$(R^{\oplus A})^{\oplus B} \approx R^{\oplus (A\times B)}$?

Question

Let $R$ be a ring and let $A$ and $B$ be sets. How can I see the following isomorphism of $R$-modules

$$(R^{\oplus A})^{\oplus B} \approx R^{\oplus (A\times B)}$$

as a consequence of the universal properties of Cartesian product of sets and coproduct of $R$-modules? The free functor from sets to $R$-modules is not right adjoint so it doesn't commute with the Cartesian product of sets, but something is happening here ...

You need to use the fact that, for sets, multiplication really is just repeated addition. — Zhen Lin, Mar 14 '16 at 22:14

Derek Elkins left SE · Accepted Answer · 2016-03-14T22:53:21.557

3

$R^{\oplus A}$ for a set $A$ is a copower, i.e. $\text{Hom}(R^{\oplus A},S) \cong \text{Hom}(A,\text{Hom}(R,S))$ (natural in $S$ and $A$, and $R$ is viewed as a 1-dimensional $R$-module on the right). If needed, show that this is equivalent to $R^{\oplus A}$ is a colimit of a discrete diagram, i.e. a coproduct. Establishing this adjunction is where you'd used the universal property of coproducts. From there, the rest is pretty straight forward: $$\begin{align} \text{Hom}((R^{\oplus A})^{\oplus B},S) & \cong \text{Hom}(B,\text{Hom}(R^{\oplus A}, S)) \\ & \cong \text{Hom}(B,\text{Hom}(A,\text{Hom}(R,S))) \\ & \cong \text{Hom}(A\times B,\text{Hom}(R,S)) \\ & \cong \text{Hom}(R^{\oplus (A\times B)},S) \end{align}$$ Then, "by Yoneda". The introduction of the cartesian product is via the adjunction giving cartesian closure. Note this proof works for any category that has copowers, which includes any category with arbitrary coproducts.

edited Mar 14 '16 at 22:53

answered Mar 14 '16 at 22:42

Derek Elkins left SE

24,967

Beautiful answer. Thanks. – Drew Armstrong Mar 15 '16 at 12:30
This answer played a role in a homework assignment here: http://www.math.miami.edu/~armstrong/762sp16/762hw3sol.pdf – Drew Armstrong Apr 12 '16 at 21:19
@DrewArmstrong Thanks for the attribution (except my name is not "David"). StackExchange can automatically give you the bibtex/amsrefs for a citation. This can ensure that you stay within the license of content on StackExchange sites, and it makes it easier for your students (or others) to find the question/answer if they so desired. – Derek Elkins left SE Apr 13 '16 at 03:19

score 0 · Answer 2 · answered Mar 14 '16 at 23:02

While Derek's answer is the formal way to prove it, here's an intuitive reason that can be made formal by induction:

$$(R^{\oplus A})^{\oplus B} = \underbrace{\left ( \underbrace{R \oplus \dots \oplus R}_{A \text{ times}} \right ) \oplus \left ( \underbrace{R \oplus \dots \oplus R}_{A \text{ times}} \right ) \oplus \dots \oplus \left ( \underbrace{R \oplus \dots \oplus R}_{A \text{ times}} \right )}_{B \text{ times}} $$

Now, you only need to see that $(A \oplus B) \oplus C \cong A \oplus (B \oplus C)$, which you should be able to prove using the universal property.

$(R^{\oplus A})^{\oplus B} \approx R^{\oplus (A\times B)}$?

2 Answers2

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