# Quotient of Direct Sum of Modules

Let $R$ be a commutative (unital) ring, $I$ an ideal of $R$, and $M$, $N$ (right) $R$-modules. For any $R$-module $A$, let $A^{(n)}$ denote $A\oplus\cdots\oplus A$ ($n$-times).

I've reached a point in a proof where I need to show that given $R^{(n)}\cong M\oplus N$, we have that $(R/I)^{(n)}\cong M/MI\oplus N/NI$.

I have shown this using the fact that $R/I\otimes_RA\cong A/AI$ for any $R$-module $A$, and that tensor products distribute across direct sums.

Is there a reasonable direct proof of this result without the use of tensor products?

Perhaps I'm missing some obvious isomorphism, but I'm struggling even to show that $R^{(n)}/R^{(n)}I\cong(R/I)^{(n)}$ directly...

There are two facts you need to show: for any $R$-modules $M$ and $N$, any ideal $I\subseteq R$, and any submodules $A\subseteq M$ and $B\subseteq N$,
1. $(M\oplus N)I=MI\oplus NI$
2. $(M\oplus N)/(A\oplus B)\cong (M/A)\oplus (N/B)$
Fact 1 lets you conclude that $R^{(n)}I=(RI)^{(n)}=I^{(n)}$, and then fact 2 implies $R^{(n)}/I^{(n)}\cong (R/I)^{(n)}$, so that $$R^{(n)}/R^{(n)}I\cong R^{(n)}/I^{(n)}\cong (R/I)^{(n)}$$ and on the other hand, fact 1 and fact 2 together show that $$(M\oplus N)/(M\oplus N)I\cong M/MI\oplus N/NI$$ so that if $R^{(n)}\cong M\oplus N$, we can conclude that $(R/I)^{(n)}\cong M/MI\oplus N/NI$.