Quotient Modules and their Direct Sum

Question

Let $R$ be a ring and $I$ be an ideal of $R$. Let $M$ be a module over $R$.Let $N_{1}$, $N_{2}$ be submodules of $M$ such that M = $N_{1}$ $\oplus$ $N_{2}$. Show that
$a)$ $M/IM$ is a module over $R/I$.
$b)$ $M/IM\simeq N_{1}/IN_{1}$ $\oplus$ $N_{2}/IN_{2}$ as an $R/I$ module.

For $a)$, I am successful in showing that if $M/IM$ is well-defined then, it is a module over $R/I$.

But How do I show that $M/IM$ is well-defined? and also, please provide me hints for part $b)$. and maybe proof will be appreciated.

Answer 1

For (a): I am not sure what you mean, but I guess it is the following: For any ideal $I \subset R$ you can prove that the set $IM = \lbrace im \mid i \in I, m \in M \rbrace$ is a submodule of $M$ (the submodule generated by $I$), which means you can consider the quotient module $M/IM$.

For (b): You can for example show that $M/IM$ has the universal property of the direct sum on the right hand side. You can also just define the isomorphism directly. Define a surjective map from $M$ Into the direct sum of the quotients and compute the kernel.

Let us consider the morphism

$f \colon N_1 \oplus N_2 \rightarrow N_1/IN_1 \oplus N_2/IN_2$, $(n_1,n_2) \mapsto (n_1 + IN_1, n_2 + IN_2)$.

Can you compute the kernel?

Answer 2

$M/IM$ is well-defined because $IM$ is a sub-module of $M$. Indeed it is stable by sum and by multiplication by elements of $R$. For b) I would start building a surjective function $M\longrightarrow N_1/IN_1 \times N_2/IN_2$, then I'd show that $IM$ is its kernel.

Answer 3

Another approach is to recall the existence of a natural isomorphism $$\varphi_M:M/MI\to M\otimes_R R/I $$ Since the scalar extension functor $-\otimes_RR/I$, having right adjoint, preverses direct sums, the assertion follows.