_B/{\bf
I}^{*n+1}_B$ are surjective. Then the morphisms ${\bf A}^*/{\bf
I}^{*n+1}_A \to {\bf B}^*/{\bf I}^{*n+1}_B$ are surjective. Taking inverse limits (which is the dual of a direct limit of injections) we get the epimorphism $f: {\bf A}^* \to {\bf B}^*$ ([@Amel 6.17 (2)]).

If ${\bf I}^*_A/{\bf I}^{*2}_A \to {\bf I}^*_B/{\bf I}^{*2}_B$ is an injective morphism, then the morphisms $\bar{S}^n ({\bf
I}^*_A/{\bf I}^{*2}_A) \to \bar{S}^n ({\bf I}^*_B/{\bf I}^{*2}_B)$ are injective and from the commutative diagram $$\xymatrix{
\bar{S}^n ({\bf I}^*_A/{\bf I}^{*2}_A) \ar@{^{(}->}[r]
\ar[d]^-{epi} & \bar{S}^n ({\bf I}^*_B/{\bf I}^{*2}_B) \ar@{=}[d]
\\ {\bf I}^{*n}_A/{\bf I}^{*n+1}_A \ar[r] & {\bf I}^{*n}_B/{\bf
I}^{*n+1}_B }$$ it is deduced that the morphisms ${\bf
I}^{*n}_A/{\bf I}^{*n+1}_A \to {\bf I}^{*n}_B/{\bf I}^{*n+1}_B$ are injective. Let $0 \neq a \in {\bf A}^*$ and let $n$ be such that $a \in {\bf I}_A^{*n}$ and $a \neq {\bf I}_A^{*n+1}$ (see [@Amel 6.17 (2)]). Then $f(a)\neq 0$, since $0 \neq \bar{a} \in {\bf I}^{*n}_A/{\bf I}^{*n+1}_A$ and $0 \neq \overline{f(a)} \in {\bf I}^{*n}_B/{\bf I}^{*n+1}_B$.

Let $G= {\rm Spec}\, A$ be an abelian $k$-group and let $k$ be an algebraically closed field of characterisic zero. Then, there exists an isomorphism of $k$-groups $$G = (\underset{X}{\prod}
G_a) \times D$$ where $D$ is a diagonalizable group.

We must prove that if $G$ is unipotent, then it is isomorphic to a direct product of additive groups.

Thus, we have that ${\bf A}^*$ is local. Let ${\bf I}^*$ be its radical. The morphism $i: G \to 1 + {\bf I}^*/{\bf I}^{*2}$, $i(g) = 1 + \overline{g-1}$ is a morphism of $k$-groups. Let us prove that it is an isomorphism. Consider ${\bf k}$ as the trivial $G$-module. Then, $E := Der(G, {\bf k}) = {\rm Hom}_{lin} (G, {\bf
k})$. The ring of functions of $1 + {\bf I}^*/{\bf I}^{*2} = {\bf
I}^*/{\bf I}^{*2}$ is $\oplus_n S^n E$. The morphism between the rings induced by $i$ is the natural morphism $\oplus_n S^n E \to
A$. Dually, we have the morphism ${\bf A}^* \to k[[{\bf E}^*]] =:
{\bf B}^*$. The induced morphism ${\bf I}^*/{\bf I}^{*2} \to {\bf
I}_B^*/{\bf I}_B^{*2} = {\bf E}^*$ is the identity morphism. By the previous lemma, ${\bf A}^* = k[[{\bf E}^*]] $ and $G= {\bf
I}^*/{\bf I}^{*2}$.

Inversion formula
=================

Let ${\bf B} \subseteq \prod_J {\bf k}$ be a quasi-coherent $k$-algebra (with or without unit). We define ${\rm Spec}\, {\bf B}$ to be the functor over the category of commutative algebras (with unit) $$({\rm Spec}\, {\bf B})(C) := \{ f \in {\rm Hom}_{k-alg} (B,C)\, \colon\, ({\rm Im} f )= C \} .$$

If $k$ is a field it is easily seen that ${\rm Spec}\, {\bf B} = {\rm Spec}\, \prod_J {\bf k}$ if and only if ${\bf B} = \oplus_J {\bf k}$.

We will say that $\oplus_J k $ is the cuasi-coherent ring of functions of ${\rm Spec}\, \prod_J {\bf k}$.

As well, $\oplus_J {\bf k}$ is characterized by being the maximal quasi-coherent $\prod_J {\bf k}$-submodule of $\prod_J {\bf k}$.

Let $G^* = {\rm Spec}\, \prod_J {\bf k}$ be a functor of discrete groups. $G^*$ operates on $G^*$ by translations on the left, then it operates on $\prod_J {\bf k}$, then on $\oplus_J k$. The mapping $w_{G^*}: \oplus_J k \to k$, $w_{G^*}((\lambda_i)) =
\sum_i \lambda_i$ is the only $G^*$-invariant linear form up to a multiplicative factor. We will say that the linear mapping $$\phi_{G^*}: \oplus_J {\bf k} \to (\prod_J {\bf k})^*
\hookrightarrow ( \oplus_J {\bf k} )^*, a \mapsto w_{G^*}(a \cdot
-)$$ is the transposed Fourier transform of $G^*$.

$G = {\rm Spec}\, A$ is a semisimple $k$-group if and only if there exists a (unique) linear mapping $w_G: A \to k$, $G$-invariant and such that $w_G(1)=1$ (see [@Amel2 2.10]). The morphism $\phi_G: A \to A^*$, $\phi_G(a) := w_G(a^* \cdot -)$ (where $*: A \to A$ is the morphism between rings induced by the morphism $G \to G, g \mapsto g^{-1}$) is called the Fourier transform of $G$ (see [@Amel3]). If $G$ is also commutative, then $A^* = \prod_J k$ and ${\rm Im}\, \phi_G = \oplus_J k =:A'$. Let $e: A \to k$ the identity element of $G$ (that is, the unit of $A^*$). The diagram $$\xymatrix{ A \ar[dr]^-{e}
\ar@{=}[rr]^-{\phi_G} & & \ar[dl]^-{w_{G^*}}  A' \\ & k & }$$ is commutative (see [@Amel3], section 2).

The transposed Fourier transform of the dual group of a semisimple abelian $k$-group is the inverse of the Fourier transform of the semisimple abelian group.

$\phi_{G^*}(\phi_G(a))(w) = w_{G^*}(\phi_G(a) \cdot w) =
w_{G^*}(\phi_G(a \cdot w)) = (a \cdot w)(e) = a(w)$ for all $a \in
A$ and $w \in A^*$.

[99]{}

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