$(\bigoplus_{A}M_{\alpha })/ I(\bigoplus_{A}M_{\alpha }) \cong \bigoplus _{A}M_{\alpha }/IM_{\alpha }$

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 $I(\bigoplus_{A}M_{\alpha })= \bigoplus_{A} IM_{\alpha }$ and $(\bigoplus_{A}M_{\alpha })/ I(\bigoplus_{A}M_{\alpha }) \cong \bigoplus _{A}M_{\alpha }/IM_{\alpha }$

I´ve already prove that $I(\bigoplus_{A}M_{\alpha })= \bigoplus_{A} IM_{\alpha }$, but I'm stuck in the second part because of the notation. Can anybody help me with that? I just need an idea. Thanks!

Answer 1

If you had proved (1) then use it: $(\bigoplus_{A}M_{\alpha })/ I(\bigoplus_{A}M_{\alpha }) =(\bigoplus_{A}M_{\alpha })/(\bigoplus_{A} IM_{\alpha }) \cong \bigoplus _{A}M_{\alpha }/IM_{\alpha }$ since $IM_{\alpha }\subset M_{\alpha }$.