$I(\oplus_AM_\alpha)=\oplus_AIM_\alpha$ and $\oplus_AM_\alpha/I(\oplus_AM_\alpha)\cong\oplus_AM_\alpha/IM_\alpha$

Question

Let $(M_\alpha )_{\alpha \in A}$ be an indexed set of left R-modules and let I be a left ideal of R. Prove that:

$I(\oplus_AM_\alpha)=\oplus_AIM_\alpha$ and $\oplus_AM_\alpha/I(\oplus_AM_\alpha)\cong\oplus_AM_\alpha/IM_\alpha$

If I can prove the first part then second part is the consequence of it. But I couldnt prove the first part. I will be very appreciate for any help.

Answer 1

Hint: each $IM_\alpha$ is an $R$-submodule of $M_\alpha$. Hence we have a map $$ \jmath = \oplus \iota_\alpha \ \colon \bigoplus_\alpha IM_\alpha\to \bigoplus_{\alpha}M_\alpha, $$

which is by construction injective. Compute its image.

For this, pick $(x_\alpha)$ in the domain with finitely of them being non-zero, of course. Then $x_\alpha = \sum_{i=1}^{n_\alpha}r_{i,\alpha} y_{i,\alpha}$ with $r_{i,\alpha} \in I, y_{i,\alpha} \in M_\alpha$. The image via $\jmath$ is the same tuple.

Suppose that the non-zero indices are $\alpha_1, \ldots, \alpha_j$, then $(x_\alpha) = \sum_{1 \leq k\leq j, 1 \leq i \leq n_{\alpha_k}}r_{i,\alpha}y_{i,\alpha_k}$. The latter is a sum of elements of $\oplus M_\alpha$ with coefficients in $I$, hence belongs to $I\oplus_\alpha M_\alpha$.

Reciprocally, if $x = \sum_{i=1}^n r_i x_i$ with $r_i \in I, x_i \in \oplus_{\alpha} M_\alpha$, then each $x_i$ is a finite sum of elements of each submodule. Adding zero elements if needed, we can assume that there exist $\alpha_1, \ldots, \alpha_k$ such that $x_i = \sum_{j}y_{ij}$ and $y_{ij} \in M_{\alpha_j}$. But then $x$ is a sum of elements $r_i y_{ij} = \jmath(r_i y_{ij})$ that belong to each $I M_{\alpha_j}$.