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I need to prove that $(R^{ \oplus A_1 })^{ \oplus A_2 } \cong R^{ \oplus (A_1 \times A_2)}$ as $R$-modules.

Recall that

if $N$ is an $R$-module and $A$ is a set, then we can construct $R$-module $N^{ \oplus A} = \{ \alpha: A \to N \ | \ \alpha(a) \neq 0$ for only finitely many $a \in A \}$.

An $R$-module structure is defined componentwise: if $r \in R, \alpha, \beta \in N^{ \oplus A}$, then

$(\alpha+\beta)(a) = \alpha(a) + \beta(a)$

$0_{N^{ \oplus A}} = 0: A \to N$ such that $\forall a \in A \ \ 0(a) = 0$

$(- \alpha)(a) = -\alpha(a)$

$(r\alpha)(a) = r(\alpha(a))$

Now we have two such $R$-modules. Since $R^{\oplus (A_1 \times A_2)}$ is the free $R$-module on a set $A_1 \times A_2$ , every function $\alpha \in R^{ \oplus (A_1 \times A_2) }$ can be written as a finite sum $\sum\nolimits_{x \in A_1, y \in A_2} r_{x,y}j_{x,y}$ such that $r_{x,y} = \alpha(x,y)$ and $j_{x,y}(x',y') = 1$ if $x' = x$ and $y' = y$ and $j_{x,y}(x',y') = 0$ is $x' \neq x$ or $y' \neq y$.

So we need to construct an $R$-module isomorphism $\phi: R^{ \oplus (A_1 \times A_2)} \to (R^{ \oplus A_1 })^{ \oplus A_2 }$.

Since $\phi$ must be an $R$-module homomorphism,

$\phi( \sum\nolimits_{x \in A_1, y \in A_2} r_{x,y}j_{x,y} ) = \sum\nolimits_{x \in A_1, y \in A_2} r_{x,y}\phi(j_{x,y})$

So we it suffices to define an image of each $j_{x,y}$. Any ideas?

Assumed background: basic properties of rings and modules, categories and universal properties( such as the one of a free $R$-module or a categorical product ) are introduced, but functors, limits and colimits are not. So this is not a duplicate of this question.

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Let $f \in R^{\oplus(A_{1} \times A_{2})}$. Define a map $\varphi_{f} \colon A_{2} \to R^{\oplus A_{1}}$ by $\alpha \mapsto h_{f}$, where $h_{f} \in R^{\oplus A_{1}}$ is the map given by $h_{f}(\beta) = f(\beta, \alpha)$. Now, define a map $\Phi \colon R^{\oplus(A_{1} \times A_{2})} \to (R^{\oplus A_{1}})^{\oplus A_{2}}$ by $\Phi(f) = \varphi_{f}$. You should check three things:

$1$) $\Phi$ is well-defined, i.e $\phi_{f}$ indeed belongs to $(R^{\oplus A_{1}})^{\oplus A_{2}}$ (part of this will involve checking that $h_{f}$ belongs to $R^{\oplus A_{1}}$)

$2$) $\Phi$ is an $R$-module homomorphism

$3$) $\Phi$ is bijective

which establishes the claim. This is a bit tedious, so you can see the advantage of the categorical point of view.

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How about checking that $X:=(R^{ \oplus A_1 })^{ \oplus A_2 }$ satisfies the universal property of the free $R$-module on $A_1 \times A_2$? Define $i:A_1\times A_2 \to X$ by $i(a_1,a_2) = \alpha_{a_1,a_2}$, where the mapping $(\alpha_{a_1,a_2}(b_2))(b_1) := 1$ if $a_i = b_i$ and vanishes otherwise. Note that this lets us decompose any $\alpha \in X$ as a unique $R$-linear combination, $\alpha = \sum_{a_1 \in A_1, \, a_2 \in A_2} (\alpha(a_2))(a_1)\alpha_{a_1,a_2}$.

Now given an arbitrary map $f:A_1 \times A_2 \to M$ into a $R$-module $M$, its only possible extension $\overline{f}:X \to M$ (that is, $R$-module homomorphism such that $f = \overline{f}i$) must be given by $$\overline{f}(\alpha) = \overline{f}\left( \sum_{a_1 \in A_1, \, a_2 \in A_2} (\alpha(a_2))(a_1)\alpha_{a_1,a_2} \right) = \sum_{a_1 \in A_1, \, a_2 \in A_2} (\alpha(a_2))(a_1) f(a_1,a_2).$$

This proves the universal property. As any two objects satisfying the universal property must be isomorphic, we are done.

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  • $\begingroup$ How do we know that defining $(\alpha_{a_1,a_2}(b_2))(b_1) := 1$ if $a_i=b_i$ lets us decompose $\alpha \in X$ as a $R$-linear combination $\alpha = \sum_{a_1 \in A_1, \, a_2 \in A_2} (\alpha(a_2))(a_1)\alpha_{a_1,a_2}$? $\endgroup$ – Al Jebr Jun 20 '19 at 21:34
  • $\begingroup$ @AlJebr In order for the maps on both sides to be equal, by definition they must yield the same value whenever you evaluate them at any $b_2 \in \oplus A_2$. This now yields maps $\oplus A_1 \to R$, which again to be equal must yield the same value when evaluated at any $b_1 \in \oplus A_1$. And this is the case, since both sides then yield $(\alpha(b_2))(b_1)$. $\endgroup$ – Alex Provost Jun 21 '19 at 2:50

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