The statement is true. Fix an isomorphism $\phi:M\otimes_RN\to R^n$, and write the inverse image of the first standard basis vector as
$$
\phi^{-1}(e_1)=\sum_{j=1}^r m_j\otimes n_j.
$$
Define
$$
\psi:N^r\longrightarrow R,
\qquad
(y_j)_j\longmapsto
\operatorname{pr}_1\!\left(\phi\left(\sum_jm_j\otimes y_j\right)\right).
$$
Since $\psi((n_j)_j)=1$, this is a split surjection. Tensor its splitting with $M$. The resulting split surjection has the form
$$
M\otimes_RN^r
\cong(M\otimes_RN)^r
\cong R^{nr}	woheadrightarrow M.
$$
Thus $M$ is a direct summand of a finite <free module>, so it is a <projective module>. Symmetry gives the same conclusion for $N$. This is <projectivity of factors of a nonzero finite free tensor product>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 101 / 1 / ii / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-101/1/ii/c/solution