$b w_i = \sum\limits_{i=1}^r a_{i,j} w_j \implies \det(bI - (a_{i,j})) w_i = 0$ in an $R$-module.

Question

Let $R$ be a commutative ring with $1$ and $M$ a finitely generated $R$ module. Then if $\{w_j : j = 1 \dots r\}$ is a set of generators for $M$ and $b \in R$, then $b w_i = \sum\limits_{j=1}^r a_{i,j} w_j$ for some $a_{i,j} \in R$. This implies $\det(bI - (a_{i,j}))w_i = 0, i = 1 \dots r$ somehow.

And that is derived knowing that $A A^* = A^* A = \det(A)I$ for any $r\times r$ matrix $A$ over $R$.

Hints?

Answer 1

The set of equations can be rewritten $b\bar{w} = A \bar{w} \implies (bI - A)\bar{w} = 0$. Now apply the linear algebra formula on $B = (bI - A)$ to get:

$$ B^*B \bar{w} = \det(B)I \bar{w}$$

But $B\bar{w} = 0$, so we get $\det(B)I\bar{w} = \bar{0}$. Done.