Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 4 / 11E / a / Solution

Let $S=R[\varphi ]$ be the commutative subring of endomorphisms generated by scalar multiplication and $\varphi$. Commutativity follows because the scalar maps commute with the $R$-linear map $\varphi$, and every element of $S$ is a polynomial in $\varphi$ with scalar coefficients. Thus determinants and the adjugate identity apply to matrices over $S$.

With the convention for the coefficients in the question, the column of generators satisfiesMultiply by $\mathrm{adj}(C)$ over $S$. Since $\mathrm{adj}(C)C=(\det C)I_n$, the endomorphism $\det C$ kills every generator $m_i$. It therefore kills every element $\sum r_i{m}_i$ of the module, by $R$-linearity. Hence it is the zero endomorphism, andThe argument does not require the generating set to be a basis; relations among the generators cause no difficulty. This is the determinant trick.