Let $R$ be a commutative ring with unity and consider the free $R$-module $R^n$. Given a matrix $A$ with coefficients in $R$, define a homomorphism $\Phi: R^n\to R^n$ by $\Phi(u) = Au$. My question is how to prove that $\det(A)R^n \subset \mathrm{Im}(\Phi)$? (We have that $\det(A)R^n = \{\det(A)u: \ u\in R^n\}$ is a submodule.)
I started doing the obvious, ie, trying to show that for any $u\in R^n$, there is some $\tilde{u}\in R^n$ such that $\det(A)u = A\tilde{u}$. Then I fixed the canonical base and tried to show that this equation (which is a linear system) always have a solution, but this approach seems more linear algebra and less abstract algebra, also it's a little messy.
Is there a nice way to prove this? Thanks.
Let $B$ be the adjoint matrix of $A$. Then $ABu=\det(A)u$ for all $u\in R^n$. Thus $Bu$ is an element of $R^n$ such that $A(Bu)=\det(A)u$.