# If $A\ne 0$ is a square matrix over a commutative ring with $\det A=0$, then its null space contains an element whose components are minors of $A$

Let $R$ denote a commutative ring and $A\ne 0$ a $n\times n$ matrix over $R$ with $\det A=0$. Then there exists a $x\in\ker A\setminus\left\{0\right\}$ such that all components of $x$ are minors of $A$ (ignoring signs).

Proof: Cramer's rule yields: $$0=Ax=\text{adj }A\cdot Ax=\det A\begin{pmatrix} x_1&&\\&\ddots&\\&&x_n \end{pmatrix}=\begin{pmatrix} \det A\det X_1&&\\&\ddots&\\&&\det A\det X_n \end{pmatrix}$$ where $X_k$ is the identity matrix with the $k$-th column replaced by $x$ and satisfies $$x_k=\det X_k$$So we've got $$\det A\cdot x_k=\det\left(AX_k\right)=\det A_k$$ where $A_k$ is $A$ with the $k$-th column replaced by $0$. How do I need to proceed?

Since $\det A=0$ we get, for example, that $a_{11}d_1+\cdots+a_{1n}d_n=0$, where $d_i$ are the cofactors (in $A$) of the elements $a_{1i}$, $1\le i\le n$. Then, by Laplace expansion, $A(d_1,\dots,d_n)^T=0$. If one of these cofactors is $\ne0$, you win.

Otherwise, assume that all $(n-1)$-minors of $A$ are zero. For simplicity consider $A'=(a_{ij})_{1\le i,j\le n-1}$. Apply the same reasoning as before and get $A'(d_1',\dots,d_{n-1}')^T=0$, where $d_i'$ are the cofactors (in $A'$) of the elements $a_{1i}$, $1\le i\le n-1$, in $A'$. Now note that $A(d_1',\dots,d_{n-1}',0)=0$. If one of these cofactors is $\ne0$, you win.

I think now you can finish the proof.