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As known, we can define dual vector space $V^{\vee}$ as $$V^{\vee}=\mathrm{Hom}_{k}(V, k)$$ where $V$ is $k$-vector space. Also, for finite abelian group $G$ we can define its Pontryagin dual $G^{\vee}$ as $$G^{\vee}=\mathrm{How}(G, \mathbb{S}^{1}).$$ Then for any given ring $R$, is there any natural way to define its dual $R^{\vee}$ as $$R^{\vee}=\mathrm{Hom}(R, R_{0})$$ for some $R_{0}$? If it is true, then what is $R_{0}$? As I think, dual concept might satisfy some double dual property : $(R^{\vee})^{\vee}$ is naturally isomorphic to $R$ for some good $R$'s. I tried $R_{0}=\mathbb{Z}$, but $\mathrm{Hom}(\mathbb{Z}/2, \mathbb{Z})=0$ so I think this is not satisfactory.

Seewoo Lee
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  • Usually you consider the dual of a module $M$ over a ring $R$ as $\operatorname{Hom}_R(M,R)$. If $M = R$, then $\operatorname{Hom}_R(R,R) = R$. This is the right analogy to agree with the vector space example. – walkar Mar 15 '17 at 05:04
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    @walkar But I think ring homomorphism and module homomorphism are different. Also, If we regard abelian group as $\mathbb{Z}$-module, then dual of abelian group will be $\mathrm{Hom}(G, \mathbb{Z})$, which is always zero for finite groups. – Seewoo Lee Mar 15 '17 at 05:57
  • This is true -- the analogy isn't perfect and demonstrates key differences between module theory and linear algebra. But the vector space example and the Pontryagin dual of an abelian group you needed some distinguished object to hom into. For rings, it's not so clear what that should be. In the vector space example, $K$ is not categorically special (not initial or final) so how do you get to a "canonical" choice for rings? – walkar Mar 15 '17 at 05:59

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For a finite-dimensional algebra $A$ over a field $k$, the linear dual $A^{\ast}$ (this is not a hom in the category) is a finite-dimensional coalgebra, and this is a contravariant equivalence of categories. For infinite-dimensional algebras we can take the "finite dual," which consists of all linear functionals $A \to k$ factoring through a finite-dimensional algebra quotient of $A$, and this is also a coalgebra (not true if we just take the linear dual), although this doesn't produce an equivalence.

For commutative rings $R$, arguably the correct "dual" object, in the sense that it produces a contravariant equivalence, is the corresponding affine scheme $\text{Spec } R$, in the sense of algebraic geometry. If we restrict our attention to finitely generated reduced algebras over an algebraically closed field $k$ then the corresponding affine scheme is closely related to the Hom object $\text{Hom}(R, k)$, by a sequence of results ending in the Nullstellensatz.

But there is no reasonable sense in which we can take the dual of a ring and get another ring-like object. The categories of finite-dimensional vector spaces and finite abelian groups have the very special property that they are equivalent to their opposites, and this is just not true for rings or for any reasonable subcategory of rings.

Qiaochu Yuan
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