The map
$$
(R/I)\otimes_RM\longrightarrow M/IM,\qquad
(r+I)\otimes m\longmapsto rm+IM
$$
is well defined and surjective. Its inverse sends $m+IM$ to $(1+I)\otimes m$; elements of $IM$ map to zero because $(1+I)\otimes am=(a+I)\otimes m=0$ for $a\in I$. Hence
$$
\boxed{(R/I)\otimes_RM\simeq M/IM.}
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 101 / 3 / a / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-101/3/a/i/solution