# Modifying $\frac{\prod_\alpha A_\alpha}{\prod_\alpha B_\alpha}\simeq \prod_\alpha\frac{A_\alpha}{B_\alpha}$ for direct sums

Let $$\{A_\alpha\}$$ be a family of $$R$$-modules, each $$B_\alpha\subset A_\alpha$$ a submodule and $$\pi_\alpha:A_\alpha\to A_\alpha/B_\alpha$$ be the canonical projection map. Then the map

$$\prod_\alpha\pi_\alpha:\prod_\alpha A_\alpha\to \prod_\alpha\frac{A_\alpha}{B_\alpha}$$

is surjective, and has kernel $$\prod_\alpha B_\alpha$$. Therefore, by the first isomorphism theorem we have

$$\frac{\prod_\alpha A_\alpha}{\prod_\alpha B_\alpha}\simeq \prod_\alpha\frac{A_\alpha}{B_\alpha}$$

What I'm curious about is how to modify this proof for direct sums. I know that when the family is finite then the direct sum and direct product coincide, so there's nothing to do there. It's when it's an infinite family where I'm uncertain. With $$\bigoplus_\alpha A_\alpha$$ then only finitely many components are non-zero, but I'm not sure if that means I'd need to alter the argument to account for this, or if it can simply be applied to direct sums as well to show that

$$\frac{\bigoplus_\alpha A_\alpha}{\bigoplus_\alpha B_\alpha}\simeq \bigoplus_\alpha\frac{A_\alpha}{B_\alpha}$$

So my question is, is an alteration to the argument above necessary for infinite direct sums?

• The argument is exactly the same, no? Rewrite everything with $\bigoplus$ in place of $\prod$. Jul 13, 2020 at 23:36
• @Batominovski Well that's what I'm not 100% sure about. Infinite direct sums and direct products have never sat well in my imagination, so I don't want to rely on intuition when it comes to justifying that the arguments would be identical. Jul 13, 2020 at 23:39

This answer here is solely for the purpose of giving this question an answer. Since the OP obtained the desired answer (see the comments under the question), I am providing a different way using category theory to show that $$P:=\frac{\prod\limits_{\alpha\in J}\,A_\alpha}{\prod\limits_{\alpha\in J}\,B_\alpha}\cong \prod_{\alpha\in J}\,\frac{A_\alpha}{B_\alpha}\text{ and }S:=\frac{\bigoplus\limits_{\alpha\in J}\,A_\alpha}{\bigoplus\limits_{\alpha\in J}\,B_\alpha}\cong \bigoplus_{\alpha\in J}\,\frac{A_\alpha}{B_\alpha}\,.$$ Explicit isomorphisms can be seen in (*) and (#).
For each $$\beta \in J$$, $$\iota_\beta:A_\beta\to \bigoplus\limits_{\alpha\in J}\,A_\alpha$$ and $$\pi_\beta: \prod\limits_{\alpha\in J}\,A_\alpha\to A_\beta$$ denote the canonical injection and the canonical projection, respectively. Let $$q:\bigoplus\limits_{\alpha\in J}\,A_\alpha\to S$$ be the quotient map. Then, $$q\circ \iota_\beta$$ vanishes on $$B_\beta$$. Therefore, $$q\circ \iota_\beta$$ factors through the quotient map $$q_\beta:A_\beta\to\dfrac{A_\beta}{B_\beta}$$. In other words, there exists a (unique) map $$i_\beta:\dfrac{A_\beta}{B_\beta}\to S$$ such that $$q\circ \iota_\beta=i_\beta\circ q_\beta\,.$$ We claim that $$S$$ together with the maps $$i_\beta:\dfrac{A_\beta}{B_\beta}\to S$$ for $$\beta\in J$$ is a categorical coproduct (direct sum) of the family $$\left(\dfrac{A_\alpha}{B_\alpha}\right)_{\alpha\in J}$$. Let $$T$$ be any $$R$$-module together with morphisms $$\tau_\beta:\dfrac{A_\beta}{B_\beta}\to T$$ for each $$\beta\in J$$. We want to show that a there exists a unique morphism $$\phi:S\to T$$ such that $$\phi\circ i_\beta=\tau_\beta$$ for each $$\beta\in J$$.
We define $$\phi\left((a_\alpha)_{\alpha\in J}+\bigoplus_{\alpha\in J}\,B_\alpha\right):=\sum_{\alpha\in J}\,\tau_\alpha\left(a_\alpha+B_\alpha\right)\text{ for all }(a_\alpha)_{\alpha\in J}\in\bigoplus_{\alpha\in J}\,A_\alpha\,.$$ It is easy to verified that $$\phi$$ is a well defined morphism, and it is the only morphism such that $$\phi\circ i_\beta=\tau_\beta$$ for all $$\beta\in J$$. We can now then conclude that $$S$$ is a coproduct of the family $$\left(\dfrac{A_\alpha}{B_\alpha}\right)_{\alpha\in J}$$. Since coproducts are unique up to isomorphism, we obtain $$S\cong \bigoplus\limits_{\alpha \in J}\,\dfrac{A_\alpha}{B_\alpha}$$, via the isomorphism $$\sigma:S\to \bigoplus\limits_{\alpha \in J}\,\dfrac{A_\alpha}{B_\alpha}$$ given by $$\sigma\left((a_\alpha)_{\alpha\in J}+\bigoplus_{\alpha\in J}\,B_\alpha\right):=\sum_{\alpha\in J}\,\bar{\iota}_\alpha\left(a_\alpha+B_\alpha\right)\text{ for all }(a_\alpha)_{\alpha\in J}\in\bigoplus_{\alpha\in J}\,A_\alpha\,,\tag{*}$$ where $$\bar{\iota}_\beta:\dfrac{A_\beta}{B_\beta}\to \bigoplus\limits_{\alpha\in J}\,\dfrac{A_\alpha}{B_\alpha}$$ is the canonical injection for each $$\beta\in J$$.
Observe now that, for every $$\beta\in J$$, $$q_\beta\circ \pi_\beta$$ vanishes on $$\prod\limits_{\alpha\in J}\,B_\alpha$$. Therefore, $$q_\beta\circ\pi_\beta$$ factors through the quotient map $$k:\prod\limits_{\alpha\in J}\,A_\alpha\to P$$. Ergo, there exists a (unique) morphism $$\varpi_\beta:P\to \dfrac{A_\beta}{B_\beta}$$ such that $$q_\beta\circ\pi_\beta=\varpi_\beta\circ k\,.$$ We claim that $$P$$ together with the morphisms $$\varpi:P\to \dfrac{A_\beta}{B_\beta}$$ is a categorical product (direct product) of the family $$\left(\dfrac{A_\alpha}{B_\alpha}\right)_{\alpha\in J}$$. Let $$Q$$ be any $$R$$-module together with morphisms $$\kappa_\beta:Q\to\dfrac{A_\beta}{B_\beta}$$ for all $$\beta\in J$$. We need to show that there exists a unique morphism $$\psi:Q\to P$$ such that $$\varpi_\beta\circ \psi=\kappa_\beta$$ for all $$\beta\in J$$.
We define $$\psi\left(x\right):=\big(\kappa_\alpha(x)\big)_{\alpha\in J}+\prod_{\alpha\in J}\,B_\alpha\text{ for all }x\in Q\,.$$ It is easily seen that $$\psi$$ is a well defined morphism, and it is the only morphism such that $$\varpi_\beta\circ \psi=\kappa_\beta$$ for all $$\beta\in J$$. We now conclude that $$P$$ is indeed a product of the family $$\left(\dfrac{A_\alpha}{B_\alpha}\right)_{\alpha\in J}$$. Since products are unique up to isomorphism, we have $$P\cong \prod\limits_{\alpha\in J}\,\dfrac{A_\alpha}{B_\alpha}$$ via the isomorphism $$\varsigma: \prod\limits_{\alpha\in J}\,\dfrac{A_\alpha}{B_\alpha}\to P$$ given by $$\varsigma\Big(\big(a_\alpha+B_\beta\big)_{\alpha\in J}\Big):=\big(a_\alpha\big)_{\alpha\in J}+\prod_{\alpha\in J}\,B_\alpha\text{ for all }\big(a_\alpha\big)_{\alpha\in J}\in \prod_{\alpha\in J}\,A_\alpha\,.\tag{#}$$